# questions

title: the-standard-deviation-of-the-numbers-22-26-28-20-24-30-is-1
slug: the-standard-deviation-of-the-numbers-22-26-28-20-24-30-is-1-41857
tags: ['Mathematics', 'Statistics', 'AP EAMCET', 'AP EAMCET 2018 (25 Apr Shift 1)']
source_db: MarksBatch2_P3.db
question:
<div class="question">The standard deviation of the numbers <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>22</mn><mo>,</mo><mo> </mo><mn>26</mn><mo>,</mo><mo> </mo><mn>28</mn><mo>,</mo><mo> </mo><mn>20</mn><mo>,</mo><mo> </mo><mn>24</mn><mo>,</mo><mo> </mo><mn>30</mn></math> is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn><mo>.</mo><mn>4</mn></math></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>3</mn><mo>.</mo><mn>24</mn></math></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>3</mn><mo>.</mo><mn>42</mn></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>3</mn><mo>.</mo><mn>42</mn></math></span> </div>

answer:
<div class="solution">We know that<p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">S</mi><mo>.</mo><mi mathvariant="normal">D</mi><mo>=</mo><msqrt><mfrac><mrow><mstyle displaystyle="true"><mo>∑</mo></mstyle><msup><mfenced><mrow><msub><mi>x</mi><mi>i</mi></msub><mo>-</mo><mi>μ</mi></mrow></mfenced><mn>2</mn></msup></mrow><mi>N</mi></mfrac></msqrt></math> </p><p>where, </p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>μ</mi><mo>→</mo></math>Mean</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>N</mi><mo>→</mo></math>Number of terms</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>x</mi><mi>i</mi></msub><mo>→</mo></math>terms</p><p>Now,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>μ</mi><mo>=</mo><mfrac><mrow><mn>22</mn><mo>+</mo><mn>26</mn><mo>+</mo><mn>28</mn><mo>+</mo><mn>20</mn><mo>+</mo><mn>24</mn><mo>+</mo><mn>30</mn></mrow><mn>6</mn></mfrac></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mi>μ</mi><mo>=</mo><mfrac><mn>150</mn><mn>6</mn></mfrac><mo>=</mo><mn>25</mn></math></p><p>Now, </p><table border="1" cellpadding="1" cellspacing="1" style="width: 500px;"> <tbody> <tr> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>x</mi><mi>i</mi></msub></math></td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mfenced><mrow><msub><mi>x</mi><mi>i</mi></msub><mo>-</mo><mi>μ</mi></mrow></mfenced><mn>2</mn></msup></math></td> </tr> <tr> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>22</mn></math></td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mfenced><mrow><mn>22</mn><mo>-</mo><mn>25</mn></mrow></mfenced><mn>2</mn></msup><mo>=</mo><mn>9</mn></math></td> </tr> <tr> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>26</mn></math></td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mfenced><mrow><mn>26</mn><mo>-</mo><mn>25</mn></mrow></mfenced><mn>2</mn></msup><mo>=</mo><mn>1</mn></math></td> </tr> <tr> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>28</mn></math></td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mfenced><mrow><mn>28</mn><mo>-</mo><mn>25</mn></mrow></mfenced><mn>2</mn></msup><mo>=</mo><mn>9</mn></math></td> </tr> <tr> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>20</mn></math></td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mfenced><mrow><mn>20</mn><mo>-</mo><mn>25</mn></mrow></mfenced><mn>2</mn></msup><mo>=</mo><mn>25</mn></math></td> </tr> <tr> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>24</mn></math></td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mfenced><mrow><mn>24</mn><mo>-</mo><mn>25</mn></mrow></mfenced><mn>2</mn></msup><mo>=</mo><mn>1</mn></math></td> </tr> <tr> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>30</mn></math></td> <td><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mfenced><mrow><mn>30</mn><mo>-</mo><mn>25</mn></mrow></mfenced><mn>2</mn></msup><mo>=</mo><mn>25</mn></math></td> </tr> </tbody></table><p>Therefore, <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∑</mo><msup><mfenced><mrow><msub><mi>x</mi><mi>i</mi></msub><mo>-</mo><mi>μ</mi></mrow></mfenced><mn>2</mn></msup><mo>=</mo><mn>70</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">S</mi><mo>.</mo><mi mathvariant="normal">D</mi><mo>=</mo><msqrt><mfrac><mn>70</mn><mn>6</mn></mfrac></msqrt><mo>=</mo><mn>3</mn><mo>.</mo><mn>42</mn></math></p></div>


title: the-standard-deviation-of-the-numbers-22-26-28-20-24-30-is
slug: the-standard-deviation-of-the-numbers-22-26-28-20-24-30-is-18055
question: <div class="question">The standard deviation of the numbers 22, $26,28,20,24,30$ is</div>
tags: ['Mathematics', 'Statistics', 'AP EAMCET', 'AP EAMCET 2018 (23 Apr Shift 1)']
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">3.42</span> </div>
source_db: MarksBatch2_P3.db
options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">2</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">2.4</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">3.24</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">3.42</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

answer:
<div class="solution">Given numbers 22, 26, 28, 20, 24, 30<br/>$$<br/>\bar{x}=\frac{22+26+28+20+24+30}{6}=\frac{150}{6}=25<br/>$$<br/>Now, $\left(x_i-\bar{x}\right)^2$ is $9,1,9,25,1,25$<br/>$$<br/>\begin{aligned}<br/>&amp; \text { Now, } S D=\sqrt{\frac{\Sigma\left(x_i-\bar{x}\right)^2}{n}} \\<br/>&amp; =\sqrt{\frac{9+1+9+25+1+25}{6}}=\sqrt{\frac{70}{6}}=3.42<br/>\end{aligned}<br/>$$</div>


title: the-standard-deviation-of-the-numbers-31-32-33-46-47-is
slug: the-standard-deviation-of-the-numbers-31-32-33-46-47-is-54931
question: <div class="question">The standard deviation of the numbers 31 , $32,33 \ldots 46,47$ is</div>
tags: ['Mathematics', 'Statistics', 'KCET', 'KCET 2021']
source_db: MarksBatch2_P3.db
options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\sqrt{\frac{17}{12}}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\sqrt{\frac{47^{2}-1}{12}}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$2 \sqrt{6}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$4 \sqrt{3}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$2 \sqrt{6}$</span> </div>

answer:
<div class="solution">Given, numbers : $31,32,33, \ldots, 46,47$. The standard deviation of $31,32,33, \ldots, 47$ will be the same as those of $1,2,3, \ldots, 17$. (decreasing each item by 30 )<br/>$\because \quad \mathrm{SD}=\sqrt{\frac{n^{2}-1}{12}}$<br/>Here, $\quad n=17$<br/>$\begin{aligned}<br/>\therefore \quad \mathrm{SD} &amp;=\sqrt{\frac{17^{2}-1}{12}}=\sqrt{\frac{288}{12}} \\<br/>&amp;=\sqrt{24}=2 \sqrt{6}<br/>\end{aligned}$</div>


title: the-standard-deviation-of-the-observations-5-5-5-5-5-is
slug: the-standard-deviation-of-the-observations-5-5-5-5-5-is-64249
question: <div class="question">The standard deviation of the observations $5,5,5,5,5$ is</div>
tags: ['Mathematics', 'Statistics', 'NDA', 'NDA 2013 (Phase 1)']
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">0</span> </div>
source_db: MarksBatch2_P3.db
options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">0</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">5</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">20</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">25</span> </li> </ul>

answer:
<div class="solution">Given observation are $5,5,5,5,5$<br/>$\bar{x}=\frac{5+5+5+5+5}{5}=\frac{25}{5}=5$<br/>Now, $\sigma^{2}=\sum_{i=1}^{5} \frac{\left(x_{i}-\bar{x}\right)^{2}}{N}$<br/>Where $\mathrm{N}=$ total number of observations.<br/>Variance $=\sigma^{2}$<br/>$\frac{(5-5)+(5-5)+(5-5)+(5-5)+(5-5)}{=}$<br/>Hence, standard deviation $=\sqrt{v a r}=0$</div>


title: the-standard-deviation-of-the-scores-505-510-515-520-595-is
slug: the-standard-deviation-of-the-scores-505-510-515-520-595-is-91612
tags: ['Mathematics', 'Statistics', 'AP EAMCET', 'AP EAMCET 2018 (23 Apr Shift 2)']
source_db: MarksBatch2_P3.db
question:
<div class="question">The standard deviation of the scores 505, 510, $515,520, \ldots \ldots ., 595$ is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$500+5 \sqrt{30}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$505+\sqrt{30}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$5 \sqrt{30}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$5+\sqrt{30}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$5 \sqrt{30}$</span> </div>

answer:
<div class="solution">Given, scores 505, 510, 515, 520, ... 595 the mean of these 19 scores $=550=\bar{X}$<br/>So,<br/>$$<br/>\begin{aligned}<br/>\sum_{i=1}^{19}\left(x_i-\bar{x}\right)^2 &amp; =2\left(45^2+40^2+35^2+\ldots+5^2\right) \\<br/>&amp; =2 \times 5^2 \times\left[1^2+2^2+3^2+\ldots+9^2\right]<br/>\end{aligned}<br/>$$<br/>So, standard derivation<br/>$$<br/>=\sqrt{\frac{\sum_{i=1}^{19}\left(x_i-\bar{x}\right)^2}{n}}=\sqrt{\frac{2 \times 5^2 \times \frac{9 \times 10 \times 19}{6}}{19}}=5 \sqrt{30}<br/>$$</div>


tags: ['Mathematics', 'Statistics', 'JEE Main']
source_db: MarksBatch2_P3.db
title:
the-standard-deviation-of-the-wages-of-85-employees-is-1540-after-one-year-each-of-them-is-given-an-increment-of-25-the-standard-deviation-of-new-wage

slug:
the-standard-deviation-of-the-wages-of-85-employees-is-1540-after-one-year-each-of-them-is-given-an-increment-of-25-the-standard-deviation-of-new-wage-33043

question:
<div class="question">The standard deviation of the wages of \( 85 \) employees is \( â¹ 15.40 \). After one year, each of them is given an increment of \( â¹ 25 \) The standard deviation of new wages (in â¹) is</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">\( 15.40 \)</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">\( 40.40 \)</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">\( 20.40 \)</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">\( 10.40 \)</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">\( 15.40 \)</span> </div>

answer:
<div class="solution"><blockquote><p>The arithmetic mean of the square of deviation of the variable values from its actual arithmetic mean is known as variance and its square root is known as standard deviation.</p><p>Standard deviation is given by formula</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi>σ</mi><mn>2</mn></msup><mo>=</mo><mfrac><mrow><mo>∑</mo><msup><mfenced><mrow><msub><mi>x</mi><mi>i</mi></msub><mo>-</mo><mover><mi>x</mi><mo>¯</mo></mover></mrow></mfenced><mn>2</mn></msup></mrow><mi>n</mi></mfrac></math></p><p>If data is changed from <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>x</mi><mi>i</mi></msub></math> to <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>x</mi><mi>i</mi></msub><mo>+</mo><mi>k</mi></math> then new standard deviation is given by <math xmlns="http://www.w3.org/1998/Math/MathML"><msup><msub><mi>σ</mi><mn>1</mn></msub><mn>2</mn></msup><mo>=</mo><mfrac><mrow><mo>∑</mo><msup><mfenced><mrow><mfenced><mrow><msub><mi>x</mi><mi>i</mi></msub><mo>+</mo><mi>k</mi></mrow></mfenced><mo>-</mo><mfenced><mrow><mover><mi>x</mi><mo>¯</mo></mover><mo>+</mo><mi>k</mi></mrow></mfenced></mrow></mfenced><mn>2</mn></msup></mrow><mi>n</mi></mfrac><mo>=</mo><msup><mi>σ</mi><mn>2</mn></msup></math></p><p>since, <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>S</mi><mi>D</mi></math> is independent of change of origin. Therefore, the standard deviation remains same i.e. <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>15</mn><mo>.</mo><mn>40</mn></math>.</p></blockquote></div>


tags: ['Mathematics', 'Statistics', 'TS EAMCET', 'TS EAMCET 2020 (11 Sep Shift 2)']
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">7</span> </div>
source_db: MarksBatch2_P3.db
title:
the-standard-deviations-of-two-sets-of-observations-x-x-i-and-y-y-i-i-1-2-100-are-respectively-5-and-6-if-x-y-are-their-means-and-i-1-100-x-i-x-y-i-y-

slug:
the-standard-deviations-of-two-sets-of-observations-x-x-i-and-y-y-i-i-1-2-100-are-respectively-5-and-6-if-x-y-are-their-means-and-i-1-100-x-i-x-y-i-y-35748

question:
<div class="question">The standard deviations of two sets of observations $X=\left\{x_i\right\}$ and $Y=\left\{y_i\right\}$ $(i=1,2, \ldots, 100)$ are respectively 5 and 6 . If $\bar{x}, \bar{y}$ are their means and $\sum_{i=1}^{100}\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)=600$, then the standard deviation of $Z=\left\{z_i / z_i=x_i-y_i\right)$ is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">12</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">6</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">7</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">10</span> </li> </ul>

answer:
<div class="solution">From the given information, we can say that<br/>$\sqrt{\frac{\sum_{i=1}^{100}\left(x_i-\bar{x}\right)^2}{100}}=5 \text { and } \sqrt{\frac{\sum_{i=1}^{100}\left(y_i-\bar{y}\right)^2}{100}}=6$<br/>$\therefore \quad \sum_{i=1}^{100}\left(x_i-\bar{x}\right)^2=2500$ and $\sum_{i=1}^{100}\left(y_i-\bar{y}\right)^2=3600$<br/>and it is also given that<br/>$\sum_{i=1}^{100}\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)=600$<br/>$\because \quad a^2+b^2-2 a b=(a-b)^2$<br/>$\sum_{i=1}^{100}\left(x_i-\bar{x}\right)^2+\sum_{i=1}^{100}\left(y_i-\bar{y}\right)^2-\sum_{i=1}^{100}\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)$<br/>$=\sum_{i=1}^{100}\left(\left(x_i-\bar{x}\right)-\left(y_i-\bar{y}\right)\right)^2$<br/>$\Rightarrow \quad \sum_{i=1}^{100}\left[\left(x_i-\bar{x}\right)-\left(y_i-\bar{y}\right)\right]^2=2500+3600-1200$<br/>$\Rightarrow \quad \sum_{i=1}^{100}\left[\left(x_i-y_i\right)-(\bar{x}-\bar{y})\right]^2=4900$<br/>$\Rightarrow \sqrt{\frac{\sum_{i=1}^{100}\left[\left(x_i-y_i\right)-(\bar{x}-\bar{y})\right]^2}{100}}=7$</div>


tags: ['Mathematics', 'Statistics', 'JEE Main']
source_db: MarksBatch2_P3.db
title:
the-standard-deviations-of-x-i-i-1-2-10-and-y-i-i-1-10-are-respectively-a-and-b-x-y-are-the-means-of-these-two-sets-of-observations-respectively-if-z--1

slug:
the-standard-deviations-of-x-i-i-1-2-10-and-y-i-i-1-10-are-respectively-a-and-b-x-y-are-the-means-of-these-two-sets-of-observations-respectively-if-z-1-31977

question:
<div class="question">The standard deviations of <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>x</mi><mi>i</mi></msub><mfenced><mrow><mi>i</mi><mo>=</mo><mn>1</mn><mo>,</mo><mn>2</mn><mo>,</mo><mo>…</mo><mo>.</mo><mn>10</mn></mrow></mfenced></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>y</mi><mi>i</mi></msub><mfenced><mrow><mi>i</mi><mo>=</mo><mn>1</mn><mo>,</mo><mo>…</mo><mn>10</mn></mrow></mfenced></math> are respectively <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>b</mi></math>. <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>x</mi><mo>¯</mo></mover><mo>,</mo><mo> </mo><mover><mi>y</mi><mo>¯</mo></mover></math> are the means of these two sets of observations respectively. If <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mi>i</mi></msub><mo>=</mo><mfenced><mrow><msub><mi>x</mi><mi>i</mi></msub><mo>-</mo><mover><mi>x</mi><mo>¯</mo></mover></mrow></mfenced><mfenced><mrow><msub><mi>y</mi><mi>i</mi></msub><mo>-</mo><mover><mi>y</mi><mo>¯</mo></mover></mrow></mfenced></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><munderover><mo>∑</mo><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mn>10</mn></munderover><msub><mi>z</mi><mi>i</mi></msub><mo>=</mo><mi>c</mi></math>, then the standard deviation of the observations <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mrow><msub><mi>x</mi><mi>i</mi></msub><mo>-</mo><msub><mi>y</mi><mi>i</mi></msub></mrow></mfenced><mo>,</mo><mo> </mo><mfenced><mrow><mi>i</mi><mo>=</mo><mn>1</mn><mo>,</mo><mn>2</mn><mo>,</mo><mo>…</mo><mn>10</mn></mrow></mfenced></math> is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><msup><mi>a</mi><mn>2</mn></msup><mo>+</mo><msup><mi>b</mi><mn>2</mn></msup><mo>+</mo><mfrac><mi>c</mi><mn>5</mn></mfrac></msqrt></math></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><msup><mi>a</mi><mn>2</mn></msup><mo>+</mo><msup><mi>b</mi><mn>2</mn></msup><mo>-</mo><mfrac><mi>c</mi><mn>5</mn></mfrac></msqrt></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><msup><mi>a</mi><mn>2</mn></msup><mo>+</mo><msup><mi>b</mi><mn>2</mn></msup><mo>-</mo><mfrac><msup><mi>c</mi><mn>2</mn></msup><mn>5</mn></mfrac></msqrt></math></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><msup><mi>a</mi><mn>2</mn></msup><mo>+</mo><msup><mi>b</mi><mn>2</mn></msup><mo>+</mo><mfrac><msup><mi>c</mi><mn>2</mn></msup><mn>5</mn></mfrac></msqrt></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><msup><mi>a</mi><mn>2</mn></msup><mo>+</mo><msup><mi>b</mi><mn>2</mn></msup><mo>-</mo><mfrac><mi>c</mi><mn>5</mn></mfrac></msqrt></math></span> </div>

answer:
<div class="solution"><p>It is given that, <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mi>i</mi></msub><mo mathvariant="italic">=</mo><mfenced><mrow><msub><mi>x</mi><mi>i</mi></msub><mo mathvariant="italic">-</mo><mover><mi>x</mi><mo mathvariant="italic">¯</mo></mover></mrow></mfenced><mfenced><mrow><msub><mi>y</mi><mi>i</mi></msub><mo mathvariant="italic">-</mo><mover><mi>y</mi><mo mathvariant="italic">¯</mo></mover></mrow></mfenced></math></p><p>And <math xmlns="http://www.w3.org/1998/Math/MathML"><munderover><mo>∑</mo><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mn>10</mn></munderover><msub><mi>z</mi><mi>i</mi></msub><mo>=</mo><mi>c</mi></math></p><p>Rewrite the above equation.</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><munderover><mo>∑</mo><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mn>10</mn></munderover><mfenced close="}" open="{"><mrow><mfenced><mrow><msub><mi>x</mi><mi>i</mi></msub><mo>-</mo><mover><mi>x</mi><mo>¯</mo></mover></mrow></mfenced><mfenced><mrow><msub><mi>y</mi><mi>i</mi></msub><mo>-</mo><mover><mi>y</mi><mo>¯</mo></mover></mrow></mfenced></mrow></mfenced><mo>=</mo><mi>c</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><munderover><mo>∑</mo><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mn>10</mn></munderover><mfenced close="}" open="{"><mrow><msub><mi>x</mi><mi>i</mi></msub><msub><mi>y</mi><mi>i</mi></msub><mo>-</mo><msub><mi>x</mi><mi>i</mi></msub><mover><mi>y</mi><mo>¯</mo></mover><mo>-</mo><mover><mi>x</mi><mo>¯</mo></mover><msub><mi>y</mi><mi>i</mi></msub><mo>+</mo><mover><mi>x</mi><mo>¯</mo></mover><mo>·</mo><mover><mi>y</mi><mo>¯</mo></mover></mrow></mfenced><mo>=</mo><mi>c</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><munderover><mrow><mo> </mo><mo>∑</mo></mrow><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mn>10</mn></munderover><msub><mi>x</mi><mi>i</mi></msub><msub><mi>y</mi><mi>i</mi></msub><mo>-</mo><munderover><mo>∑</mo><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mn>10</mn></munderover><msub><mi>x</mi><mi>i</mi></msub><mover><mi>y</mi><mo>¯</mo></mover><mo>-</mo><munderover><mo>∑</mo><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mn>10</mn></munderover><mover><mi>x</mi><mo>¯</mo></mover><msub><mi>y</mi><mi>i</mi></msub><mo>+</mo><munderover><mo>∑</mo><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mn>10</mn></munderover><mover><mi>x</mi><mo>¯</mo></mover><mo>·</mo><mover><mi>y</mi><mo>¯</mo></mover><mo>=</mo><mi>c</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><munderover><mo>∑</mo><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mn>10</mn></munderover><msub><mi>x</mi><mi>i</mi></msub><msub><mi>y</mi><mi>i</mi></msub><mo>-</mo><mn>10</mn><mover><mi>x</mi><mo>¯</mo></mover><mo>·</mo><mover><mi>y</mi><mo>¯</mo></mover><mo>-</mo><mn>10</mn><mover><mi>x</mi><mo>¯</mo></mover><mo>·</mo><mover><mi>y</mi><mo>¯</mo></mover><mo>+</mo><mn>10</mn><mover><mi>x</mi><mo>¯</mo></mover><mo>·</mo><mover><mi>y</mi><mo>¯</mo></mover><mo>=</mo><mi>c</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><munderover><mo>∑</mo><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mn>10</mn></munderover><msub><mi>x</mi><mi>i</mi></msub><msub><mi>y</mi><mi>i</mi></msub><mo>-</mo><mn>10</mn><mover><mi>x</mi><mo>¯</mo></mover><mo>·</mo><mover><mi>y</mi><mo>¯</mo></mover><mo>=</mo><mi>c</mi></math></p><p>Now,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>σ</mi><mfenced><mrow><msub><mi>x</mi><mn>1</mn></msub><mo>-</mo><msub><mi>y</mi><mn>1</mn></msub></mrow></mfenced></msub><mo>=</mo><msqrt><mfrac><mn>1</mn><mn>10</mn></mfrac><mo>∑</mo><msup><mfenced><mrow><msub><mi>x</mi><mi>i</mi></msub><mo>-</mo><msub><mi>y</mi><mi>i</mi></msub></mrow></mfenced><mn>2</mn></msup><mo>-</mo><mfenced><mrow><mover><mi>x</mi><mo>¯</mo></mover><mo>-</mo><mover><mi>y</mi><mo>¯</mo></mover></mrow></mfenced><msup><mrow></mrow><mn>2</mn></msup></msqrt></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><msqrt><mfenced><mrow><mfrac><mn>1</mn><mn>10</mn></mfrac><mo>∑</mo><msubsup><mi>x</mi><mi>i</mi><mn>2</mn></msubsup><mo>+</mo><mfrac><mn>1</mn><mn>10</mn></mfrac><mo>∑</mo><msubsup><mi>y</mi><mi>i</mi><mn>2</mn></msubsup><mo>-</mo><mfrac><mn>1</mn><mn>5</mn></mfrac><mo>∑</mo><msub><mi>x</mi><mi>i</mi></msub><msub><mi>y</mi><mi>i</mi></msub><mo>-</mo><msup><mover><mi>x</mi><mo>¯</mo></mover><mn>2</mn></msup><mo>-</mo><msup><mover><mi>y</mi><mo>¯</mo></mover><mn>2</mn></msup><mo>+</mo><mn>2</mn><mover><mrow><mi>x</mi><mi>y</mi></mrow><mo>¯</mo></mover></mrow></mfenced></msqrt></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><msqrt><mfenced><mrow><mfrac><mn>1</mn><mn>10</mn></mfrac><mo>∑</mo><msubsup><mi>x</mi><mi>i</mi><mn>2</mn></msubsup><mo>-</mo><msup><mover><mi>x</mi><mo>¯</mo></mover><mn>2</mn></msup></mrow></mfenced><mo>+</mo><mfenced><mrow><mfrac><mn>1</mn><mn>10</mn></mfrac><mo>∑</mo><msubsup><mi>y</mi><mi>i</mi><mn>2</mn></msubsup><mo>-</mo><msup><mover><mi>y</mi><mo>¯</mo></mover><mn>2</mn></msup></mrow></mfenced><mo>-</mo><mfrac><mn>1</mn><mn>5</mn></mfrac><mfenced><mrow><mo>∑</mo><msub><mi>x</mi><mi>i</mi></msub><msub><mi>y</mi><mi>i</mi></msub><mo>-</mo><mn>10</mn><mover><mrow><mi>x</mi><mi>y</mi></mrow><mo>¯</mo></mover></mrow></mfenced></msqrt></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><msqrt><msup><mi>a</mi><mn>2</mn></msup><mo>+</mo><msup><mi>b</mi><mn>2</mn></msup><mo>-</mo><mfrac><mi>c</mi><mn>5</mn></mfrac></msqrt></math></p></div>


tags: ['Mathematics', 'Statistics', 'AP EAMCET', 'AP EAMCET 2019 (21 Apr Shift 2)']
source_db: MarksBatch2_P3.db
title:
the-standard-deviations-of-x-i-i-1-2-10-and-y-i-i-1-10-are-respectively-a-and-b-x-y-are-the-means-of-these-two-sets-of-observations-respectively-if-z-

slug:
the-standard-deviations-of-x-i-i-1-2-10-and-y-i-i-1-10-are-respectively-a-and-b-x-y-are-the-means-of-these-two-sets-of-observations-respectively-if-z-62833

question:
<div class="question">The standard deviations of <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>x</mi><mi>i</mi></msub><mfenced><mrow><mi>i</mi><mo>=</mo><mn>1</mn><mo>,</mo><mn>2</mn><mo>,</mo><mo>…</mo><mo>.</mo><mn>10</mn></mrow></mfenced></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>y</mi><mi>i</mi></msub><mfenced><mrow><mi>i</mi><mo>=</mo><mn>1</mn><mo>,</mo><mo>…</mo><mn>10</mn></mrow></mfenced></math> are respectively <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>a</mi></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>b</mi></math>. <math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>x</mi><mo>¯</mo></mover><mo>,</mo><mo> </mo><mover><mi>y</mi><mo>¯</mo></mover></math> are the means of these two sets of observations respectively. If <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mi>i</mi></msub><mo>=</mo><mfenced><mrow><msub><mi>x</mi><mi>i</mi></msub><mo>-</mo><mover><mi>x</mi><mo>¯</mo></mover></mrow></mfenced><mfenced><mrow><msub><mi>y</mi><mi>i</mi></msub><mo>-</mo><mover><mi>y</mi><mo>¯</mo></mover></mrow></mfenced></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><munderover><mo>∑</mo><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mn>10</mn></munderover><msub><mi>z</mi><mi>i</mi></msub><mo>=</mo><mi>c</mi></math>, then the standard deviation of the observations <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mrow><msub><mi>x</mi><mi>i</mi></msub><mo>-</mo><msub><mi>y</mi><mi>i</mi></msub></mrow></mfenced><mo>,</mo><mo> </mo><mfenced><mrow><mi>i</mi><mo>=</mo><mn>1</mn><mo>,</mo><mn>2</mn><mo>,</mo><mo>…</mo><mn>10</mn></mrow></mfenced></math> is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><msup><mi>a</mi><mn>2</mn></msup><mo>+</mo><msup><mi>b</mi><mn>2</mn></msup><mo>+</mo><mfrac><mi>c</mi><mn>5</mn></mfrac></msqrt></math></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><msup><mi>a</mi><mn>2</mn></msup><mo>+</mo><msup><mi>b</mi><mn>2</mn></msup><mo>-</mo><mfrac><mi>c</mi><mn>5</mn></mfrac></msqrt></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><msup><mi>a</mi><mn>2</mn></msup><mo>+</mo><msup><mi>b</mi><mn>2</mn></msup><mo>-</mo><mfrac><msup><mi>c</mi><mn>2</mn></msup><mn>5</mn></mfrac></msqrt></math></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><msup><mi>a</mi><mn>2</mn></msup><mo>+</mo><msup><mi>b</mi><mn>2</mn></msup><mo>+</mo><mfrac><msup><mi>c</mi><mn>2</mn></msup><mn>5</mn></mfrac></msqrt></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><msqrt><msup><mi>a</mi><mn>2</mn></msup><mo>+</mo><msup><mi>b</mi><mn>2</mn></msup><mo>-</mo><mfrac><mi>c</mi><mn>5</mn></mfrac></msqrt></math></span> </div>

answer:
<div class="solution"><p>It is given that, <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>z</mi><mi>i</mi></msub><mo mathvariant="italic">=</mo><mfenced><mrow><msub><mi>x</mi><mi>i</mi></msub><mo mathvariant="italic">-</mo><mover><mi>x</mi><mo mathvariant="italic">¯</mo></mover></mrow></mfenced><mfenced><mrow><msub><mi>y</mi><mi>i</mi></msub><mo mathvariant="italic">-</mo><mover><mi>y</mi><mo mathvariant="italic">¯</mo></mover></mrow></mfenced></math></p><p>And <math xmlns="http://www.w3.org/1998/Math/MathML"><munderover><mo>∑</mo><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mn>10</mn></munderover><msub><mi>z</mi><mi>i</mi></msub><mo>=</mo><mi>c</mi></math></p><p>Rewrite the above equation.</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><munderover><mo>∑</mo><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mn>10</mn></munderover><mfenced close="}" open="{"><mrow><mfenced><mrow><msub><mi>x</mi><mi>i</mi></msub><mo>-</mo><mover><mi>x</mi><mo>¯</mo></mover></mrow></mfenced><mfenced><mrow><msub><mi>y</mi><mi>i</mi></msub><mo>-</mo><mover><mi>y</mi><mo>¯</mo></mover></mrow></mfenced></mrow></mfenced><mo>=</mo><mi>c</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><munderover><mo>∑</mo><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mn>10</mn></munderover><mfenced close="}" open="{"><mrow><msub><mi>x</mi><mi>i</mi></msub><msub><mi>y</mi><mi>i</mi></msub><mo>-</mo><msub><mi>x</mi><mi>i</mi></msub><mover><mi>y</mi><mo>¯</mo></mover><mo>-</mo><mover><mi>x</mi><mo>¯</mo></mover><msub><mi>y</mi><mi>i</mi></msub><mo>+</mo><mover><mi>x</mi><mo>¯</mo></mover><mo>·</mo><mover><mi>y</mi><mo>¯</mo></mover></mrow></mfenced><mo>=</mo><mi>c</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><munderover><mrow><mo> </mo><mo>∑</mo></mrow><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mn>10</mn></munderover><msub><mi>x</mi><mi>i</mi></msub><msub><mi>y</mi><mi>i</mi></msub><mo>-</mo><munderover><mo>∑</mo><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mn>10</mn></munderover><msub><mi>x</mi><mi>i</mi></msub><mover><mi>y</mi><mo>¯</mo></mover><mo>-</mo><munderover><mo>∑</mo><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mn>10</mn></munderover><mover><mi>x</mi><mo>¯</mo></mover><msub><mi>y</mi><mi>i</mi></msub><mo>+</mo><munderover><mo>∑</mo><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mn>10</mn></munderover><mover><mi>x</mi><mo>¯</mo></mover><mo>·</mo><mover><mi>y</mi><mo>¯</mo></mover><mo>=</mo><mi>c</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><munderover><mo>∑</mo><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mn>10</mn></munderover><msub><mi>x</mi><mi>i</mi></msub><msub><mi>y</mi><mi>i</mi></msub><mo>-</mo><mn>10</mn><mover><mi>x</mi><mo>¯</mo></mover><mo>·</mo><mover><mi>y</mi><mo>¯</mo></mover><mo>-</mo><mn>10</mn><mover><mi>x</mi><mo>¯</mo></mover><mo>·</mo><mover><mi>y</mi><mo>¯</mo></mover><mo>+</mo><mn>10</mn><mover><mi>x</mi><mo>¯</mo></mover><mo>·</mo><mover><mi>y</mi><mo>¯</mo></mover><mo>=</mo><mi>c</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>⇒</mo><mo> </mo><munderover><mo>∑</mo><mrow><mi>i</mi><mo>=</mo><mn>1</mn></mrow><mn>10</mn></munderover><msub><mi>x</mi><mi>i</mi></msub><msub><mi>y</mi><mi>i</mi></msub><mo>-</mo><mn>10</mn><mover><mi>x</mi><mo>¯</mo></mover><mo>·</mo><mover><mi>y</mi><mo>¯</mo></mover><mo>=</mo><mi>c</mi></math></p><p>Now,</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>σ</mi><mfenced><mrow><msub><mi>x</mi><mn>1</mn></msub><mo>-</mo><msub><mi>y</mi><mn>1</mn></msub></mrow></mfenced></msub><mo>=</mo><msqrt><mfrac><mn>1</mn><mn>10</mn></mfrac><mo>∑</mo><msup><mfenced><mrow><msub><mi>x</mi><mi>i</mi></msub><mo>-</mo><msub><mi>y</mi><mi>i</mi></msub></mrow></mfenced><mn>2</mn></msup><mo>-</mo><mfenced><mrow><mover><mi>x</mi><mo>¯</mo></mover><mo>-</mo><mover><mi>y</mi><mo>¯</mo></mover></mrow></mfenced><msup><mrow></mrow><mn>2</mn></msup></msqrt></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><msqrt><mfenced><mrow><mfrac><mn>1</mn><mn>10</mn></mfrac><mo>∑</mo><msubsup><mi>x</mi><mi>i</mi><mn>2</mn></msubsup><mo>+</mo><mfrac><mn>1</mn><mn>10</mn></mfrac><mo>∑</mo><msubsup><mi>y</mi><mi>i</mi><mn>2</mn></msubsup><mo>-</mo><mfrac><mn>1</mn><mn>5</mn></mfrac><mo>∑</mo><msub><mi>x</mi><mi>i</mi></msub><msub><mi>y</mi><mi>i</mi></msub><mo>-</mo><msup><mover><mi>x</mi><mo>¯</mo></mover><mn>2</mn></msup><mo>-</mo><msup><mover><mi>y</mi><mo>¯</mo></mover><mn>2</mn></msup><mo>+</mo><mn>2</mn><mover><mrow><mi>x</mi><mi>y</mi></mrow><mo>¯</mo></mover></mrow></mfenced></msqrt></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><msqrt><mfenced><mrow><mfrac><mn>1</mn><mn>10</mn></mfrac><mo>∑</mo><msubsup><mi>x</mi><mi>i</mi><mn>2</mn></msubsup><mo>-</mo><msup><mover><mi>x</mi><mo>¯</mo></mover><mn>2</mn></msup></mrow></mfenced><mo>+</mo><mfenced><mrow><mfrac><mn>1</mn><mn>10</mn></mfrac><mo>∑</mo><msubsup><mi>y</mi><mi>i</mi><mn>2</mn></msubsup><mo>-</mo><msup><mover><mi>y</mi><mo>¯</mo></mover><mn>2</mn></msup></mrow></mfenced><mo>-</mo><mfrac><mn>1</mn><mn>5</mn></mfrac><mfenced><mrow><mo>∑</mo><msub><mi>x</mi><mi>i</mi></msub><msub><mi>y</mi><mi>i</mi></msub><mo>-</mo><mn>10</mn><mover><mrow><mi>x</mi><mi>y</mi></mrow><mo>¯</mo></mover></mrow></mfenced></msqrt></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><msqrt><msup><mi>a</mi><mn>2</mn></msup><mo>+</mo><msup><mi>b</mi><mn>2</mn></msup><mo>-</mo><mfrac><mi>c</mi><mn>5</mn></mfrac></msqrt></math></p></div>


tags: ['Chemistry', 'Electrochemistry', 'MHT CET', 'MHT CET 2007']
source_db: MarksBatch2_P3.db
title:
the-standard-e-red-values-of-a-b-and-c-are-068-v-254-v-050-v-respectively-the-order-of-their-reducing-power-is

slug:
the-standard-e-red-values-of-a-b-and-c-are-068-v-254-v-050-v-respectively-the-order-of-their-reducing-power-is-93283

question:
<div class="question">The standard $E_{\text {red }}^{\circ}$ values of $A, B$ and $C$ are $+0.68 \mathrm{~V},-2.54 \mathrm{~V},-0.50 \mathrm{~V}$ respectively. The<br/>order of their reducing power is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$A&gt;B&gt;C$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$A&gt;C&gt;B$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$C&gt;B&gt;A$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$B&gt;C&gt;A$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$B&gt;C&gt;A$</span> </div>

answer:
<div class="solution">Reducing character is based upon higher negative value of reduction electrode potential.<br/>Thus, order of reducing character is<br/>$$<br/>B&gt;C&gt;A<br/>$$</div>


tags: ['Chemistry', 'Electrochemistry', 'JEE Main', 'JEE Main 2019 (12 Jan Shift 1)']
source_db: MarksBatch2_P3.db
title:
the-standard-electrode-potential-e-o-and-its-temperature-coefficient-de-dt-for-a-cell-are-2-v-and-5-10-4-v-k-1-at-300-k-respectively-the-reaction-is-z

slug:
the-standard-electrode-potential-e-o-and-its-temperature-coefficient-de-dt-for-a-cell-are-2-v-and-5-10-4-v-k-1-at-300-k-respectively-the-reaction-is-z-47967

question:
<div class="question"><p>The standard electrode potential <math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi mathvariant="normal">E</mi><mo>o</mo></msup></math> and its temperature coefficient <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced><mfrac><mi>dE</mi><mi>dT</mi></mfrac></mfenced></math> for a cell are <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn><mo> </mo><mi mathvariant="normal">V</mi></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mn>5</mn><msup><mo>×10</mo><mrow><mo>-</mo><mn>4</mn></mrow></msup><mi mathvariant="normal"> </mi><mo> </mo><mi mathvariant="normal">V</mi><mo> </mo><msup><mi mathvariant="normal">K</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math> at <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>300 </mn><mi mathvariant="normal">K</mi></math>, respectively. The reaction is <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>Zn</mi><mo> </mo><mfenced separators="|"><mi mathvariant="normal">s</mi></mfenced><mo>+</mo><msup><mi>Cu</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup><mo> </mo><mfenced separators="|"><mi>aq</mi></mfenced><mo>→</mo><msup><mi>Zn</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup><mo> </mo><mfenced separators="|"><mi>aq</mi></mfenced><mo>+</mo><mi>Cu</mi><mo> </mo><mfenced separators="|"><mi mathvariant="normal">s</mi></mfenced></math>. The standard reaction enthalpy <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced separators="|"><mrow><msub><mi mathvariant="normal">Δ</mi><mi mathvariant="normal">r</mi></msub><msup><mi mathvariant="normal">H</mi><mo>-</mo></msup></mrow></mfenced></math> at <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>300</mn><mo> </mo><mi mathvariant="normal">K</mi></math> in <math><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><msup><mrow><mi mathvariant="normal">l</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math> is </p><p><math><mo>[</mo></math>Use <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">R</mi><mo>=</mo><mn>8</mn><mo> </mo><mi mathvariant="normal">J</mi><mo> </mo><msup><mi mathvariant="normal">K</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup><mo> </mo><msup><mi>mol</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">F</mi><mo>=96,500 </mo><mi mathvariant="normal">C</mi><mo> </mo><msup><mi>mol</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup><mo>]</mo></math></p></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><math><mo>-</mo><mn>412.8</mn></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math><mn>206.4</mn></math></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math><mo>-</mo><mn>384.0</mn></math></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math><mn>192.0</mn></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math><mo>-</mo><mn>412.8</mn></math></span> </div>

answer:
<div class="solution"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">Δ</mi><mtext>G</mtext><mo>=</mo><mo>-</mo><msub><mi>nFE</mi><mi>cell</mi></msub><mspace linebreak="newline"></mspace><mo>=</mo><mo>-</mo><mn>2</mn><mo>×</mo><mn>96500</mn><mo>×</mo><mn>2</mn><mspace linebreak="newline"></mspace><mo>=</mo><mo>-</mo><mn>386</mn><mi mathvariant="normal"> </mi><mi>kJ</mi></math><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">Δ</mi><mtext>S</mtext><mo>=</mo><mi>nF</mi><mfrac><mi>dE</mi><mi>dT</mi></mfrac><mspace linebreak="newline"></mspace><mo>=</mo><mn>2</mn><mo>×</mo><mn>96500</mn><mo>×</mo><mo>-</mo><mn>5</mn><mo>×</mo><msup><mn>10</mn><mrow><mo>-</mo><mn>4</mn></mrow></msup><mo> </mo><mi mathvariant="normal">J</mi><mo> </mo><mi>degree</mi><mo> </mo><msup><mi>celcious</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mo>-</mo><mn>96</mn><mo>.5 </mo><mi mathvariant="normal">J</mi></math><br/><br/>At <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>298 </mn><mi mathvariant="normal">K</mi></math>,<br/><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>TΔ</mi><mtext>S</mtext><mo>=</mo><mn>298</mn><mo>×</mo><mfenced separators="|"><mrow><mo>-</mo><mn>96</mn><mo>.5</mo><mo> </mo><mi mathvariant="normal">J</mi></mrow></mfenced><mspace linebreak="newline"></mspace><mo>=</mo><mo>-</mo><mn>28</mn><mo>.757</mo><mi mathvariant="normal"> </mi><mi>kJ</mi></math><br/>At constant temperature <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced separators="|"><mrow><mi mathvariant="normal">T</mi><mo>=</mo><mn>248</mn><mo> </mo><mi mathvariant="normal">K</mi></mrow></mfenced></math> and pressure,<br/><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ΔG</mi><mo>=</mo><mi>ΔH</mi><mo>-</mo><mi>TΔ</mi><mtext>S</mtext></math><br/><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ΔH</mi><mo>=</mo><mi>ΔG</mi><mo>+</mo><mi>TΔ</mi><mtext>S</mtext></math><br/><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mo>-</mo><mn>386</mn><mo>-</mo><mn>28.757</mn><mspace linebreak="newline"></mspace><mo>=</mo><mn>-414.75</mn><mo>≃</mo><mn>-412.8</mn></math></div>


tags: ['Chemistry', 'Electrochemistry', 'NEET', 'NEET 2019 (Odisha)']
source_db: MarksBatch2_P3.db
title:
the-standard-electrode-potential-e-values-of-al-3-al-ag-ag-k-k-and-cr-3-cr-are-166-v-080-v-293-v-and-074-v-respectively-the-correct-decreasing-order-o

slug:
the-standard-electrode-potential-e-values-of-al-3-al-ag-ag-k-k-and-cr-3-cr-are-166-v-080-v-293-v-and-074-v-respectively-the-correct-decreasing-order-o-63121

question:
<div class="question">The standard electrode potential $\left(\mathrm{E}^{-}\right)$values of $\left(\mathrm{Al}^{3+} / \mathrm{Al}, \mathrm{Ag}^{+} / \mathrm{Ag}, \mathrm{K}^{+} / \mathrm{K}\right.$ and $\mathrm{Cr}^{3+} / \mathrm{Cr}$ are $-1.66 \mathrm{~V}, 0.80 \mathrm{~V}, 2.93 \mathrm{~V}$ and $-0.74 \mathrm{~V}$, respectively. The correct decreasing order of reducing power of the metal is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\mathrm{Ag}&gt;\mathrm{Cr}&gt;\mathrm{Al}&gt;\mathrm{K}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$\mathrm{K}&gt;\mathrm{Al}&gt;\mathrm{Cr}&gt;\mathrm{Ag}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\mathrm{K}&gt;\mathrm{Al}&gt;\mathrm{Ag}&gt;\mathrm{Cr}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\mathrm{Al}&gt;\mathrm{K}&gt;\mathrm{Ag}&gt;\mathrm{Cr}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\mathrm{K}&gt;\mathrm{Al}&gt;\mathrm{Cr}&gt;\mathrm{Ag}$</span> </div>

answer:
<div class="solution">More negative the value of standard reduction potential, higher is the reduction power.<br/>i.e. Reducing power $\propto \frac{1}{\text { standard reduction potential }}$<br/>Thus, the correct decreasing order of reducing power of the metal is<br/>$$<br/>\begin{aligned}<br/>&amp; \begin{array}{llll}<br/>\mathrm{K} &amp; &gt; &amp; \mathrm{Al} &amp; &gt;<br/>\end{array} \\<br/>&amp; \left(\mathrm{E}_{\mathrm{K}^{+} / \mathrm{K}}^0=-2.93 \mathrm{~V}\right) \quad\left(\mathrm{E}_{\mathrm{Al}^{3+} / \mathrm{Al}}^0=-1.66 \mathrm{~V}\right) \\<br/>&amp; \mathrm{Cr}&gt;\mathrm{Ag} \\<br/>&amp; \left(\mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^0=-0.74 \mathrm{~V}\right) \quad\left(\mathrm{E}_{\mathrm{Ag}^{+}}^0 / \mathrm{Ag}=0.80 \mathrm{~V}\right) \\<br/>&amp;<br/>\end{aligned}<br/>$$</div>


tags: ['Chemistry', 'Electrochemistry', 'COMEDK', 'COMEDK 2015']
source_db: MarksBatch2_P3.db
title:
the-standard-electrode-potential-for-daniell-cell-is-11-v-what-is-the-standard-gibbs-energy-for-the-reaction

slug:
the-standard-electrode-potential-for-daniell-cell-is-11-v-what-is-the-standard-gibbs-energy-for-the-reaction-88899

question:
<div class="question">The standard electrode potential for Daniell cell is $1.1 \mathrm{~V}$. What is the standard Gibbs energy for the reaction?</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$212.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$-2123 \mathrm{~kJ} \mathrm{~mol}^{-1}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$106.15 \mathrm{~kJ} \mathrm{~mol}^{-1}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$-106.15 \mathrm{~kJ} \mathrm{~mol}^{-1}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$-2123 \mathrm{~kJ} \mathrm{~mol}^{-1}$</span> </div>

answer:
<div class="solution">Net reaction occuring in a Daniell cell is<br/>$$<br/>\begin{aligned}<br/>\mathrm{Zn}(s) &amp;+\mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s) \\<br/>\Delta G &amp;=-n F E^{\circ}=-2 \times 96500 \times 1.1 \\<br/>&amp;=-212300 \mathrm{~J} \mathrm{~mol}{ }^{-1}=-212.3 \mathrm{~kJ} \mathrm{~mol}^{-1}<br/>\end{aligned}<br/>$$</div>


tags: ['Chemistry', 'Electrochemistry', 'KCET', 'KCET 2009']
source_db: MarksBatch2_P3.db
title:
the-standard-electrode-potential-for-the-halfcell-reactions-are-zn-2-2-e-zn-e-076-v-fe-2-2-e-fe-e-044-v-t-h-ee-m-f-o-f-t-h-ece-ll-re-a-c-t-i-o-n-mathr

slug:
the-standard-electrode-potential-for-the-halfcell-reactions-are-zn-2-2-e-zn-e-076-v-fe-2-2-e-fe-e-044-v-t-h-ee-m-f-o-f-t-h-ece-ll-re-a-c-t-i-o-n-mathr-25219

question:
<div class="question">The standard electrode potential for the half-cell reactions are<br/>$$<br/>\begin{aligned}<br/>&amp;\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} ; \mathrm{E}^{\circ}=-0.76 \mathrm{~V} \\<br/>&amp;\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} ; \mathrm{E}^{\circ}=-0.44 \mathrm{~V}<br/>\end{aligned}<br/>$$<br/>The emf of the cell reaction,<br/>$\mathrm{Fe}^{2+}+\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Fe}$ is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$-0.32 \mathrm{~V}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$-1.20 \mathrm{~V}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$+1.20 \mathrm{~V}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$+0.32 \mathrm{~V}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$+0.32 \mathrm{~V}$</span> </div>

answer:
<div class="solution">$\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} ; \mathrm{E}^{\circ}=-0.76 \mathrm{~V}$ $\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} ; \mathrm{E}^{\circ}=-0.44 \mathrm{~V}$ <br/>Cell reaction is <br/>$\begin{aligned} \mathrm{Fe}^{2+}+\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Fe} \\ \mathrm{E}_{\text {cell }} &amp;=\mathrm{E}_{\text {cathode }}-\mathrm{E}_{\text {anode }} \\ &amp;=-0.44-(-0.76) \\ &amp;=-0.44+0.76 \\ &amp;=0.32 \mathrm{~V} \end{aligned}$</div>


title: the-standard-electrode-potential-in-v-values-for-al-3-al-tl-3-t-1-are-respectively
slug: the-standard-electrode-potential-in-v-values-for-al-3-al-tl-3-t-1-are-respectively-92808
tags: ['Chemistry', 'Electrochemistry', 'AP EAMCET', 'AP EAMCET 2023 (18 May Shift 2)']
source_db: MarksBatch2_P3.db
question:
<div class="question">The standard electrode potential (in V) values for $\mathrm{Al}^{3+}$ / $\mathrm{Al}, \mathrm{Tl}^{3+} / \mathrm{T} 1$ are respectively</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$-1.66,-1.26$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$+1.66,+1.26$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$-1.66,+1.26$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$+1.66,-1.26$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$-1.66,+1.26$</span> </div>

answer:
<div class="solution">($\mathrm{Al}^{3+}$ is much more stable than $\mathrm{Al}$ and due to electropositive nature, the standard reduction potential of $\mathrm{Al}^{3+} / \mathrm{Al}$ is negative.<br/>$\mathrm{Tl}$ is not as electropositive as $\mathrm{Al}$ and therefore $\mathrm{Tl}^{3+}$ can easily be reduced to Tl because of which its SEP value is positive.</div>


tags: ['Chemistry', 'Electrochemistry', 'JEE Main', 'JEE Main 2023 (29 Jan Shift 1)']
source_db: MarksBatch2_P3.db
title:
the-standard-electrode-potential-m-3-m-2-for-v-cr-mn-co-are-0-26-v-0-41-v-1-57-v-and-1-97-v-respectively-the-metal-ions-which-can-liberate-h-2-from-a-

slug:
the-standard-electrode-potential-m-3-m-2-for-v-cr-mn-co-are-0-26-v-0-41-v-1-57-v-and-1-97-v-respectively-the-metal-ions-which-can-liberate-h-2-from-a-69628

question:
<div class="question">The standard electrode potential (<math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi mathvariant="normal">M</mi><mrow><mn>3</mn><mo>+</mo></mrow></msup><mo>/</mo><msup><mi mathvariant="normal">M</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup></math>) for <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">V</mi><mo>,</mo><mo> </mo><mi>Cr</mi><mo>,</mo><mo> </mo><mi>Mn</mi></math> &amp; <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>Co</mi></math> are <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mn>0</mn><mo>.</mo><mn>26</mn><mo> </mo><mi mathvariant="normal">V</mi><mo>,</mo><mo> </mo><mo>-</mo><mo> </mo><mn>0</mn><mo>.</mo><mn>41</mn><mo> </mo><mi mathvariant="normal">V</mi><mo>,</mo><mo> </mo><mo>+</mo><mn>1</mn><mo>.</mo><mn>57</mn><mo> </mo><mi mathvariant="normal">V</mi></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>+</mo><mn>1</mn><mo>.</mo><mn>97</mn><mo> </mo><mi mathvariant="normal">V</mi></math>, respectively. The metal ions which can liberate <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi mathvariant="normal">H</mi><mn>2</mn></msub></math> from a dilute acid are</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi mathvariant="normal">V</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi>Mn</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi>Cr</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi>Co</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup></math></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi mathvariant="normal">V</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi>Cr</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi>Mn</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi>Co</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi mathvariant="normal">V</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi>Cr</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup></math></span> </div>

answer:
<div class="solution"><p>The <math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi mathvariant="normal">E</mi><mo>⊝</mo></msup></math> values for the redox couple (<math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi mathvariant="normal">M</mi><mrow><mn>3</mn><mo>+</mo></mrow></msup><mo>/</mo><msup><mi mathvariant="normal">M</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup></math>) shows that <math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi>Mn</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi>Co</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup></math> ions are the strongest oxidising agents in aqueous solution.</p><p>The ions <math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi>Ti</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup><mo>,</mo><mo> </mo><msup><mi mathvariant="normal">V</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup><mo> </mo><mi>and</mi><mo> </mo><msup><mi>Cr</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup></math> are strong reducing agents and will liberate hydrogen from a dilute acid.</p><p>So, the Correct answer is <math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi mathvariant="normal">V</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup><mo> </mo><mi>and</mi><mo> </mo><msup><mi>Cr</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup></math>.</p></div>


tags: ['Chemistry', 'Electrochemistry', 'KVPY', 'KVPY 2013 (SB/SX)']
source_db: MarksBatch2_P3.db
title:
the-standard-electrode-potential-of-zn-2-zn-is-076-v-and-that-of-ca-2-cu-is-034-v-the-emf-v-and-the-free-energy-change-kj-mol-1-respectively-for-a-dan

slug:
the-standard-electrode-potential-of-zn-2-zn-is-076-v-and-that-of-ca-2-cu-is-034-v-the-emf-v-and-the-free-energy-change-kj-mol-1-respectively-for-a-dan-83793

question:
<div class="question">The standard electrode potential of $\mathrm{Zn}^{2+} / \mathrm{Zn}$ is $-0.76 \mathrm{~V}$ and that of $\mathrm{Ca}^{2+} / \mathrm{Cu}$ is $0.34 \mathrm{~V}$. The emf $(\mathrm{V})$ and the free energy change $\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$, respectively, for a Daniel cell will be</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$-0.42$ and 81</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$1.1$ and -213</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$-1.1$ and 213</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$0.42$ and $-81$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$1.1$ and -213</span> </div>

answer:
<div class="solution">$\mathrm{E}_{\mathrm{Cell}}^{0}=\mathrm{E}^{0} \mathrm{Cu}^{2+}\left|\mathrm{Cu}+\mathrm{E}^{0} \mathrm{Zn}\right| \mathrm{Zn}^{2+}$<br/>$\mathrm{E}_{\mathrm{cell}}^{0}=0.34+0.76=+1.1 \mathrm{~V}$<br/>$\Delta G=-2 \times 96500 \times 1.1=-213 V$</div>


tags: ['Chemistry', 'Electrochemistry', 'JEE Main']
source_db: MarksBatch2_P3.db
title:
the-standard-electrode-potentials-e-0-of-f-e-3-f-e-2-and-f-e-2-f-e-at-300-k-are-077-v-and-044-v-respectively-the-e-0-of-fe-3-fe-at-the-same-temperatur

slug:
the-standard-electrode-potentials-e-0-of-f-e-3-f-e-2-and-f-e-2-f-e-at-300-k-are-077-v-and-044-v-respectively-the-e-0-of-fe-3-fe-at-the-same-temperatur-85835

question:
<div class="question">The standard electrode potentials, <math><msup><mrow><mi mathvariant="normal">E</mi></mrow><mrow><mn>0</mn></mrow></msup></math> of <math><mrow><mrow><mi mathvariant="normal">F</mi><msup><mrow><mi mathvariant="normal">e</mi></mrow><mrow><mn>3</mn><mo>+</mo></mrow></msup></mrow><mo>/</mo><mrow><mi mathvariant="normal">F</mi><msup><mrow><mi mathvariant="normal">e</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup></mrow></mrow></math> and <math><mrow><mrow><mi mathvariant="normal">F</mi><msup><mrow><mi mathvariant="normal">e</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup></mrow><mo>/</mo><mrow><mi mathvariant="normal">F</mi><mi mathvariant="normal">e</mi></mrow></mrow></math> at 300 K are +0.77 V and -0.44 V, respectively. The <math><msup><mrow><mi mathvariant="normal">E</mi></mrow><mrow><mn>0</mn></mrow></msup></math> of <math> <mrow> <msup> <mrow> <mtext>Fe</mtext></mrow> <mtext>3</mtext> </msup> <msup> <mrow></mrow> <mtext>+</mtext> </msup> <mtext>/Fe</mtext></mrow> </math> at the same temperature is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">1.21 V</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">0.33 V</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">-0.036 V</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">0.036</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">-0.036 V</span> </div>

answer:
<div class="solution"><math> <mrow> <msup> <mrow> <mtext>Fe</mtext></mrow> <mrow> <mtext>3+</mtext></mrow> </msup> <mtext>â</mtext><mtext>+</mtext><mtext>â</mtext><msup> <mtext>e</mtext> <mo>â</mo> </msup> <mo>â</mo><msup> <mrow> <mtext>Fe</mtext></mrow> <mrow> <mtext>2+</mtext></mrow> </msup> <mtext>;</mtext><mtext>â</mtext><mtext>â</mtext><mtext>â</mtext><msubsup> <mrow> <mtext>ÎG</mtext></mrow> <mtext>1</mtext> <mtext>o</mtext> </msubsup> <mtext>â</mtext><mtext>=</mtext><mtext>â</mtext><mo>â</mo><mtext>1</mtext><mtext>â</mtext><mtext>Ã</mtext><mtext>â</mtext><mtext>F</mtext><mtext>â</mtext><mtext>Ã</mtext><mtext>â</mtext><mtext>0</mtext><mtext>.77</mtext></mrow> </math><br/><math> <mrow> <msup> <mrow> <mtext>Fe</mtext></mrow> <mrow> <mtext>2+</mtext></mrow> </msup> <mtext>â</mtext><mtext>+</mtext><mtext>â</mtext><mtext>2</mtext><mtext>â</mtext><msup> <mtext>e</mtext> <mo>â</mo> </msup> <mo>â</mo><mtext>Fe;</mtext><mtext>â</mtext><mtext>â</mtext><mtext>â</mtext><msubsup> <mrow> <mtext>ÎG</mtext></mrow> <mtext>2</mtext> <mtext>o</mtext> </msubsup> <mtext>â</mtext><mtext>=</mtext><mtext>â</mtext><mtext>2</mtext><mtext>â</mtext><mtext>Ã</mtext><mtext>â</mtext><mtext>F</mtext><mtext>â</mtext><mtext>Ã</mtext><mtext>â</mtext><mtext>0</mtext><mtext>.44</mtext></mrow> </math><br/><math> <mrow> <msup> <mrow> <mtext>Fe</mtext></mrow> <mrow> <mtext>3+</mtext></mrow> </msup> <mtext>â</mtext><mtext>+</mtext><mtext>â</mtext><mn>3</mn><mtext>â</mtext><msup> <mtext>e</mtext> <mo>â</mo> </msup> <mo>â</mo><mtext>Fe;</mtext><mtext>â</mtext><mtext>â</mtext><mtext>â</mtext><msubsup> <mrow> <mtext>ÎG</mtext></mrow> <mtext>3</mtext> <mtext>o</mtext> </msubsup> <mo>=</mo><mo>â</mo><mn>3</mn><mo>Ã</mo><mtext>F</mtext><mo>Ã</mo><msup> <mtext>E</mtext> <mtext>o</mtext> </msup> </mrow> </math><br/><math> <mrow> <mtext>=0</mtext><mtext>.11</mtext><mtext>â</mtext><mtext>V</mtext></mrow> </math><br/><math> <mrow> <mo>∴</mo><mtext> </mtext><msup> <mtext>E</mtext> <mtext>o</mtext> </msup> <mo>=</mo><mfrac> <mrow> <mo>−</mo><mn>0.11</mn></mrow> <mn>3</mn> </mfrac> <mo>=</mo><mo>−</mo><mn>0.036</mn><mtext> </mtext><mtext>V</mtext></mrow></math></div>


tags: ['Chemistry', 'Electrochemistry', 'JEE Main', 'JEE Main 2014 (09 Apr Online)']
source_db: MarksBatch2_P3.db
title:
the-standard-electrode-potentials-e-m-m-o-of-four-metals-a-b-c-and-d-are-1-2-v-0-6-v-0-85-v-and-0-76-v-respectively-the-sequence-of-deposition-of-meta-1

slug:
the-standard-electrode-potentials-e-m-m-o-of-four-metals-a-b-c-and-d-are-1-2-v-0-6-v-0-85-v-and-0-76-v-respectively-the-sequence-of-deposition-of-meta-1-37610

question:
<div class="question">The standard electrode potentials <math> <mrow> <mfenced close=")" open="(" separators=","> <mrow> <msubsup> <mtext>E</mtext> <mrow> <msup> <mtext>M</mtext> <mo>+</mo> </msup> <mo>/</mo> <mtext>M</mtext> </mrow> <mtext>o</mtext> </msubsup> </mrow> </mfenced> </mrow> </math> of four metals <math> <mrow> <mtext>A,</mtext><mtext> </mtext><mtext>B,</mtext><mtext> </mtext><mtext>C</mtext></mrow> </math> and <math> <mtext>D</mtext> </math> are <math> <mrow> <mo>−</mo><mtext>1</mtext><mtext>.2</mtext><mtext> </mtext><mtext>V,</mtext></mrow> </math><math> <mrow> <mtext>0</mtext><mtext>.6</mtext><mtext> </mtext><mtext>V</mtext></mrow> </math>, <math> <mrow> <mtext>0</mtext><mtext>.85</mtext><mtext> </mtext><mtext>V</mtext></mrow> </math> and <math> <mrow> <mo>−</mo><mtext>0</mtext><mtext>.76</mtext><mtext> </mtext><mtext>V</mtext></mrow> </math>, respectively. The sequence of deposition of metals on applying potential is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">B</mi><mo> </mo><mo>&gt;</mo><mo> </mo><mi mathvariant="normal">D</mi><mo> </mo><mo>&gt;</mo><mo> </mo><mi mathvariant="normal">C</mi><mo> </mo><mo>&gt;</mo><mo> </mo><mi mathvariant="normal">A</mi></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">A</mi><mo> </mo><mo>&gt;</mo><mo> </mo><mi mathvariant="normal">C</mi><mo> </mo><mo>&gt;</mo><mo> </mo><mi mathvariant="normal">B</mi><mo> </mo><mo>&gt;</mo><mo> </mo><mi mathvariant="normal">D</mi></math></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">C</mi><mo> </mo><mo>&gt;</mo><mo> </mo><mi mathvariant="normal">B</mi><mo> </mo><mo>&gt;</mo><mo> </mo><mi mathvariant="normal">D</mi><mo> </mo><mo>&gt;</mo><mo> </mo><mi mathvariant="normal">A</mi></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">D</mi><mo> </mo><mo>&gt;</mo><mo> </mo><mi mathvariant="normal">A</mi><mo> </mo><mo>&gt;</mo><mo> </mo><mi mathvariant="normal">B</mi><mo> </mo><mo>&gt;</mo><mo> </mo><mi mathvariant="normal">C</mi></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">C</mi><mo> </mo><mo>&gt;</mo><mo> </mo><mi mathvariant="normal">B</mi><mo> </mo><mo>&gt;</mo><mo> </mo><mi mathvariant="normal">D</mi><mo> </mo><mo>&gt;</mo><mo> </mo><mi mathvariant="normal">A</mi></math></span> </div>

answer:
<div class="solution"><blockquote><p>In complete cation electrolysis at cathode, the species having higher reduction potential is preferably reduced at cathode<br/><math> <mrow> <msubsup> <mtext>E</mtext> <mtext>C</mtext> <mo>∘</mo> </msubsup> <mo>&gt;</mo> <msubsup> <mtext>E</mtext> <mtext>B</mtext> <mo>∘</mo> </msubsup> <mo>&gt;</mo> <msubsup> <mtext>E</mtext> <mtext>D</mtext> <mo>∘</mo> </msubsup> <mo>&gt;</mo> <msubsup> <mtext>E</mtext> <mtext>A</mtext> <mo>∘</mo> </msubsup> </mrow> </math><br/>Hence,  <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">C</mi><mo> </mo><mo>&gt;</mo><mo> </mo><mi mathvariant="normal">B</mi><mo> </mo><mo>&gt;</mo><mo> </mo><mi mathvariant="normal">D</mi><mo> </mo><mo>&gt;</mo><mo> </mo><mi mathvariant="normal">A</mi></math><br/>will be the preference of deposition of these metals on competing.</p></blockquote></div>


tags: ['Chemistry', 'Electrochemistry', 'JEE Main']
source_db: MarksBatch2_P3.db
title:
the-standard-electrode-potentials-e-m-m-o-of-four-metals-a-b-c-and-d-are-1-2-v-0-6-v-0-85-v-and-0-76-v-respectively-the-sequence-of-deposition-of-meta

slug:
the-standard-electrode-potentials-e-m-m-o-of-four-metals-a-b-c-and-d-are-1-2-v-0-6-v-0-85-v-and-0-76-v-respectively-the-sequence-of-deposition-of-meta-71557

question:
<div class="question">The standard electrode potentials <math> <mrow> <mfenced close=")" open="(" separators=","> <mrow> <msubsup> <mtext>E</mtext> <mrow> <msup> <mtext>M</mtext> <mo>+</mo> </msup> <mo>/</mo> <mtext>M</mtext> </mrow> <mtext>o</mtext> </msubsup> </mrow> </mfenced> </mrow> </math> of four metals <math> <mrow> <mtext>A,</mtext><mtext> </mtext><mtext>B,</mtext><mtext> </mtext><mtext>C</mtext></mrow> </math>
 and <math> <mtext>D</mtext> </math>
 are <math> <mrow> <mo>−</mo><mtext>1</mtext><mtext>.2</mtext><mtext> </mtext><mtext>V,</mtext></mrow> </math>, <math> <mrow> <mtext>0</mtext><mtext>.6</mtext><mtext> </mtext><mtext>V</mtext></mrow> </math>, <math> <mrow> <mtext>0</mtext><mtext>.85</mtext><mtext> </mtext><mtext>V</mtext></mrow> </math>
 and <math> <mrow> <mo>−</mo><mtext>0</mtext><mtext>.76</mtext><mtext> </mtext><mtext>V</mtext></mrow> </math>, respectively. The sequence of deposition of metals on applying potential is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">B &gt; D &gt; C &gt; A</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">A &gt; C &gt; B &gt; D</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">C &gt; B &gt; D &gt; A</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">D &gt; A &gt; B &gt; C</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">C &gt; B &gt; D &gt; A</span> </div>

answer:
<div class="solution">In complete cation electrolysis at cathode, the species having higher reduction potential is preferably reduced at cathode<br/><math> <mrow> <msubsup> <mtext>E</mtext> <mtext>C</mtext> <mo>â</mo> </msubsup> <mo>&gt;</mo> <msubsup> <mtext>E</mtext> <mtext>B</mtext> <mo>â</mo> </msubsup> <mo>&gt;</mo> <msubsup> <mtext>E</mtext> <mtext>D</mtext> <mo>â</mo> </msubsup> <mo>&gt;</mo> <msubsup> <mtext>E</mtext> <mtext>A</mtext> <mo>â</mo> </msubsup> </mrow> </math><br/>Hence,  C &gt; B &gt; D &gt; A<br/>will be the preference of deposition of these metals on competeting.</div>


tags: ['Chemistry', 'Electrochemistry', 'JEE Main']
source_db: MarksBatch2_P3.db
title:
the-standard-electrode-potentials-of-z-n-2-z-n-and-a-g-a-g-are-0763-v-and-0799-v-respectively-the-standard-potential-of-the-cell-is

slug:
the-standard-electrode-potentials-of-z-n-2-z-n-and-a-g-a-g-are-0763-v-and-0799-v-respectively-the-standard-potential-of-the-cell-is-89357

question:
<div class="question">The standard electrode potentials of $Z n^{2+} / Z n$ and $A g^{+} / A g$ are $-0.763 \mathrm{~V}$ and $+0.799 \mathrm{~V}$ respectively. The standard potential of the cell is</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$1.56 \mathrm{~V}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$0.036 \mathrm{~V}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$-1.562 V$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$0.799 \mathrm{~V}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$1.56 \mathrm{~V}$</span> </div>

answer:
<div class="solution">$E_{\text {cell }}^o=E_{\text {cathode }}^o-E_{\text {thode }}^o=0.799-(-0.763)=1.562 \mathrm{~V}$</div>


tags: ['Chemistry', 'Electrochemistry', 'JEE Main']
source_db: MarksBatch2_P3.db
title:
the-standard-electrode-potentials-of-z-n-2-z-n-c-u-2-c-u-a-n-d-a-g-a-g-are-respectively-076-034-and-08-v-the-following-cells-were-constructed-i-z-n-z--1

slug:
the-standard-electrode-potentials-of-z-n-2-z-n-c-u-2-c-u-a-n-d-a-g-a-g-are-respectively-076-034-and-08-v-the-following-cells-were-constructed-i-z-n-z-1-57325

question:
<div class="question">The standard electrode potentials of<br/> <br/> <math><msup><mrow><mi mathvariant="normal">Z</mi><mi mathvariant="normal">n</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>|</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">Z</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal"> </mi><mo>,</mo><mi mathvariant="normal"> </mi><msup><mrow><mi mathvariant="normal">C</mi><mi mathvariant="normal">u</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>|</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">C</mi><mi mathvariant="normal">u</mi><mi mathvariant="normal"> </mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal">d</mi><mi mathvariant="normal"> </mi><msup><mrow><mi mathvariant="normal">A</mi><mi mathvariant="normal">g</mi></mrow><mrow><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>|</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">A</mi><mi mathvariant="normal">g</mi></math> are respectively - 0.76, 0.34 and 0.8 V. The following cells were constructed<br/> <br/>
I -  <math><mi mathvariant="normal">Z</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal"> </mi><mo>|</mo><mi mathvariant="normal"> </mi><msup><mrow><mi mathvariant="normal">Z</mi><mi mathvariant="normal">n</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>|</mo><mo>|</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal"> </mi><msup><mrow><mi mathvariant="normal">C</mi><mi mathvariant="normal">u</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>|</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">C</mi><mi mathvariant="normal">u</mi></math><br/> <br/>
II -  <math><mi mathvariant="normal">Z</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal"> </mi><mo>|</mo><mi mathvariant="normal"> </mi><msup><mrow><mi mathvariant="normal">Z</mi><mi mathvariant="normal">n</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>|</mo><mo>|</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal"> </mi><msup><mrow><mi mathvariant="normal">A</mi><mi mathvariant="normal">g</mi></mrow><mrow><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>|</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">A</mi><mi mathvariant="normal">g</mi></math><br/> <br/>
III - <math><mi mathvariant="normal">C</mi><mi mathvariant="normal">u</mi><mi mathvariant="normal"> </mi><mo>|</mo><mi mathvariant="normal"> </mi><msup><mrow><mi mathvariant="normal">C</mi><mi mathvariant="normal">u</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>|</mo><mo>|</mo><mi mathvariant="normal"> </mi><msup><mrow><mi mathvariant="normal">A</mi><mi mathvariant="normal">g</mi></mrow><mrow><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>|</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">A</mi><mi mathvariant="normal">g</mi></math><br/> <br/>
What is the correct order of <math><msubsup><mrow><mtext>E</mtext></mrow><mrow><mi mathvariant="normal">c</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">l</mi><mi mathvariant="normal">l</mi></mrow><mrow><mo>Â°</mo></mrow></msubsup></math> of these cells?</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">II &gt; III &gt; I</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">II &gt; I &gt; III</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">I &gt; II &gt; III</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">III &gt; I &gt; II</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">II &gt; I &gt; III</span> </div>

answer:
<div class="solution">Given,<br/><br/><math> <msup> <mtext>Zn</mtext> <mrow> <mn>2</mn><mo>+</mo></mrow> </msup> <mtext>â</mtext><mtext>+</mtext><mtext>â</mtext><mtext>â</mtext><mtext>â</mtext><msup> <mtext>2e</mtext> <mo>â</mo> </msup> <mtext>  </mtext><mo>â</mo><mo> </mo><mtext>Zn</mtext><mo> </mo><mo>,</mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><msup> <mtext>E</mtext> <mo>Â°</mo> </msup> <mo>=</mo><mo>â</mo><mn>0.76</mn><mo> </mo><mtext>V</mtext> </math><br/><br/><math> <msup> <mtext>Cu</mtext> <mrow> <mn>2</mn><mo>+</mo></mrow> </msup> <mtext> +</mtext><mtext>â</mtext><mtext>â</mtext><msup> <mtext>2e</mtext> <mo>â</mo> </msup> <mtext> </mtext><mo>â</mo><mo> </mo><mtext>Cu</mtext><mo> </mo><mo>,</mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><msup> <mtext>E</mtext> <mo>Â°</mo> </msup> <mo> </mo><mo>=</mo><mo> </mo><mn>0.34</mn><mo> </mo><mtext>V</mtext> </math><br/><br/><math> <msup> <mtext>Ag</mtext> <mo>+</mo> </msup> <mtext>  +</mtext><mtext>â</mtext><mtext>â</mtext><msup> <mtext>e</mtext> <mo>â</mo> </msup> <mo>â</mo><mo> </mo><mtext>Ag </mtext><mo>,</mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><msup> <mtext>E</mtext> <mo>Â°</mo> </msup> <mo> </mo><mo>=</mo><mo> </mo><mn>0.8</mn><mo> </mo><mtext>V</mtext> </math><br/><br/>Cell reaction of (I) is<br/><br/><math><mi mathvariant="normal">Z</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal"> </mi><mo>+</mo><mi mathvariant="normal"> </mi><msup><mrow><mi mathvariant="normal">C</mi><mi mathvariant="normal">u</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>â</mo><mi mathvariant="normal"> </mi><msup><mrow><mi mathvariant="normal">Z</mi><mi mathvariant="normal">n</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>+</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">C</mi><mi mathvariant="normal">u</mi></math><br/><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><msubsup><mi mathvariant="normal">E</mi><mi>cell</mi><mi mathvariant="normal">o</mi></msubsup><mo>=</mo><msubsup><mi mathvariant="normal">E</mi><mi>red</mi><mi mathvariant="normal">o</mi></msubsup><mo>(</mo><mi>cathode</mi><mo>)</mo><mo>-</mo><msubsup><mi mathvariant="normal">E</mi><mi>red</mi><mi mathvariant="normal">o</mi></msubsup><mo>(</mo><mi>anode</mi><mo>)</mo></math><br/><math> <mrow> <mtable columnalign="left" equalcolumns="true" equalrows="true"> <mtr columnalign="left"> <mtd columnalign="left"> <mrow> <mo>=</mo><mn>0.34</mn><mo>â</mo><mfenced> <mrow> <mo>â</mo><mn>0.76</mn></mrow> </mfenced></mrow> </mtd> </mtr> <mtr columnalign="left"> <mtd columnalign="left"> <mrow> <mo>=</mo><mn>1.10</mn><mtext>â</mtext><mtext>V</mtext></mrow> </mtd> </mtr> </mtable></mrow> </math><br/><br/>Cell reaction of (II) is<br/><br/><math> <mrow> <mtext>Zn </mtext><mo>+</mo><msup> <mrow> <mtext> 2Ag</mtext></mrow> <mo>+</mo> </msup> <mtext> </mtext><mo>â</mo><msup> <mrow> <mtext> Zn</mtext></mrow> <mrow> <mn>2</mn><mo>+</mo></mrow> </msup> <mtext> </mtext><mo>+</mo><mtext> 2Ag</mtext></mrow> </math><br/><br/><math> <mrow> <msubsup> <mtext>E</mtext> <mrow> <mtext>Cell</mtext></mrow> <mi>o</mi> </msubsup> <mo>=</mo><mo>+</mo><mn>0.80</mn><mo>â</mo><mfenced> <mrow> <mo>â</mo><mn>0.76</mn></mrow> </mfenced><mo>=</mo><mo>+</mo><mn>1.56</mn><mtext>â</mtext><mtext>V</mtext></mrow> </math><br/><br/>Cell reaction of (III) is<br/><br/><math> <mrow> <mtext>Cu </mtext><mo>+</mo><msup> <mrow> <mtext> 2Ag</mtext></mrow> <mo>+</mo> </msup> <mtext> </mtext><mo>â</mo><msup> <mrow> <mtext> Cu</mtext></mrow> <mrow> <mn>2</mn><mo>+</mo></mrow> </msup> <mtext> </mtext><mo>+</mo><mtext> 2Ag</mtext></mrow> </math><br/><br/><math> <mrow> <msubsup> <mtext>E</mtext> <mrow> <mtext>Cell</mtext></mrow> <mi>o</mi> </msubsup> <mo>=</mo><mo>+</mo><mn>0.80</mn><mo>â</mo><mfenced> <mrow> <mn>0.34</mn></mrow> </mfenced><mo>=</mo><mo>+</mo><mn>0.46</mn><mtext>â</mtext><mtext>V</mtext></mrow> </math><br/><br/>So, the correct order of <math><msubsup><mrow><mtext>E</mtext></mrow><mrow><mi mathvariant="normal">c</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">l</mi><mi mathvariant="normal">l</mi></mrow><mrow><mo>Â°</mo></mrow></msubsup></math> of these cell is<br/><br/>II &gt; I &gt; III.</div>


tags: ['Chemistry', 'Electrochemistry', 'NEET']
source_db: MarksBatch2_P3.db
title:
the-standard-electrode-potentials-of-z-n-2-z-n-c-u-2-c-u-a-n-d-a-g-a-g-are-respectively-076-034-and-08-v-the-following-cells-were-constructed-i-z-n-z-

slug:
the-standard-electrode-potentials-of-z-n-2-z-n-c-u-2-c-u-a-n-d-a-g-a-g-are-respectively-076-034-and-08-v-the-following-cells-were-constructed-i-z-n-z-40658

question:
<div class="question">The standard electrode potentials of<br/> <br/> <math><msup><mrow><mi mathvariant="normal">Z</mi><mi mathvariant="normal">n</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>|</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">Z</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal"> </mi><mo>,</mo><mi mathvariant="normal"> </mi><msup><mrow><mi mathvariant="normal">C</mi><mi mathvariant="normal">u</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>|</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">C</mi><mi mathvariant="normal">u</mi><mi mathvariant="normal"> </mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal">d</mi><mi mathvariant="normal"> </mi><msup><mrow><mi mathvariant="normal">A</mi><mi mathvariant="normal">g</mi></mrow><mrow><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>|</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">A</mi><mi mathvariant="normal">g</mi></math> are respectively - 0.76, 0.34 and 0.8 V. The following cells were constructed<br/> <br/>
I -  <math><mi mathvariant="normal">Z</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal"> </mi><mo>|</mo><mi mathvariant="normal"> </mi><msup><mrow><mi mathvariant="normal">Z</mi><mi mathvariant="normal">n</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>|</mo><mo>|</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal"> </mi><msup><mrow><mi mathvariant="normal">C</mi><mi mathvariant="normal">u</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>|</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">C</mi><mi mathvariant="normal">u</mi></math><br/> <br/>
II -  <math><mi mathvariant="normal">Z</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal"> </mi><mo>|</mo><mi mathvariant="normal"> </mi><msup><mrow><mi mathvariant="normal">Z</mi><mi mathvariant="normal">n</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>|</mo><mo>|</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal"> </mi><msup><mrow><mi mathvariant="normal">A</mi><mi mathvariant="normal">g</mi></mrow><mrow><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>|</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">A</mi><mi mathvariant="normal">g</mi></math><br/> <br/>
III - <math><mi mathvariant="normal">C</mi><mi mathvariant="normal">u</mi><mi mathvariant="normal"> </mi><mo>|</mo><mi mathvariant="normal"> </mi><msup><mrow><mi mathvariant="normal">C</mi><mi mathvariant="normal">u</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>|</mo><mo>|</mo><mi mathvariant="normal"> </mi><msup><mrow><mi mathvariant="normal">A</mi><mi mathvariant="normal">g</mi></mrow><mrow><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>|</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">A</mi><mi mathvariant="normal">g</mi></math><br/> <br/>
What is the correct order of <math><msubsup><mrow><mtext>E</mtext></mrow><mrow><mi mathvariant="normal">c</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">l</mi><mi mathvariant="normal">l</mi></mrow><mrow><mo>Â°</mo></mrow></msubsup></math> of these cells?</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">II &gt; III &gt; I</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">II &gt; I &gt; III</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">I &gt; II &gt; III</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">III &gt; I &gt; II</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">II &gt; I &gt; III</span> </div>

answer:
<div class="solution">Given,<br/><br/><math> <msup> <mtext>Zn</mtext> <mrow> <mn>2</mn><mo>+</mo></mrow> </msup> <mtext>â</mtext><mtext>+</mtext><mtext>â</mtext><mtext>â</mtext><mtext>â</mtext><msup> <mtext>2e</mtext> <mo>â</mo> </msup> <mtext>  </mtext><mo>â</mo><mo> </mo><mtext>Zn</mtext><mo> </mo><mo>,</mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><msup> <mtext>E</mtext> <mo>Â°</mo> </msup> <mo>=</mo><mo>â</mo><mn>0.76</mn><mo> </mo><mtext>V</mtext> </math><br/><br/><math> <msup> <mtext>Cu</mtext> <mrow> <mn>2</mn><mo>+</mo></mrow> </msup> <mtext> +</mtext><mtext>â</mtext><mtext>â</mtext><msup> <mtext>2e</mtext> <mo>â</mo> </msup> <mtext> </mtext><mo>â</mo><mo> </mo><mtext>Cu</mtext><mo> </mo><mo>,</mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><msup> <mtext>E</mtext> <mo>Â°</mo> </msup> <mo> </mo><mo>=</mo><mo> </mo><mn>0.34</mn><mo> </mo><mtext>V</mtext> </math><br/><br/><math> <msup> <mtext>Ag</mtext> <mo>+</mo> </msup> <mtext>  +</mtext><mtext>â</mtext><mtext>â</mtext><msup> <mtext>e</mtext> <mo>â</mo> </msup> <mo>â</mo><mo> </mo><mtext>Ag </mtext><mo>,</mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><mo> </mo><msup> <mtext>E</mtext> <mo>Â°</mo> </msup> <mo> </mo><mo>=</mo><mo> </mo><mn>0.8</mn><mo> </mo><mtext>V</mtext> </math><br/><br/>Cell reaction of (I) is<br/><br/><math><mi mathvariant="normal">Z</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal"> </mi><mo>+</mo><mi mathvariant="normal"> </mi><msup><mrow><mi mathvariant="normal">C</mi><mi mathvariant="normal">u</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>â</mo><mi mathvariant="normal"> </mi><msup><mrow><mi mathvariant="normal">Z</mi><mi mathvariant="normal">n</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mi mathvariant="normal"> </mi><mo>+</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal">C</mi><mi mathvariant="normal">u</mi></math><br/><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><msubsup><mi mathvariant="normal">E</mi><mi>cell</mi><mi mathvariant="normal">o</mi></msubsup><mo>=</mo><msubsup><mi mathvariant="normal">E</mi><mi>red</mi><mi mathvariant="normal">o</mi></msubsup><mo>(</mo><mi>cathode</mi><mo>)</mo><mo>-</mo><msubsup><mi mathvariant="normal">E</mi><mi>red</mi><mi mathvariant="normal">o</mi></msubsup><mo>(</mo><mi>anode</mi><mo>)</mo></math><br/><math> <mrow> <mtable columnalign="left" equalcolumns="true" equalrows="true"> <mtr columnalign="left"> <mtd columnalign="left"> <mrow> <mo>=</mo><mn>0.34</mn><mo>â</mo><mfenced> <mrow> <mo>â</mo><mn>0.76</mn></mrow> </mfenced></mrow> </mtd> </mtr> <mtr columnalign="left"> <mtd columnalign="left"> <mrow> <mo>=</mo><mn>1.10</mn><mtext>â</mtext><mtext>V</mtext></mrow> </mtd> </mtr> </mtable></mrow> </math><br/><br/>Cell reaction of (II) is<br/><br/><math> <mrow> <mtext>Zn </mtext><mo>+</mo><msup> <mrow> <mtext> 2Ag</mtext></mrow> <mo>+</mo> </msup> <mtext> </mtext><mo>â</mo><msup> <mrow> <mtext> Zn</mtext></mrow> <mrow> <mn>2</mn><mo>+</mo></mrow> </msup> <mtext> </mtext><mo>+</mo><mtext> 2Ag</mtext></mrow> </math><br/><br/><math> <mrow> <msubsup> <mtext>E</mtext> <mrow> <mtext>Cell</mtext></mrow> <mi>o</mi> </msubsup> <mo>=</mo><mo>+</mo><mn>0.80</mn><mo>â</mo><mfenced> <mrow> <mo>â</mo><mn>0.76</mn></mrow> </mfenced><mo>=</mo><mo>+</mo><mn>1.56</mn><mtext>â</mtext><mtext>V</mtext></mrow> </math><br/><br/>Cell reaction of (III) is<br/><br/><math> <mrow> <mtext>Cu </mtext><mo>+</mo><msup> <mrow> <mtext> 2Ag</mtext></mrow> <mo>+</mo> </msup> <mtext> </mtext><mo>â</mo><msup> <mrow> <mtext> Cu</mtext></mrow> <mrow> <mn>2</mn><mo>+</mo></mrow> </msup> <mtext> </mtext><mo>+</mo><mtext> 2Ag</mtext></mrow> </math><br/><br/><math> <mrow> <msubsup> <mtext>E</mtext> <mrow> <mtext>Cell</mtext></mrow> <mi>o</mi> </msubsup> <mo>=</mo><mo>+</mo><mn>0.80</mn><mo>â</mo><mfenced> <mrow> <mn>0.34</mn></mrow> </mfenced><mo>=</mo><mo>+</mo><mn>0.46</mn><mtext>â</mtext><mtext>V</mtext></mrow> </math><br/><br/>So, the correct order of <math><msubsup><mrow><mtext>E</mtext></mrow><mrow><mi mathvariant="normal">c</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">l</mi><mi mathvariant="normal">l</mi></mrow><mrow><mo>Â°</mo></mrow></msubsup></math> of these cell is<br/><br/>II &gt; I &gt; III.</div>


tags: ['Chemistry', 'Electrochemistry', 'JEE Main']
source_db: MarksBatch2_P3.db
title:
the-standard-emf-of-a-cell-involving-one-electron-change-is-found-to-be-0591-v-at-2-5-c-the-equilibrium-constant-of-the-reaction-is-f-96-500-c-mol-1-r-1

slug:
the-standard-emf-of-a-cell-involving-one-electron-change-is-found-to-be-0591-v-at-2-5-c-the-equilibrium-constant-of-the-reaction-is-f-96-500-c-mol-1-r-1-21431

question:
<div class="question">The standard e.m.f of a cell, involving one electron change is found to be $0.591 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction is $\left(\mathrm{F}=96,500 \mathrm{C} \mathrm{mol}^{-1}: \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$1.0 \times 10^1$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$1.0 \times 10^{30}$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$1.0 \times 10^{10}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$1.0 \times 10^5$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$1.0 \times 10^{10}$<br/></span> </div>

answer:
<div class="solution">$\therefore \log \mathrm{K}_{\mathrm{eq}}=\frac{\mathrm{nE}^{\circ}}{0.0591}<br/>=\frac{1 \times 0.591}{0.0591}$ $\Rightarrow \mathrm{K}_{\mathrm{eq}}=10^{10}$</div>


tags: ['Chemistry', 'Electrochemistry', 'JEE Main', 'JEE Main 2004']
source_db: MarksBatch2_P3.db
title:
the-standard-emf-of-a-cell-involving-one-electron-change-is-found-to-be-0591-v-at-2-5-c-the-equilibrium-constant-of-the-reaction-is-f-96-500-c-mol-1-r

slug:
the-standard-emf-of-a-cell-involving-one-electron-change-is-found-to-be-0591-v-at-2-5-c-the-equilibrium-constant-of-the-reaction-is-f-96-500-c-mol-1-r-89549

question:
<div class="question">The standard e.m.f of a cell, involving one electron change is found to be $0.591 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction is $\left(\mathrm{F}=96,500 \mathrm{C} \mathrm{mol}^{-1}: \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$1.0 \times 10^1$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$1.0 \times 10^{30}$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$1.0 \times 10^{10}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$1.0 \times 10^5$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$1.0 \times 10^{10}$<br/></span> </div>

answer:
<div class="solution">$\therefore \log \mathrm{K}_{\mathrm{eq}}=\frac{\mathrm{nE}^{\circ}}{0.0591}<br/>=\frac{1 \times 0.591}{0.0591}$ $\Rightarrow \mathrm{K}_{\mathrm{eq}}=10^{10}$</div>


tags: ['Chemistry', 'Electrochemistry', 'NEET']
source_db: MarksBatch2_P3.db
title:
the-standard-emf-of-a-cell-involving-one-electron-change-is-found-to-be-0591-v-at-2-5-o-c-the-equilibrium-constant-of-the-reaction-is-f-96500-c-m-o-l-

slug:
the-standard-emf-of-a-cell-involving-one-electron-change-is-found-to-be-0591-v-at-2-5-o-c-the-equilibrium-constant-of-the-reaction-is-f-96500-c-m-o-l-10225

question:
<div class="question">The standard e.m.f of a cell, involving one electron change is found to be 0.591 V at <math><mn>2</mn><msup><mrow><mn>5 </mn></mrow><mrow><mi mathvariant="normal">o</mi></mrow></msup><mi mathvariant="normal">C</mi></math> . The equilibrium constant of the reaction is <math><mfenced separators="|"><mrow><mi mathvariant="normal">F</mi><mo>=</mo><mn>96500 </mn><mi mathvariant="normal"></mi><mi mathvariant="normal">C</mi><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><msup><mrow><mi mathvariant="normal">l</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup><mo>,</mo><mi mathvariant="normal">R</mi><mo>=</mo><mn>8.314 </mn><mi mathvariant="normal"></mi><mi mathvariant="normal">J</mi><msup><mrow><mi mathvariant="normal">K</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><msup><mrow><mi mathvariant="normal">l</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></mrow></mfenced></math></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math><mn>1.0</mn><mo>Ã</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>30</mn></mrow></msup></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math><mn>1.0</mn><mo>Ã</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>15</mn></mrow></msup></math></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math><mn>1.0</mn><mo>Ã</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>5</mn></mrow></msup></math></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><math><mn>1.0</mn><mo>Ã</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>10</mn></mrow></msup></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math><mn>1.0</mn><mo>Ã</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>10</mn></mrow></msup></math></span> </div>

answer:
<div class="solution">Relation between <math><msub><mrow><mi mathvariant="normal">K</mi></mrow><mrow><mi mathvariant="normal">e</mi><mi mathvariant="normal">q</mi></mrow></msub></math> and <math><msub><mrow><mi mathvariant="normal">E</mi></mrow><mrow><mi mathvariant="normal">c</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">l</mi><mi mathvariant="normal">l</mi></mrow></msub></math> is<br/><math> <mrow> <msubsup> <mtext>E</mtext> <mrow> <mtext>cell</mtext></mrow> <mtext>o</mtext> </msubsup> <mo>=</mo><mfrac> <mrow> <mn>2.303</mn><mtext>RT</mtext></mrow> <mrow> <mtext>nF</mtext></mrow> </mfrac> <msub> <mrow> <mtext>logK</mtext></mrow> <mrow> <mtext>eq</mtext></mrow> </msub> </mrow></math><br/><math> <mrow> <msubsup> <mtext>E</mtext> <mrow> <mtext>cell</mtext></mrow> <mtext>o</mtext> </msubsup> <mo>=</mo><mfrac> <mrow> <mn>0.0591</mn></mrow> <mtext>n</mtext> </mfrac> <msub> <mrow> <mtext>logK</mtext></mrow> <mrow> <mtext>eq</mtext></mrow> </msub> </mrow></math><br/><math> <mrow> <mn>0.591</mn><mo>=</mo><mfrac> <mrow> <mn>0.0591</mn></mrow> <mn>1</mn> </mfrac> <msub> <mrow> <mtext>logK</mtext></mrow> <mrow> <mtext>eq</mtext></mrow> </msub> </mrow></math><br/><math> <mrow> <msub> <mrow> <mtext>logK</mtext></mrow> <mrow> <mtext>eq</mtext></mrow> </msub> <mo>=</mo><mn>10</mn></mrow></math><br/><math> <mrow> <msub> <mtext>K</mtext> <mrow> <mtext>eq</mtext></mrow> </msub> <mo>=</mo><mn>1</mn><mo>×</mo><msup> <mrow> <mn>10</mn></mrow> <mrow> <mn>10</mn></mrow> </msup> </mrow></math></div>


tags: ['Chemistry', 'Electrochemistry', 'JEE Main']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">10</span> </div>
source_db: MarksBatch2_P3.db
title:
the-standard-emf-of-a-cell-involving-one-electron-change-is-found-to-be-0591-v-at-25-the-equilibrium-constant-of-the-reaction-is-10-x-find-the-value-o

slug:
the-standard-emf-of-a-cell-involving-one-electron-change-is-found-to-be-0591-v-at-25-the-equilibrium-constant-of-the-reaction-is-10-x-find-the-value-o-57026

question:
<div class="question">The standard EMF. of a cell, involving one electron change is found to be <math> <mrow> <mtext>0.591 V</mtext> </mrow> </math> at <math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>25</mtext><mi mathvariant="normal">â</mi></math>. The equilibrium constant of the reaction is <math> <mrow> <msup> <mrow> <mtext>10</mtext></mrow> <mtext>x</mtext> </msup> </mrow> </math>. Find the value of <math> <mtext>x</mtext> </math>.<br/> <math> <mrow> <mfenced close=")" open="(" separators=","> <mrow> <mtext>F</mtext> <mo>=</mo> <msup> <mtext>96500 C  equivalent</mtext><mn>-1</mn> </msup> <mtext>, R</mtext> <mo>=</mo> <msup> <mtext>8.314 J K</mtext> <mn>-1</mn> </msup> <msup> <mtext> mol</mtext> <mn>-1</mn> </msup> </mrow> </mfenced> </mrow> </math></div>

answer:
<div class="solution"><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ÎG</mi><mo>=</mo><mi>ÎG</mi><mo>Â°</mo><mo>â</mo><mfrac><mrow><mn>0</mn><mo>.591</mo></mrow><mi mathvariant="normal">n</mi></mfrac><mi>logQ</mi></math><br/><br/>at equilibrium<br/><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ÎG</mi><mo>=</mo><mn>0</mn><mo>,</mo><msub><mi mathvariant="normal">E</mi><mi>cell</mi></msub><mo>=</mo><mn>0</mn><mo>,</mo><mtext>â</mtext><mi mathvariant="normal">Q</mi><mo>=</mo><msub><mi mathvariant="normal">K</mi><mi mathvariant="normal">C</mi></msub><mo>,</mo><mi>ÎG</mi><mo>Â°</mo><mo>=</mo><mo>â</mo><mfrac><mrow><mn>0</mn><mo>.591</mo></mrow><mn>1</mn></mfrac><mi>logQ</mi></math><br/><br/><math> <mrow> <msub> <mtext>E</mtext> <mrow> <mtext> cell</mtext></mrow> </msub> <mtext>â</mtext><mtext>=</mtext><mtext>â</mtext><msubsup> <mtext>E</mtext> <mrow> <mtext> cell</mtext></mrow> <mtext>o</mtext> </msubsup> <mo>â</mo><mfrac> <mrow> <mtext>0</mtext><mtext>.0591</mtext></mrow> <mtext>n</mtext> </mfrac> <msub> <mrow> <mtext>log K</mtext></mrow> <mtext>c</mtext> </msub> </mrow> </math><br/><br/><math> <mrow> <mn>0</mn><mo> </mo><mo>=</mo><mo> </mo><mtext>0</mtext><mtext>.591</mtext><mo>â</mo><mfrac> <mrow> <mtext>0</mtext><mtext>.0591</mtext></mrow> <mn>1</mn> </mfrac> <mo> </mo><msub> <mrow> <mtext>log K</mtext></mrow> <mtext>c</mtext> </msub> </mrow> </math><br/><br/><math> <mrow> <mo>⇒</mo><mo> </mo><mo> </mo><mo>−</mo><mtext>0</mtext><mtext>.591</mtext><mo> </mo><mo>=</mo><mo> </mo><mo>−</mo><mtext>0</mtext><mtext>.0591</mtext><mo> </mo><msub> <mrow> <mtext>log K</mtext></mrow> <mtext>c</mtext> </msub> </mrow></math><br/><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>â    </mo><msub><mtext>log K</mtext><mtext>c</mtext></msub><mo> = </mo><mfrac><mtext>0.591</mtext><mtext>0.0591</mtext></mfrac><mo> = </mo><mn>10</mn></math><br/><br/><math> <mrow> <mo>â´     </mo> <msub> <mtext>K</mtext> <mtext>c</mtext> </msub> <mo> = </mo> <mtext>antilog 10</mtext> <mo> = </mo> <mn>1</mn> <mo>Ã</mo> <msup> <mtext>10</mtext> <mtext>10</mtext> </msup> </mrow> </math>.</div>


tags: ['Chemistry', 'Electrochemistry', 'KCET', 'KCET 2011']
source_db: MarksBatch2_P3.db
title:
the-standard-emf-of-a-galvanic-cell-involving-2-moles-of-electrons-in-its-redox-reaction-is-059-v-the-equilibrium-constant-for-the-redox-reaction-of-t

slug:
the-standard-emf-of-a-galvanic-cell-involving-2-moles-of-electrons-in-its-redox-reaction-is-059-v-the-equilibrium-constant-for-the-redox-reaction-of-t-13672

question:
<div class="question">The standard emf of a galvanic cell involving 2 moles of electrons in its redox reaction is $0.59 \mathrm{~V}$. The equilibrium constant for the redox reaction of the cell is</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$10^{20}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$10^{5}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">10</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$10^{10}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$10^{20}$</span> </div>

answer:
<div class="solution">In a galvanic cell, $\Delta G^{\circ}=-n F E^{\circ}$<br/>Relationship between $\Delta G^{\circ}$ and equilibrium constant<br/>$$<br/>\Delta G^{\circ}=-2.303 R T \log K_{p}<br/>$$<br/>From Eqs. (i) and (ii), we get<br/>$$<br/>\begin{gathered}<br/>-2.303 R T \log K_{p}=-n F E^{\circ} \\<br/>\log K_{p}=\frac{n F E^{\circ}}{2.303 R T}=\frac{n E^{\circ}}{\frac{2.303 R T}{F}} \\<br/>=\frac{2 \times 0.59}{0.059}=20 \\<br/>K_{p}=\mathrm{AL}(20) \quad(\mathrm{AL}=\text { Antilog }) \\<br/>K_{p}=10^{20}<br/>\end{gathered}<br/>$$</div>


tags: ['Chemistry', 'Electrochemistry', 'NEET', 'NEET 2004']
source_db: MarksBatch2_P3.db
title:
the-standard-emf-of-a-galvanic-cell-involving-cell-reaction-with-n-2-is-found-to-be-0295-v-at-2-5-c-the-equilibrium-constant-of-the-reaction-would-be-

slug:
the-standard-emf-of-a-galvanic-cell-involving-cell-reaction-with-n-2-is-found-to-be-0295-v-at-2-5-c-the-equilibrium-constant-of-the-reaction-would-be-25894

question:
<div class="question">The standard e.m.f of a galvanic cell involving cell reaction with $\mathrm{n}=2$ is found to be $0.295 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction would be:<br/>(Given $\mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1}, \mathrm{R}=8.314 \mathrm{~J}$ $\mathrm{K}^{-1} \mathrm{~mol}^{-1}$</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$2.0 \times 10^{11}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$4.0 \times 10^{12}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$1.0 \times 10^2$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$1.0 \times 10^{10}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$1.0 \times 10^{10}$</span> </div>

answer:
<div class="solution">$\mathrm{E}=\mathrm{E}^{\circ}-\frac{0.0591}{n} \log _{10} \mathrm{Q}$ at $25^{\circ} \mathrm{C}$<br/>At equilibrium, $\mathrm{E}=0, \mathrm{Q}=\mathrm{K}$<br/>$$<br/>0=\mathrm{E}^{\circ}-\frac{0.0591}{n} \log _{10} \mathrm{~K}<br/>$$<br/>or, $\mathrm{K}=$ Antilog $\left[\frac{n \mathrm{E}^{\circ}}{0.0591}\right]$<br/>or, $\mathrm{K}=$ Antilog $\left[\frac{2 \times 0.295}{0.0591}\right]$<br/>$$<br/>\begin{aligned}<br/>&amp; =\text { Antilog }\left[\frac{0.590}{0.0591}\right] \\<br/>&amp; =\text { Antilog } 10=1 \times 10^{10} .<br/>\end{aligned}<br/>$$<br/>Related Theory<br/>The cell potential is the difference between the electrode potentials (reduction potentials) of the cathode and anode. It is called the standard electromotive force (emf) of the cell when no current is drawn through the cells.</div>


tags: ['Chemistry', 'Electrochemistry', 'VITEEE', 'VITEEE 2022']
source_db: MarksBatch2_P3.db
title:
the-standard-emf-of-a-galvanic-cell-involving-cell-reaction-with-n2-is-found-to-be-0295-mathrmv-at-25circ-mathrmc-the-equilibrium-constant-of-the-reac

slug:
the-standard-emf-of-a-galvanic-cell-involving-cell-reaction-with-n2-is-found-to-be-0295-mathrmv-at-25circ-mathrmc-the-equilibrium-constant-of-the-reac-68583

question:
<div class="question">The standard e.m.f. of a galvanic cell involving cell reaction with \(n=2\) is found to be \(0.295 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). The equilibrium constant of the reaction would be (Given \(\mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1}\); \(\mathrm{R}=8.314 \mathrm{~J}^{-1} \mathrm{~mol}^{-1}\))</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">\(2.0 \times 10^{11}\)</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">\(4.0 \times 10^{12}\)</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">\(1.0 \times 10^2\)</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">\(1.0 \times 10^{10}\)</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">\(1.0 \times 10^{10}\)</span> </div>

answer:
<div class="solution">\(\mathrm{E}^{\circ}=\frac{0.0591}{\mathrm{n}} \log _{10} \mathrm{~K}\)<br/>Here, \(\mathrm{n}=2, \mathrm{E}^{\circ}=0.295\)<br/>\(\therefore \log _{10} \mathrm{~K}=\frac{2 \times 0.295}{0.0591}=10 \text { or } \mathrm{K}=10^{10}\)</div>


tags: ['Chemistry', 'Chemical Equilibrium', 'KCET', 'KCET 2010']
source_db: MarksBatch2_P3.db
title:
the-standard-emf-of-galvanic-cell-involving-3-moles-of-electrons-in-its-redox-reaction-is-059-v-the-equilibrium-constant-for-the-reaction-of-the-cell-

slug:
the-standard-emf-of-galvanic-cell-involving-3-moles-of-electrons-in-its-redox-reaction-is-059-v-the-equilibrium-constant-for-the-reaction-of-the-cell-17405

question:
<div class="question">The standard emf of galvanic cell involving 3 moles of electrons in its redox reaction is $0.59 \mathrm{~V}$. The equilibrium constant for the reaction of the cell is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$10^{25}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$10^{20}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$10^{15}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$10^{30}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$10^{30}$</span> </div>

answer:
<div class="solution">$\begin{aligned} \mathrm{E}_{\text {cell }}^{\circ} &amp;=\frac{0.0591}{\mathrm{n}} \log \mathrm{K}_{\mathrm{c}} \\ 0.59 &amp;=\frac{0.0591}{3} \log \mathrm{K}_{\mathrm{c}} \\ \frac{0.59 \times 3}{0.059} &amp;=\log \mathrm{K}_{\mathrm{c}} \\ \therefore \quad \log \mathrm{K}_{\mathrm{c}} &amp;=30 \\ \mathrm{~K}_{\mathrm{c}} &amp;=\text { antilog } 30 \\ &amp;=10^{30} \end{aligned}$</div>


tags: ['Chemistry', 'Electrochemistry', 'COMEDK', 'COMEDK 2014']
source_db: MarksBatch2_P3.db
title:
the-standard-emf-of-the-cell-zn-zn-2-ag-ag-is-156-v-if-the-standard-reduction-potential-of-ag-is-08-v-the-standard-oxidation-potential-of-zn-is

slug:
the-standard-emf-of-the-cell-zn-zn-2-ag-ag-is-156-v-if-the-standard-reduction-potential-of-ag-is-08-v-the-standard-oxidation-potential-of-zn-is-61602

question:
<div class="question">The standard emf of the cell $\mathrm{Zn}\left|\mathrm{Zn}^{2+}\right|\left|\mathrm{Ag}^{+}\right| \mathrm{Ag}$ is $1.56 \mathrm{~V}$. If the standard reduction potential of $\mathrm{Ag}$ is $0.8 \mathrm{~V}$, the standard oxidation potential of $\mathrm{Zn}$ is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$-0.76 \mathrm{~V}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$+0.76 \mathrm{~V}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$-2.36 \mathrm{~V}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$+2.36 \mathrm{~V}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$+0.76 \mathrm{~V}$</span> </div>

answer:
<div class="solution">For the cell, $\mathrm{Zn}\left|\mathrm{Zn}^{2+} \| \mathrm{Ag}^{+}\right| \mathrm{Ag}$<br/>$E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}$<br/>$1.56 \mathrm{~V}=E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}-E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{0}$<br/>$E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}$<br/>$\therefore \quad E_{\mathrm{Zn} / \mathrm{Zn}^{2+}}^{\circ}=-E_{\mathrm{Zn}^{2}+/ \mathrm{Zn}}^{\circ}$<br/>$=-(-0.76)=+0.76 \mathrm{~V}$</div>


tags: ['Chemistry', 'Thermodynamics (C)', 'JIPMER', 'JIPMER 2014']
source_db: MarksBatch2_P3.db
title:
the-standard-enthalpies-of-formation-of-a-nh-3-b-co-2-c-hi-and-d-so-2-are-4619-3934-2494-and-2969-kj-mol-1-respectively-the-increasing-order-of-their-

slug:
the-standard-enthalpies-of-formation-of-a-nh-3-b-co-2-c-hi-and-d-so-2-are-4619-3934-2494-and-2969-kj-mol-1-respectively-the-increasing-order-of-their-36191

question:
<div class="question">The standard enthalpies of formation of $A\left(\mathrm{NH}_3\right), B\left(\mathrm{CO}_2\right), C(\mathrm{HI})$ and $D\left(\mathrm{SO}_2\right)$ are -46.19 , $-393.4,+24.94$ and $-296.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The increasing order of their stability is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$B \lt D \lt A \lt C$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$C \lt A \lt D \lt B$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$D \lt B \lt C \lt A$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$A \lt C \lt D \lt B$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$C \lt A \lt D \lt B$</span> </div>

answer:
<div class="solution">Lower the value of standard enthalpy of formation, higher the stability. The increasing order of standard enthalpies of formation of given molecules is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jipmer/916kvmdrDRQl6SvE_hkoOis3MLMK7jagk6bei05Oo9o.original.fullsize.png"/><br/>Thus, the increasing order of stability is<br/>$C \lt A \lt D \lt B$</div>


tags: ['Chemistry', 'Thermodynamics (C)', 'JEE Main', 'JEE Main 2021 (17 Mar Shift 1)']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">230</span> </div>
source_db: MarksBatch2_P3.db
title:
the-standard-enthalpies-of-formation-of-al-2-o-3-and-cao-are-1675-kj-mol-1-and-635-kj-mol-1-respectively-for-the-reaction-3-cao-2-al-3-ca-al-2-o-3-the

slug:
the-standard-enthalpies-of-formation-of-al-2-o-3-and-cao-are-1675-kj-mol-1-and-635-kj-mol-1-respectively-for-the-reaction-3-cao-2-al-3-ca-al-2-o-3-the-96372

question:
<div class="question"><p>The standard enthalpies of formation of <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>Al</mi><mn>2</mn></msub><msub><mi mathvariant="normal">O</mi><mn>3</mn></msub></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>CaO</mi></math> are <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mn>1675</mn><mo> </mo><mi>kJ</mi><msup><mrow><mo> </mo><mi>mol</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mn>635</mn><mo> </mo><mi>kJ</mi><msup><mrow><mo> </mo><mi>mol</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math> respectively.</p><p>For the reaction</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>3</mn><mi>CaO</mi><mo>+</mo><mn>2</mn><mi>Al</mi><mo>→</mo><mn>3</mn><mi>Ca</mi><mo>+</mo><msub><mi>Al</mi><mn>2</mn></msub><msub><mi mathvariant="normal">O</mi><mn>3</mn></msub></math> the standard reaction enthalpy <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi mathvariant="normal">Δ</mi><mi mathvariant="normal">r</mi></msub><msup><mi mathvariant="normal">H</mi><mn>0</mn></msup><mo>=</mo></math> ________ <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>kJ</mi><mo>.</mo></math></p><p>(Round off to the Nearest Integer).</p></div>

answer:
<div class="solution"><p>Given reaction:</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>3</mn><mi>CaO</mi><mo>+</mo><mi>Al</mi><mo>→</mo><msub><mi>Al</mi><mn>2</mn></msub><msub><mi mathvariant="normal">O</mi><mn>3</mn></msub><mo>+</mo><mn>3</mn><mi>Ca</mi></math></p><p>Now, <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi mathvariant="normal">Δ</mi><mi mathvariant="normal">r</mi></msub><mi mathvariant="normal">H</mi><mo>°</mo><mo>=</mo><msub><mi>ΣΔ</mi><mi mathvariant="normal">f</mi></msub><msubsup><mi mathvariant="normal">H</mi><msub><mrow></mrow><mtext>Products </mtext></msub><mo>º</mo></msubsup><mo>-</mo><msub><mi>ΣΔ</mi><mi mathvariant="normal">f</mi></msub><msubsup><mi mathvariant="normal">H</mi><mi>Reactants</mi><mo>º</mo></msubsup></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mo>[</mo><mn>1</mn><mo>×</mo><mo>(</mo><mo>-</mo><mn>1675</mn><mo>)</mo><mo>+</mo><mn>3</mn><mo>×</mo><mn>0</mn><mo>]</mo><mo>-</mo><mo>[</mo><mn>3</mn><mo>×</mo><mo>(</mo><mo>-</mo><mn>635</mn><mo>)</mo><mo>+</mo><mn>2</mn><mo>×</mo><mn>0</mn><mo>]</mo></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mo>+</mo><mn>230</mn><mo> </mo><mi>kJ</mi><msup><mrow><mo> </mo><mi>mol</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></p></div>


tags: ['Chemistry', 'Thermodynamics (C)', 'NEET']
source_db: MarksBatch2_P3.db
title:
the-standard-enthalpies-of-formation-of-c-o-2-g-h-2-o-l-and-glucose-s-at-25c-are-400-k-j-m-o-l-300-k-j-m-o-l-and-1300-k-j-m-o-l-respectively-the-stand

slug:
the-standard-enthalpies-of-formation-of-c-o-2-g-h-2-o-l-and-glucose-s-at-25c-are-400-k-j-m-o-l-300-k-j-m-o-l-and-1300-k-j-m-o-l-respectively-the-stand-10227

question:
<div class="question">The standard enthalpies of formation of <math><mi mathvariant="normal">C</mi><msub><mrow><mi mathvariant="normal">O</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>(</mo><mi mathvariant="normal">g</mi><mo>)</mo><mo>,</mo><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mi mathvariant="normal">O</mi><mo>(</mo><mi mathvariant="normal">l</mi><mo>)</mo></math> and glucose (s) at <math><mn>25Â°C</mn></math> are <math><mn>â400 </mn> <mrow><mrow><mi mathvariant="normal">k</mi><mi mathvariant="normal">J</mi></mrow><mo>/</mo><mrow><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">l</mi></mrow></mrow></math>, <math><mn>â300 </mn><mrow><mrow><mi mathvariant="normal">k</mi><mi mathvariant="normal">J</mi></mrow><mo>/</mo><mrow><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">l</mi></mrow></mrow></math> and <math><mn>â1300 </mn><mrow><mrow><mi mathvariant="normal">k</mi><mi mathvariant="normal">J</mi></mrow><mo>/</mo><mrow><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">l</mi></mrow></mrow></math> , respectively. The standard enthalpy of combustion per gram of glucose at <math><mn>25Â°C</mn></math> is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">+ 2900 kJ</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">â2900 kJ</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">â16.11 kJ</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">+ 16.11 kJ</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">â16.11 kJ</span> </div>

answer:
<div class="solution"><math> <mrow> <msub> <mtext>C</mtext> <mn>6</mn> </msub> <msub> <mtext>H</mtext> <mrow> <mn>12</mn></mrow> </msub> <msub> <mtext>O</mtext> <mn>6</mn> </msub> <mo>+</mo><mn>6</mn><msub> <mrow> <mtext>CO</mtext></mrow> <mn>2</mn> </msub> <mo>â</mo><mn>6</mn><msub> <mtext>O</mtext> <mn>2</mn> </msub> <mo>+</mo><mn>6</mn><msub> <mtext>H</mtext> <mn>2</mn> </msub> <msub> <mrow> <mtext>O</mtext></mrow></msub></mrow></math><br/><math><mi mathvariant="normal">C</mi><msub><mrow><mi mathvariant="normal">O</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>â</mo><mo>-</mo><mn>400</mn></math> <math><mi mathvariant="normal">k</mi><mi mathvariant="normal">J</mi><mo>/</mo><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">l</mi></math><br/><math><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mi mathvariant="normal">O</mi><mo>=</mo><mo>-</mo><mn>300</mn></math> <math><mi mathvariant="normal">k</mi><mi mathvariant="normal">J</mi><mo>/</mo><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">l</mi></math><br/>Glucose <math><mo>=</mo><mo>-</mo><mn>1300</mn></math> <math><mi mathvariant="normal">k</mi><mi mathvariant="normal">J</mi><mo>/</mo><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">l</mi></math><br/><math><mi mathvariant="normal">Î</mi><mi mathvariant="normal">H</mi><mo>=</mo><mi mathvariant="normal">Î</mi><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mi mathvariant="normal">P</mi></mrow></msub><mo>-</mo><mi mathvariant="normal">Î</mi><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mi mathvariant="normal">R</mi></mrow></msub><mo>=</mo><mn>6</mn><mi mathvariant="normal">Î</mi><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mi mathvariant="normal">f</mi><mo>(</mo><mi mathvariant="normal">C</mi><msub><mrow><mi mathvariant="normal">O</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>)</mo></mrow></msub><mo>+</mo><mn>6</mn><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mi mathvariant="normal">f</mi><mo>(</mo><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mi mathvariant="normal">O</mi><mo>)</mo></mrow></msub><mo>-</mo><mi mathvariant="normal">Î</mi><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mi mathvariant="normal">f</mi><mo>(</mo><msub><mrow><mi mathvariant="normal">C</mi></mrow><mrow><mn>6</mn></mrow></msub><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>12</mn></mrow></msub><msub><mrow><mi mathvariant="normal">O</mi></mrow><mrow><mn>6</mn></mrow></msub><mo>)</mo></mrow></msub></math><br/><math><mo>=</mo><mfenced separators="|"><mrow><mn>6</mn><mo>Ã</mo><mo>-</mo><mn>400</mn></mrow></mfenced><mo>+</mo><mn>6</mn><mfenced separators="|"><mrow><mo>-</mo><mn>300</mn></mrow></mfenced><mo>-</mo><mfenced separators="|"><mrow><mo>-</mo><mn>1300</mn></mrow></mfenced><mo>=</mo><mo>-</mo><mn>2400</mn><mo>-</mo><mn>1800</mn><mo>+</mo><mn>1300</mn><mo>=</mo><mo>-</mo><mn>2900</mn></math> <math><mi mathvariant="normal">k</mi><mi mathvariant="normal">J</mi><mo>/</mo><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">l</mi></math><br/><math><mi mathvariant="normal">Î</mi><mi mathvariant="normal">H</mi><mo>=</mo><mfrac><mrow><mo>-</mo><mn>2900</mn></mrow><mrow><mn>180</mn></mrow></mfrac><mo>=</mo><mo>-</mo><mn>16.11</mn></math> <math><mi mathvariant="normal">k</mi><mi mathvariant="normal">J</mi><mo>/</mo><mi mathvariant="normal">g</mi></math></div>


tags: ['Chemistry', 'Thermodynamics (C)', 'AP EAMCET', 'AP EAMCET 2023 (19 May Shift 1)']
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">82</span> </div>
source_db: MarksBatch2_P3.db
title:
the-standard-enthalpy-of-atomization-of-ethane-according-to-the-equation-c-2-h-6-g-2-c-g-6-h-g-is-622-kj-mol-1-if-standard-mean-c-h-bond-dissociation-

slug:
the-standard-enthalpy-of-atomization-of-ethane-according-to-the-equation-c-2-h-6-g-2-c-g-6-h-g-is-622-kj-mol-1-if-standard-mean-c-h-bond-dissociation-95403

question:
<div class="question">The standard enthalpy of atomization of ethane according to the equation $\mathrm{C}_2 \mathrm{H}_6(\mathrm{~g}) \rightarrow 2 \mathrm{C}(\mathrm{g})+6 \mathrm{H}(\mathrm{g})$ is $622 \mathrm{~kJ} \mathrm{~mol}^{-1}$. If standard mean $\mathrm{C}-\mathrm{H}$ bond dissociation enthalpy is $90 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$, the standard mean dissociation enthalpy of $\mathrm{C}-\mathrm{C}$ bond (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">540</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">90</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">85</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">82</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

answer:
<div class="solution">$\begin{aligned} &amp; \Delta_{\mathrm{a}} \mathrm{H}^{\circ}=\Delta_{\mathrm{C}-\mathrm{H}} \mathrm{H}^{\circ}+\Delta_{\mathrm{C}-\mathrm{C}} \mathrm{H}^{\circ} \\ &amp; 622=[6 \times 90]+\left[\Delta_{\mathrm{C}-\mathrm{C}} \mathrm{H}^{\circ}\right]=540+\Delta_{\mathrm{C}-\mathrm{C}} \dot{\mathrm{H}}^{\circ} \\ &amp; \Rightarrow \Delta_{\mathrm{C}-\mathrm{C}} \mathrm{H}^{\circ}=622-540=82 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$</div>


tags: ['Chemistry', 'Thermodynamics (C)', 'JEE Main', 'JEE Main 2014 (12 Apr Online)']
source_db: MarksBatch2_P3.db
title:
the-standard-enthalpy-of-formation-f-h-298-for-methane-ch-4-is-749-kj-mol-1-in-order-to-calculate-the-average-energy-given-out-in-the-formation-of-a-c

slug:
the-standard-enthalpy-of-formation-f-h-298-for-methane-ch-4-is-749-kj-mol-1-in-order-to-calculate-the-average-energy-given-out-in-the-formation-of-a-c-73340

question:
<div class="question">The standard enthalpy of formation $\left(\Delta_f \mathrm{H}^{\circ}{ }_{298}\right)$ for methane, $\mathrm{CH}_4$ is $-74.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$. In order to calculate the average energy given out in the formation of a $\mathrm{C}-\mathrm{H}$ bond from this it is necessary to know which one of the following?</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>The dissociation energy of the hydrogen molecule, $\mathrm{H}_2$.<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>The first four ionisation energies of carbon.<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>The dissociation energy of $\mathrm{H}_2$ and enthalpy and sublimation of carbon (graphite).<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>The first four ionisation energies of carbon and electron affinity of hydrogen.</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>The dissociation energy of the hydrogen molecule, $\mathrm{H}_2$.<br/></span> </div>

answer:
<div class="solution">To calculate average enthalpy of $\mathrm{C}-\mathrm{H}$ bond in methane following informations are needed<br/>(i) dissociation energy of $\mathrm{H}_2$ i.e.<br/>$$<br/>\frac{1}{2} \mathrm{H}_2(g) \longrightarrow \mathrm{H}(\mathrm{g}) ; \Delta \mathrm{H}=x(\text { suppose } \quad)<br/>$$<br/>(ii) Sublimation energy of $\mathrm{C}$ (graphite) to $\mathrm{C}(g)$ $\mathrm{C}($ graphite $) \longrightarrow \mathrm{C}(g) ; \Delta \mathrm{H}=y$ (Suppose) Given<br/>$\mathrm{C}($ graphite $)+2 \mathrm{H}_2(\mathrm{~g}) \longrightarrow \mathrm{CH}_4(\mathrm{~g}) ; \Delta \mathrm{H}=75 \mathrm{~kJ} \mathrm{~mol}^{-1}$</div>


tags: ['Chemistry', 'Thermodynamics (C)', 'JEE Main', 'JEE Main 2006']
answer: <div class="solution">No Solution Available</div>
source_db: MarksBatch2_P3.db
title:
the-standard-enthalpy-of-formation-f-h-at-298-k-for-methane-ch-4-g-is-748-kj-mol-1-the-additional-information-required-to-determine-the-average-energy

slug:
the-standard-enthalpy-of-formation-f-h-at-298-k-for-methane-ch-4-g-is-748-kj-mol-1-the-additional-information-required-to-determine-the-average-energy-80034

question:
<div class="question">The standard enthalpy of formation $\left(\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\right)$ at $298 \mathrm{~K}$ for methane, $\mathrm{CH}_4(\mathrm{~g})$, is $-74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The additional information required to determine the average energy for $\mathrm{C}-\mathrm{H}$ bond formation would be</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>the dissociation energy of $\mathrm{H}_2$ and enthalpy of sublimation of carbon<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>latent heat of vapourization of methane<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>the first four ionization energies of carbon and electron gain enthalpy of hydrogen<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>the dissociation energy of hydrogen molecule, $\mathrm{H}_2$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>the dissociation energy of $\mathrm{H}_2$ and enthalpy of sublimation of carbon<br/></span> </div>


title: the-standard-enthalpy-of-formation-for-carbon-is-zero-for
slug: the-standard-enthalpy-of-formation-for-carbon-is-zero-for-12592
tags: ['Chemistry', 'Thermodynamics (C)', 'NEET']
answer: <div class="solution">Correct Option is : 2 <br/> C(Graphite)</div>
source_db: MarksBatch2_P3.db
question:
<div class="question"><img src="https://cdn-question-pool.getmarks.app/modules/dbq/cheas77dtry/cuGEOVCByawLX79M0WFMdr1sKhjrItYZURwLlF7BKe0JRP0xEA.png"/>The standard enthalpy of formation for carbon is zero for-</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">C(Diamond)</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">C(Graphite)</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">C(g)</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">All of these</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">C(Graphite)</span> </div>


title: the-standard-enthalpy-of-formation-is-zero-for-1
slug: the-standard-enthalpy-of-formation-is-zero-for-1-61179
tags: ['Chemistry', 'Thermodynamics (C)', 'NEET']
answer: <div class="solution">Correct Option is : 1 <br/> I<sub>2</sub>(s)</div>
source_db: MarksBatch2_P3.db
question:
<div class="question"><img src="https://cdn-question-pool.getmarks.app/modules/dbq/cheas77dtry/cuGEOVCByawLX79M0WFMdr1sKhjrItYZURwLlF7BKe0JRP0xEA.png"/>The standard enthalpy of formation is zero for-</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">I<sub>2</sub>(s)</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">H<sub>2</sub>O(l)</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">C(diamond)</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">SO<sub>2</sub>(g)</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">I<sub>2</sub>(s)</span> </div>


title: the-standard-enthalpy-of-formation-is-zero-for
slug: the-standard-enthalpy-of-formation-is-zero-for-47641
tags: ['Chemistry', 'Thermodynamics (C)', 'NEET']
answer: <div class="solution">Correct Option is : 3 <br/> H<sub>2</sub>(g)</div>
source_db: MarksBatch2_P3.db
question:
<div class="question"><img src="https://cdn-question-pool.getmarks.app/modules/dbq/cheas77dtry/cuGEOVCByawLX79M0WFMdr1sKhjrItYZURwLlF7BKe0JRP0xEA.png"/>The standard enthalpy of formation is zero for-</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">H<sub>2</sub>(l)</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">CH<sub>4</sub>(g)</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">H<sub>2</sub>(g)</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">HBr(g)</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">H<sub>2</sub>(g)</span> </div>


tags: ['Chemistry', 'Thermodynamics (C)', 'AP EAMCET', 'AP EAMCET 2023 (18 May Shift 1)']
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">$183$</span> </div>
source_db: MarksBatch2_P3.db
title:
the-standard-enthalpy-of-formation-of-co-2-g-cao-s-and-caco-3-s-are-393-634-1210-kj-mol-1-respectively-if-all-the-substances-are-in-standard-state-the

slug:
the-standard-enthalpy-of-formation-of-co-2-g-cao-s-and-caco-3-s-are-393-634-1210-kj-mol-1-respectively-if-all-the-substances-are-in-standard-state-the-67223

question:
<div class="question">The standard enthalpy of formation of $\mathrm{CO}_2(\mathrm{~g})$, $\mathrm{CaO}(\mathrm{s})$ and $\mathrm{CaCO}_3(\mathrm{~s})$ are $-393,-634,-1210 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. If all the substances are in standard state, the standard enthalpy of decomposition of calcium carbonate to $\mathrm{CaO}(\mathrm{s})$ and $\mathrm{CO}_2$ (g) (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$969$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$183$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$-969$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$-183$</span> </li> </ul>

answer:
<div class="solution">$\mathrm{CaCO}_3(\mathrm{~s}) \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$<br/>$\Delta \mathrm{H}_{\mathrm{r}}^{\circ}=\left[\Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{CaO})+\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{CO}_2\right)\right]-\left[\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{CaCO}_3\right)\right]$<br/>$\begin{aligned} &amp; =[(-634)+(-393)]-[-1210] \\ &amp; =+183 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$</div>


tags: ['Chemistry', 'Thermodynamics (C)', 'TS EAMCET', 'TS EAMCET 2018 (05 May Shift 1)']
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">-1058</span> </div>
source_db: MarksBatch2_P3.db
title:
the-standard-enthalpy-of-formation-of-co-g-co-2-g-n-2-o-g-and-n-2-o-4-g-are-respectively-10-393-81-and-10-kj-mol-1-enthalpy-change-in-kj-of-the-follow

slug:
the-standard-enthalpy-of-formation-of-co-g-co-2-g-n-2-o-g-and-n-2-o-4-g-are-respectively-10-393-81-and-10-kj-mol-1-enthalpy-change-in-kj-of-the-follow-81715

question:
<div class="question">The standard enthalpy of formation of $\mathrm{CO}(g)$, $\mathrm{CO}_2(g), \mathrm{N}_2 \mathrm{O}(g)$ and $\mathrm{N}_2 \mathrm{O}_4(g)$ are respectively -10 , $-393,81$ and $-10 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Enthalpy change (in $\mathrm{kJ}$ ) of the following reaction is<br/>$$<br/>\mathrm{N}_2 \mathrm{O}_4(\mathrm{~g})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{N}_2 \mathrm{O}(\mathrm{g})+3 \mathrm{CO}_2(\mathrm{~g})<br/>$$</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">-1058</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">+1058</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">-957</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">+957</span> </li> </ul>

answer:
<div class="solution">Given, standard enthalpy of formation of $\mathrm{CO}=-10 \mathrm{~kJ} / \mathrm{mol}$<br/>Standard enthalpy of formation of $\mathrm{CO}_2$<br/>$$<br/>=-393 \mathrm{~kJ} / \mathrm{mol}<br/>$$<br/><br/>Standard enthalpy of formation of $\mathrm{N}_2 \mathrm{O}$<br/>$$<br/>=81 \mathrm{~kJ} / \mathrm{mol}<br/>$$<br/><br/>Standard enthalpy of formation of $\mathrm{N}_2 \mathrm{O}_4$<br/>$$<br/>=-10 \mathrm{~kJ} / \mathrm{mol}<br/>$$<br/>$$<br/>\begin{gathered}<br/>\mathrm{N}_2 \mathrm{O}_4+3 \mathrm{CO} \longrightarrow \mathrm{N}_2 \mathrm{O}+3 \mathrm{CO}_2 \\<br/>\Delta H(\text { enthalpy change })=\Sigma \Delta H_{\mathrm{N}_2 \mathrm{O}}+3 \Delta H_{\mathrm{CO}_2} \\<br/>-\Sigma \Delta H_{\mathrm{N}_2 \mathrm{O}_4}+3 \Delta H_{\mathrm{CO}} \\<br/>=[\Sigma 81+(3 \times-393)-\Sigma(-10+(3 \times-10)] \\<br/>\Delta H=-1058 \mathrm{~kJ} / \mathrm{mol}<br/>\end{gathered}<br/>$$</div>


tags: ['Chemistry', 'Thermodynamics (C)', 'JEE Main']
source_db: MarksBatch2_P3.db
title:
the-standard-enthalpy-of-formation-of-gaseous-h-2-o-at-298-k-is-241-82-kj-mo-l-1-calculate-h-f-o-at-373-k-given-the-following-values-of-the-molar-heat

slug:
the-standard-enthalpy-of-formation-of-gaseous-h-2-o-at-298-k-is-241-82-kj-mo-l-1-calculate-h-f-o-at-373-k-given-the-following-values-of-the-molar-heat-70781

question:
<div class="question">The standard enthalpy of formation of gaseous <math><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mi mathvariant="normal">O</mi></math> at <math> <mrow> <mtext>298</mtext><mtext>â</mtext><mtext>K</mtext></mrow> </math> is <math> <mrow> <mo>â</mo><mtext>241</mtext><mtext>.82</mtext><mtext>â</mtext><mtext>kJ</mtext></mrow> </math> <math><mtext>mo</mtext><msup><mrow><mtext>l</mtext></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math>. Calculate <math> <mrow> <msubsup> <mrow> <mtext>ÎH</mtext></mrow> <mtext>f</mtext> <mtext>o</mtext> </msubsup> </mrow> </math> at <math> <mrow> <mtext>373</mtext><mtext>â</mtext><mtext>K,</mtext></mrow> </math> given the following values of the molar heat capacities at constant pressure.<br/>
Molar heat capacity of <math><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mfenced separators="|"><mrow><mi mathvariant="normal">g</mi></mrow></mfenced><mo>=</mo><mn>33.58</mn><mtext> J</mtext><msup><mrow><mtext>K</mtext></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><msup><mrow><mi mathvariant="normal">l</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math><br/>
Molar heat capacity of <math><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mfenced separators="|"><mrow><mi mathvariant="normal">g</mi></mrow></mfenced><mo>=</mo><mn>28.84</mn><mtext> J</mtext><msup><mrow><mtext>K</mtext></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><msup><mrow><mi mathvariant="normal">l</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math><br/>
Molar heat capacity of <math><msub><mrow><mi mathvariant="normal">O</mi></mrow><mrow><mn>2</mn></mrow></msub><mfenced separators="|"><mrow><mi mathvariant="normal">g</mi></mrow></mfenced><mo>=</mo><mn>29.37</mn><mtext> J</mtext><msup><mrow><mtext>K</mtext></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><msup><mrow><mi mathvariant="normal">l</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math><br/>
Assume that the heat capacities are independent of temperature.</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><math><mo>-</mo><mn>242.6</mn><mtext> kJ </mtext><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><msup><mrow><mi mathvariant="normal">l</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math><mo>-</mo><mn>485.2</mn><mtext>Â kJÂ </mtext><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><msup><mrow><mi mathvariant="normal">l</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math><mo>-</mo><mn>121.3</mn><mtext>Â kJÂ </mtext><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><msup><mrow><mi mathvariant="normal">l</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math><mo>-</mo><mn>286.4</mn><mtext>Â kJÂ </mtext><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><msup><mrow><mi mathvariant="normal">l</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math><mo>-</mo><mn>242.6</mn><mtext> kJ </mtext><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><msup><mrow><mi mathvariant="normal">l</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></span> </div>

answer:
<div class="solution">The reaction is <math><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mfenced separators="|"><mrow><mi mathvariant="normal">g</mi></mrow></mfenced><mo>+</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><msub><mrow><mi mathvariant="normal">O</mi></mrow><mrow><mn>2</mn></mrow></msub><mfenced separators="|"><mrow><mi mathvariant="normal">g</mi></mrow></mfenced><mo>â</mo><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mi mathvariant="normal">O</mi><mfenced separators="|"><mrow><mi mathvariant="normal">g</mi></mrow></mfenced></math><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">Î</mi><msubsup><mi mathvariant="normal">C</mi><mi mathvariant="normal">p</mi><mi mathvariant="normal">o</mi></msubsup><mo>=</mo><msubsup><mi mathvariant="normal">C</mi><mrow><mi mathvariant="normal">p</mi><mo>,</mo><mi mathvariant="normal">m</mi></mrow><mi mathvariant="normal">o</mi></msubsup><mfenced separators="|"><mrow><msub><mi mathvariant="normal">H</mi><mn>2</mn></msub><mi mathvariant="normal">O</mi><mo>,</mo><mi mathvariant="normal">g</mi></mrow></mfenced><mo>-</mo><mfenced close="}" open="{" separators="|"><mrow><msubsup><mi mathvariant="normal">C</mi><mrow><mi mathvariant="normal">p</mi><mo>,</mo><mi mathvariant="normal">m</mi></mrow><mi mathvariant="normal">o</mi></msubsup><mfenced separators="|"><mrow><msub><mi mathvariant="normal">H</mi><mn>2</mn></msub><mo>,</mo><mi mathvariant="normal">g</mi></mrow></mfenced><mo>+</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><msubsup><mi mathvariant="normal">C</mi><mrow><mi mathvariant="normal">p</mi><mo>,</mo><mi mathvariant="normal">m</mi></mrow><mi mathvariant="normal">o</mi></msubsup><mfenced separators="|"><msub><mi mathvariant="normal">O</mi><mn>2</mn></msub></mfenced></mrow></mfenced></math><br/><math><mo>=</mo><mn>33.58</mn><mo>-</mo><mfenced close="}" open="{" separators="|"><mrow><mn>28.84</mn><mo>+</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mfenced separators="|"><mrow><mn>29.37</mn></mrow></mfenced></mrow></mfenced></math><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mo>-</mo><mn>9.94</mn><msup><mtext> JK</mtext><mrow><mo>-</mo><mn>1</mn></mrow></msup><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><msup><mi mathvariant="normal">l</mi><mrow><mo>-</mo><mn>1</mn></mrow></msup></math><br/>Using Kirchhoff's equation,<br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">Î</mi><mi mathvariant="normal">H</mi><mo>Â°</mo><mfenced separators="|"><mrow><mn>373</mn><mo> </mo><mi mathvariant="normal">K</mi></mrow></mfenced><mo>=</mo><mi mathvariant="normal">Î</mi><mi mathvariant="normal">H</mi><mo>Â°</mo><mfenced separators="|"><mrow><mn>298</mn><mo> </mo><mi mathvariant="normal">K</mi></mrow></mfenced><mo>+</mo><mfenced separators="|"><mrow><msub><mi mathvariant="normal">T</mi><mn>2</mn></msub><mo>-</mo><msub><mi mathvariant="normal">T</mi><mn>1</mn></msub></mrow></mfenced><mi mathvariant="normal">Î</mi><msubsup><mi mathvariant="normal">C</mi><mi mathvariant="normal">p</mi><mi mathvariant="normal">o</mi></msubsup></math><br/><math><mo>=</mo><mo>-</mo><mn>241.82</mn><mo>+</mo><mfenced separators="|"><mrow><mn>373</mn><mo>-</mo><mn>298</mn></mrow></mfenced><mo>Ã</mo><mfenced separators="|"><mrow><mo>-</mo><mn>9.94</mn></mrow></mfenced></math><br/><math><mo>=</mo><mo>-</mo><mn>242.6</mn><mtext> kJ mo</mtext><msup><mrow><mtext>l</mtext></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math></div>


tags: ['Chemistry', 'Thermodynamics (C)', 'JEE Main', 'JEE Main 2014 (09 Apr Online)']
source_db: MarksBatch2_P3.db
title:
the-standard-enthalpy-of-formation-of-nh-3-is-46-0-kj-mol-if-bond-enthalpy-of-h-2-is-436-kj-mol-and-that-of-n-2-is-712-kj-mol-the-average-bond-enthalp

slug:
the-standard-enthalpy-of-formation-of-nh-3-is-46-0-kj-mol-if-bond-enthalpy-of-h-2-is-436-kj-mol-and-that-of-n-2-is-712-kj-mol-the-average-bond-enthalp-39474

question:
<div class="question">The standard enthalpy of formation of<math xmlns="http://www.w3.org/1998/Math/MathML"><mo> </mo><msub><mi>NH</mi><mn>3</mn></msub></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mo> </mo><mn>46</mn><mo>.</mo><mn>0</mn><mo> </mo><mi>kJ</mi><mo>/</mo><mi>mol</mi></math>. If bond enthalpy of <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi mathvariant="normal">H</mi><mn>2</mn></msub></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mn>436</mn><mo> </mo><mi>kJ</mi><mo>/</mo><mi>mol</mi></math> and that of <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi mathvariant="normal">N</mi><mn>2</mn></msub></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mn>712</mn><mo> </mo><mi>kJ</mi><mo>/</mo><mi>mol</mi></math> , the average bond enthalpy of <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">N</mi><mo>-</mo><mi mathvariant="normal">H</mi></math> bond in <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>NH</mi><mn>3</mn></msub></math> is :</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mo> </mo><mn>964</mn><mo> </mo><mi>kJ</mi><mo>/</mo><mi>mol</mi></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mo> </mo><mn>1102</mn><mo> </mo><mi>kJ</mi><mo>/</mo><mi>mol</mi></math></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>+</mo><mo> </mo><mn>1056</mn><mo> </mo><mi>kJ</mi><mo>/</mo><mi>mol</mi></math></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>+</mo><mo> </mo><mn>352</mn><mo> </mo><mi>kJ</mi><mo>/</mo><mi>mol</mi></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>+</mo><mo> </mo><mn>352</mn><mo> </mo><mi>kJ</mi><mo>/</mo><mi>mol</mi></math></span> </div>

answer:
<div class="solution">For The Reaction<br/><br/><math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <msub> <mtext>N</mtext> <mn>2</mn> </msub> <mfenced close=")" open="(" separators=","> <mrow> <mtext>g</mtext> </mrow> </mfenced> <mo>+</mo> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <msub> <mtext>H</mtext> <mn>2</mn> </msub> <mfenced close=")" open="(" separators=","> <mrow> <mtext>g</mtext> </mrow> </mfenced> <mo>â</mo> <msub> <mtext>NH</mtext> <mn>3</mn> </msub> <mfenced close=")" open="(" separators=","> <mrow> <mtext>g</mtext> </mrow> </mfenced> <mo>;</mo> <mtext>Î</mtext> <msubsup> <mtext>H</mtext> <mtext>f</mtext> <mo>â</mo> </msubsup> <mo>=</mo> <mo>-</mo> <mn>4</mn> <mn>6</mn> <msup> <mo>KJ mol</mo> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> </mrow> </math><br/><br/><math> <mrow> <mtext>Î</mtext> <msubsup> <mo>H</mo> <mtext>f</mtext> <mo>â</mo> </msubsup> <mo>=</mo> <mo>-</mo> <mn>4</mn> <mn>6</mn> <mo>=</mo> <mo>â</mo> <mrow> <mtext>B.E</mtext> <mo>-</mo> <mo>â</mo> <mtext>B.E</mtext> </mrow> </mrow> </math><br/>                              Reactants      Product<br/><br/><math> <mrow> <mo>â</mo> <mtext>B.E</mtext> <mo>=</mo> <mo>â</mo> <mtext>B.E</mtext> <mo>+</mo> <mn>4</mn> <mn>6</mn> </mrow> </math><br/>Product        Reactants<br/><br/>               <math> <mrow> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mfenced close=")" open="(" separators=","> <mrow> <mn>7</mn> <mn>1</mn> <mn>2</mn> </mrow> </mfenced> <mo>+</mo> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mfenced close=")" open="(" separators=","> <mrow> <mn>4</mn> <mn>3</mn> <mn>6</mn> </mrow> </mfenced> <mo>+</mo> <mn>4</mn> <mn>6</mn> </mrow> </math><br/><br/>               = 356 + 654 + 46 = 1056<br/><br/>Averge B.E for N-H Bond <math> <mrow> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mn>0</mn> <mn>5</mn> <mn>6</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>=</mo> <mn>3</mn> <mn>5</mn> <mn>2</mn> <msup> <mo>KJ mol</mo> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> </mrow> </math></div>


tags: ['Chemistry', 'Thermodynamics (C)', 'JEE Main', 'JEE Main 2010']
source_db: MarksBatch2_P3.db
title:
the-standard-enthalpy-of-formation-of-nh-3-is-460-kj-mol-1-if-the-enthalpy-of-formation-of-h-2-from-its-atoms-is-436-kj-mol-1-and-that-of-n-2-is-712-k

slug:
the-standard-enthalpy-of-formation-of-nh-3-is-460-kj-mol-1-if-the-enthalpy-of-formation-of-h-2-from-its-atoms-is-436-kj-mol-1-and-that-of-n-2-is-712-k-18024

question:
<div class="question">The standard enthalpy of formation of $\mathrm{NH}_3$ is $-46.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$. If the enthalpy of formation of $\mathrm{H}_2$ from its atoms is $-436 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and that of $\mathrm{N}_2$ is $-712 \mathrm{~kJ} \mathrm{~mol}^{-1}$, the average bond enthalpy of $\mathrm{N}-\mathrm{H}$ bond in $\mathrm{NH}_3$ is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$-964 \mathrm{~kJ} \mathrm{~mol}^{-1}$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$+352 \mathrm{~kJ} \mathrm{~mol}^{-1}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$+1056 \mathrm{~kJ} \mathrm{~mol}^{-1}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$-1102 \mathrm{~kJ} \mathrm{~mol}^{-1}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>$+352 \mathrm{~kJ} \mathrm{~mol}^{-1}$<br/></span> </div>

answer:
<div class="solution">Enthalpy of formation of $\mathrm{NH}_3=-46 \mathrm{~kJ} / \mathrm{mole}$<br/>$$<br/>\therefore \mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3 \quad \Delta \mathrm{H}_{\mathrm{f}}=-2 \times 46 \mathrm{~kJ} \mathrm{~mol}<br/>$$<br/>Bond breaking is endothermic and Bond formation is exothermic<br/>Assuming ' $\mathrm{x}$ ' is the bond energy of $\mathrm{N}-\mathrm{H}$ bond $\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$<br/>$$<br/>\begin{aligned}<br/>&amp; \therefore 712+(3 \times 436)-6 \mathrm{x}=-46 \times 2 \\<br/>&amp; \therefore \mathrm{x}=352 \mathrm{~kJ} / \mathrm{mol}<br/>\end{aligned}<br/>$$</div>


tags: ['Chemistry', 'Thermodynamics (C)', 'JEE Main']
source_db: MarksBatch2_P3.db
title:
the-standard-enthalpy-of-formation-of-nh-3-is-460-kjmol-if-bond-enthalpy-of-h-2-is-436-kjmol-and-that-of-n-2-is-712-kjmol-the-average-bond-enthalpy-of

slug:
the-standard-enthalpy-of-formation-of-nh-3-is-460-kjmol-if-bond-enthalpy-of-h-2-is-436-kjmol-and-that-of-n-2-is-712-kjmol-the-average-bond-enthalpy-of-44124

question:
<div class="question">The standard enthalpy of formation of NH<sub>3</sub> is - 46.0 kJ/mol. If bond enthalpy of H<sub>2</sub>  is - 436 kJ/mol and that of N<sub>2</sub> is - 712 kJ/mol , the average bond enthalpy of N - H bond in NH<sub>3</sub> is :</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">- 964 kJ/mol</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">- 1102 kJ/mol</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">+ 1056 kJ/mol</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">+ 352 kJ/mol</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">+ 352 kJ/mol</span> </div>

answer:
<div class="solution">For The Reaction<br/><br/><math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <msub> <mtext>N</mtext> <mn>2</mn> </msub> <mfenced close=")" open="(" separators=","> <mrow> <mtext>g</mtext> </mrow> </mfenced> <mo>+</mo> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <msub> <mtext>H</mtext> <mn>2</mn> </msub> <mfenced close=")" open="(" separators=","> <mrow> <mtext>g</mtext> </mrow> </mfenced> <mo>â</mo> <msub> <mtext>NH</mtext> <mn>3</mn> </msub> <mfenced close=")" open="(" separators=","> <mrow> <mtext>g</mtext> </mrow> </mfenced> <mo>;</mo> <mtext>Î</mtext> <msubsup> <mtext>H</mtext> <mtext>f</mtext> <mo>â</mo> </msubsup> <mo>=</mo> <mo>-</mo> <mn>4</mn> <mn>6</mn> <msup> <mo>KJ mol</mo> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> </mrow> </math><br/><br/><math> <mrow> <mtext>Î</mtext> <msubsup> <mo>H</mo> <mtext>f</mtext> <mo>â</mo> </msubsup> <mo>=</mo> <mo>-</mo> <mn>4</mn> <mn>6</mn> <mo>=</mo> <mo>â</mo> <mrow> <mtext>B.E</mtext> <mo>-</mo> <mo>â</mo> <mtext>B.E</mtext> </mrow> </mrow> </math><br/>                              Reactants      Product<br/><br/><math> <mrow> <mo>â</mo> <mtext>B.E</mtext> <mo>=</mo> <mo>â</mo> <mtext>B.E</mtext> <mo>+</mo> <mn>4</mn> <mn>6</mn> </mrow> </math><br/>Product        Reactants<br/><br/>               <math> <mrow> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mfenced close=")" open="(" separators=","> <mrow> <mn>7</mn> <mn>1</mn> <mn>2</mn> </mrow> </mfenced> <mo>+</mo> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mfenced close=")" open="(" separators=","> <mrow> <mn>4</mn> <mn>3</mn> <mn>6</mn> </mrow> </mfenced> <mo>+</mo> <mn>4</mn> <mn>6</mn> </mrow> </math><br/><br/>               = 356 + 654 + 46 = 1056<br/><br/>Averge B.E for N-H Bond <math> <mrow> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mn>0</mn> <mn>5</mn> <mn>6</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>=</mo> <mn>3</mn> <mn>5</mn> <mn>2</mn> <msup> <mo>KJ mol</mo> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </msup> </mrow> </math></div>


tags: ['Chemistry', 'Thermodynamics (C)', 'NEET']
source_db: MarksBatch2_P3.db
title:
the-standard-enthalpy-of-formation-of-octane-c-8-h-18-is-250-kjmol-1-the-enthalpy-of-formation-of-c-o-2-g-and-h-2-o-l-are-394-kjmol-1-and-286-kjmol-1-

slug:
the-standard-enthalpy-of-formation-of-octane-c-8-h-18-is-250-kjmol-1-the-enthalpy-of-formation-of-c-o-2-g-and-h-2-o-l-are-394-kjmol-1-and-286-kjmol-1-80587

question:
<div class="question">The standard enthalpy of formation of octane <math><mo>(</mo><msub><mrow><mi mathvariant="normal">C</mi></mrow><mrow><mn>8</mn></mrow></msub><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>18</mn></mrow></msub><mo>)</mo></math> is <math> <mrow> <mo>â</mo><mn>250</mn><mtext>â</mtext><msup> <mrow> <mtext>kJmol</mtext></mrow> <mrow> <mo>â</mo><mn>1</mn></mrow> </msup> </mrow> </math> . The enthalpy of formation of <math><mi mathvariant="normal">C</mi><msub><mrow><mi mathvariant="normal">O</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>(</mo><mi mathvariant="normal">g</mi><mo>)</mo></math> and <math><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mi mathvariant="normal">O</mi><mo>(</mo><mi mathvariant="normal">l</mi><mo>)</mo></math> are <math> <mrow> <mo>â</mo><mn>394</mn><mtext>â</mtext><msup> <mrow> <mtext>kJmol</mtext></mrow> <mrow> <mo>â</mo><mn>1</mn></mrow> </msup> </mrow> </math> and <math> <mrow> <mo>â</mo><mn>286</mn><mtext>â</mtext><msup> <mrow> <mtext>kJmol</mtext></mrow> <mrow> <mo>â</mo><mn>1</mn></mrow> </msup> </mrow> </math> respectively. The enthalpy of combustion of <math><msub><mrow><mi mathvariant="normal">C</mi></mrow><mrow><mn>8</mn></mrow></msub><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>18</mn></mrow></msub><mo>(</mo><mi mathvariant="normal">g</mi><mo>)</mo></math> is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math> <mrow> <mo>−</mo><mn>5200</mn><mtext> </mtext><msup> <mrow> <mtext>kJmol</mtext></mrow> <mrow> <mo>−</mo><mn>1</mn></mrow> </msup> </mrow></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math> <mrow> <mo>−</mo><mn>5726</mn><mtext> </mtext><msup> <mrow> <mtext>kJmol</mtext></mrow> <mrow> <mo>−</mo><mn>1</mn></mrow> </msup> </mrow></math></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><math> <mrow> <mo>−</mo><mn>5476</mn><mtext> </mtext><msup> <mrow> <mtext>kJmol</mtext></mrow> <mrow> <mo>−</mo><mn>1</mn></mrow> </msup> </mrow></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math> <mrow> <mo>−</mo><mn>5310</mn><mtext> </mtext><msup> <mrow> <mtext>kJmol</mtext></mrow> <mrow> <mo>−</mo><mn>1</mn></mrow> </msup> </mrow></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math> <mrow> <mo>−</mo><mn>5476</mn><mtext> </mtext><msup> <mrow> <mtext>kJmol</mtext></mrow> <mrow> <mo>−</mo><mn>1</mn></mrow> </msup> </mrow></math></span> </div>

answer:
<div class="solution"><math><msub><mrow><mi mathvariant="normal">C</mi></mrow><mrow><mn>8</mn></mrow></msub><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>18</mn></mrow></msub><mo>(</mo><mi mathvariant="normal">g</mi><mo>)</mo><mo>+</mo><mfrac><mrow><mn>25</mn></mrow><mrow><mn>2</mn></mrow></mfrac><msub><mrow><mi mathvariant="normal">O</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>(</mo><mi mathvariant="normal">g</mi><mo>)</mo><mo>â</mo><mn>8</mn><mi mathvariant="normal">C</mi><msub><mrow><mi mathvariant="normal">O</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>(</mo><mi mathvariant="normal">g</mi><mo>)</mo><mo>+</mo><mn>9</mn><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mi mathvariant="normal">O</mi><mo>(</mo><mi mathvariant="normal">l</mi><mo>)</mo></math><br/><br/><math><mi mathvariant="normal">Î</mi><mi mathvariant="normal">H</mi><mo>Â°</mo><mo>=</mo><mn>8</mn><mo>Ã</mo><mo>(</mo><mo>-</mo><mn>394</mn><mo>)</mo><mo>+</mo><mn>9</mn><mo>Ã</mo><mfenced separators="|"><mrow><mo>-</mo><mn>286</mn></mrow></mfenced><mo>-</mo><mo>(</mo><mo>-</mo><mn>250</mn><mo>)</mo></math> <math><mo>=</mo><mo>-</mo><mn>5476</mn></math></div>


tags: ['Chemistry', 'Thermodynamics (C)', 'JEE Main']
source_db: MarksBatch2_P3.db
title:
the-standard-enthalpy-of-neutralization-of-strong-acid-and-strong-base-is-573-k-j-equiv-1-if-the-enthalpy-of-neutralization-of-the-first-proton-of-aqu-1

slug:
the-standard-enthalpy-of-neutralization-of-strong-acid-and-strong-base-is-573-k-j-equiv-1-if-the-enthalpy-of-neutralization-of-the-first-proton-of-aqu-1-84708

question:
<div class="question">The standard enthalpy of neutralization of strong acid and strong base is <math><mo>-</mo><mn>57.3</mn></math> <math><mi mathvariant="normal">k</mi><mi mathvariant="normal">J</mi></math> <math> <mrow> <msup> <mrow> <mtext>equiv</mtext></mrow> <mrow> <mo>â</mo><mn>1</mn></mrow> </msup> <mo>.</mo></mrow> </math> If the enthalpy of neutralization of the first proton of aqueous <math><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mi mathvariant="normal">S</mi></math> is <math><mo>-</mo><mn>33.7</mn></math> <math><mi mathvariant="normal">k</mi><mi mathvariant="normal">J</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><msup><mrow><mi mathvariant="normal">l</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math> then the <math> <mrow> <msub> <mrow> <mfenced> <mrow> <msub> <mrow> <mtext>pK</mtext></mrow> <mtext>a</mtext> </msub> </mrow> </mfenced></mrow> <mn>1</mn> </msub> </mrow> </math> of <math><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mi mathvariant="normal">S</mi></math> is</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><math><mfenced separators="|"><mrow><mfrac><mrow><mn>23.6</mn><mo>Ã</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>3</mn></mrow></msup><mo>-</mo><mi mathvariant="normal">T</mi><mi mathvariant="normal">Î</mi><mi mathvariant="normal">S</mi><mo>Â°</mo></mrow><mrow><mn>2.303</mn><mi mathvariant="normal">R</mi><mi mathvariant="normal">T</mi></mrow></mfrac></mrow></mfenced></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math><mfenced separators="|"><mrow><mfrac><mrow><mn>2.303</mn><mi mathvariant="normal">R</mi><mi mathvariant="normal">T</mi></mrow><mrow><mn>23.6</mn><mo>-</mo><mi mathvariant="normal">T</mi><mi mathvariant="normal">Î</mi><mi mathvariant="normal">S</mi><mo>Â°</mo></mrow></mfrac></mrow></mfenced></math></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math><mfenced separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">T</mi><mi mathvariant="normal">Î</mi><mi mathvariant="normal">S</mi><mo>Â°</mo><mo>-</mo><mn>23.6</mn></mrow><mrow><mi mathvariant="normal">R</mi><mi mathvariant="normal">T</mi></mrow></mfrac></mrow></mfenced></math></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math><mn>2.303</mn><mfenced separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">T</mi><mi mathvariant="normal">Î</mi><mi mathvariant="normal">S</mi><mo>Â°</mo><mo>-</mo><mn>23.6</mn></mrow><mrow><mi mathvariant="normal">R</mi><mi mathvariant="normal">T</mi></mrow></mfrac></mrow></mfenced></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math><mfenced separators="|"><mrow><mfrac><mrow><mn>23.6</mn><mo>Ã</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>3</mn></mrow></msup><mo>-</mo><mi mathvariant="normal">T</mi><mi mathvariant="normal">Î</mi><mi mathvariant="normal">S</mi><mo>Â°</mo></mrow><mrow><mn>2.303</mn><mi mathvariant="normal">R</mi><mi mathvariant="normal">T</mi></mrow></mfrac></mrow></mfenced></math></span> </div>

answer:
<div class="solution"><math><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mi mathvariant="normal">S</mi><mo>â</mo><msup><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mo>+</mo></mrow></msup><mo>+</mo><mi mathvariant="normal">H</mi><msup><mrow><mi mathvariant="normal">S</mi></mrow><mrow><mo>-</mo></mrow></msup></math><br/><math><mi mathvariant="normal">Î</mi><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mi mathvariant="normal">i</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">n</mi></mrow></msub><mo>=</mo><mn>23.6</mn></math> <math><mi mathvariant="normal">k</mi><mi mathvariant="normal">J</mi></math><br/><math><mo>â´</mo><mi mathvariant="normal">Î</mi><mi mathvariant="normal">G</mi><mo>Â°</mo><mo>=</mo><mn>23.6</mn><mo>Ã</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>3</mn></mrow></msup><mo>-</mo><mi mathvariant="normal">T</mi><mi mathvariant="normal">Î</mi><mi mathvariant="normal">S</mi><mo>Â°</mo><mo>=</mo><mo>-</mo><mn>2.303</mn><mi mathvariant="normal"></mi><mi mathvariant="normal">R</mi><mi mathvariant="normal">T</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">l</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">g</mi><mi mathvariant="normal"></mi><msub><mrow><mi mathvariant="normal">K</mi></mrow><mrow><msub><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>1</mn></mrow></msub></mrow></msub></math><br/><math><mo>â´</mo><msub><mrow><mi mathvariant="normal">P</mi></mrow><mrow><mi mathvariant="normal">K</mi><msub><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>1</mn></mrow></msub></mrow></msub><mo>=</mo><mfrac><mrow><mn>23.6</mn><mo>Ã</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>3</mn></mrow></msup><mo>-</mo><mi mathvariant="normal">T</mi><mi mathvariant="normal">Î</mi><mi mathvariant="normal">S</mi><mo>Â°</mo></mrow><mrow><mn>2.303</mn><mi mathvariant="normal"></mi><mi mathvariant="normal">R</mi><mi mathvariant="normal">T</mi></mrow></mfrac></math></div>


tags: ['Chemistry', 'Thermodynamics (C)', 'NEET']
source_db: MarksBatch2_P3.db
title:
the-standard-enthalpy-of-neutralization-of-strong-acid-and-strong-base-is-573-k-j-equiv-1-if-the-enthalpy-of-neutralization-of-the-first-proton-of-aqu

slug:
the-standard-enthalpy-of-neutralization-of-strong-acid-and-strong-base-is-573-k-j-equiv-1-if-the-enthalpy-of-neutralization-of-the-first-proton-of-aqu-69104

question:
<div class="question">The standard enthalpy of neutralization of strong acid and strong base is <math><mo>-</mo><mn>57.3</mn></math> <math><mi mathvariant="normal">k</mi><mi mathvariant="normal">J</mi></math> <math> <mrow> <msup> <mrow> <mtext>equiv</mtext></mrow> <mrow> <mo>â</mo><mn>1</mn></mrow> </msup> <mo>.</mo></mrow> </math> If the enthalpy of neutralization of the first proton of aqueous <math><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mi mathvariant="normal">S</mi></math> is <math><mo>-</mo><mn>33.7</mn></math> <math><mi mathvariant="normal">k</mi><mi mathvariant="normal">J</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><msup><mrow><mi mathvariant="normal">l</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math> then the <math> <mrow> <msub> <mrow> <mfenced> <mrow> <msub> <mrow> <mtext>pK</mtext></mrow> <mtext>a</mtext> </msub> </mrow> </mfenced></mrow> <mn>1</mn> </msub> </mrow> </math> of <math><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mi mathvariant="normal">S</mi></math> is</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><math><mfenced separators="|"><mrow><mfrac><mrow><mn>23.6</mn><mo>Ã</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>3</mn></mrow></msup><mo>-</mo><mi mathvariant="normal">T</mi><mi mathvariant="normal">Î</mi><mi mathvariant="normal">S</mi><mo>Â°</mo></mrow><mrow><mn>2.303</mn><mi mathvariant="normal">R</mi><mi mathvariant="normal">T</mi></mrow></mfrac></mrow></mfenced></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math><mfenced separators="|"><mrow><mfrac><mrow><mn>2.303</mn><mi mathvariant="normal">R</mi><mi mathvariant="normal">T</mi></mrow><mrow><mn>23.6</mn><mo>-</mo><mi mathvariant="normal">T</mi><mi mathvariant="normal">Î</mi><mi mathvariant="normal">S</mi><mo>Â°</mo></mrow></mfrac></mrow></mfenced></math></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math><mfenced separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">T</mi><mi mathvariant="normal">Î</mi><mi mathvariant="normal">S</mi><mo>Â°</mo><mo>-</mo><mn>23.6</mn></mrow><mrow><mi mathvariant="normal">R</mi><mi mathvariant="normal">T</mi></mrow></mfrac></mrow></mfenced></math></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math><mn>2.303</mn><mfenced separators="|"><mrow><mfrac><mrow><mi mathvariant="normal">T</mi><mi mathvariant="normal">Î</mi><mi mathvariant="normal">S</mi><mo>Â°</mo><mo>-</mo><mn>23.6</mn></mrow><mrow><mi mathvariant="normal">R</mi><mi mathvariant="normal">T</mi></mrow></mfrac></mrow></mfenced></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math><mfenced separators="|"><mrow><mfrac><mrow><mn>23.6</mn><mo>Ã</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>3</mn></mrow></msup><mo>-</mo><mi mathvariant="normal">T</mi><mi mathvariant="normal">Î</mi><mi mathvariant="normal">S</mi><mo>Â°</mo></mrow><mrow><mn>2.303</mn><mi mathvariant="normal">R</mi><mi mathvariant="normal">T</mi></mrow></mfrac></mrow></mfenced></math></span> </div>

answer:
<div class="solution"><math><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mi mathvariant="normal">S</mi><mo>â</mo><msup><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mo>+</mo></mrow></msup><mo>+</mo><mi mathvariant="normal">H</mi><msup><mrow><mi mathvariant="normal">S</mi></mrow><mrow><mo>-</mo></mrow></msup></math><br/><math><mi mathvariant="normal">Î</mi><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mi mathvariant="normal">i</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">n</mi></mrow></msub><mo>=</mo><mn>23.6</mn></math> <math><mi mathvariant="normal">k</mi><mi mathvariant="normal">J</mi></math><br/><math><mo>â´</mo><mi mathvariant="normal">Î</mi><mi mathvariant="normal">G</mi><mo>Â°</mo><mo>=</mo><mn>23.6</mn><mo>Ã</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>3</mn></mrow></msup><mo>-</mo><mi mathvariant="normal">T</mi><mi mathvariant="normal">Î</mi><mi mathvariant="normal">S</mi><mo>Â°</mo><mo>=</mo><mo>-</mo><mn>2.303</mn><mi mathvariant="normal"></mi><mi mathvariant="normal">R</mi><mi mathvariant="normal">T</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">l</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">g</mi><mi mathvariant="normal"></mi><msub><mrow><mi mathvariant="normal">K</mi></mrow><mrow><msub><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>1</mn></mrow></msub></mrow></msub></math><br/><math><mo>â´</mo><msub><mrow><mi mathvariant="normal">P</mi></mrow><mrow><mi mathvariant="normal">K</mi><msub><mrow><mi mathvariant="normal">a</mi></mrow><mrow><mn>1</mn></mrow></msub></mrow></msub><mo>=</mo><mfrac><mrow><mn>23.6</mn><mo>Ã</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>3</mn></mrow></msup><mo>-</mo><mi mathvariant="normal">T</mi><mi mathvariant="normal">Î</mi><mi mathvariant="normal">S</mi><mo>Â°</mo></mrow><mrow><mn>2.303</mn><mi mathvariant="normal"></mi><mi mathvariant="normal">R</mi><mi mathvariant="normal">T</mi></mrow></mfrac></math></div>


tags: ['Chemistry', 'Thermodynamics (C)', 'JEE Main', 'JEE Main 2022 (25 Jun Shift 1)']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">300</span> </div>
source_db: MarksBatch2_P3.db
title:
the-standard-entropy-change-for-the-reaction-4-fe-s-3-o-2-g-2-fe-2-o-3-s-is-550-j-k-1-at-298-k-given-the-standard-enthalpy-change-for-the-reaction-is-

slug:
the-standard-entropy-change-for-the-reaction-4-fe-s-3-o-2-g-2-fe-2-o-3-s-is-550-j-k-1-at-298-k-given-the-standard-enthalpy-change-for-the-reaction-is-61292

question:
<div class="question"><p>The standard entropy change for the reaction</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>4</mn><mi>Fe</mi><mfenced><mi mathvariant="normal">s</mi></mfenced><mo>+</mo><mn>3</mn><msub><mi mathvariant="normal">O</mi><mn>2</mn></msub><mfenced><mi mathvariant="normal">g</mi></mfenced><mo>→</mo><mn>2</mn><msub><mi>Fe</mi><mn>2</mn></msub><msub><mi mathvariant="normal">O</mi><mn>3</mn></msub><mfenced><mi mathvariant="normal">s</mi></mfenced></math> is <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mn>550</mn><mo> </mo><mi mathvariant="normal">J</mi><msup><mrow><mo> </mo><mi mathvariant="normal">K</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math> at <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>298</mn><mo> </mo><mi mathvariant="normal">K</mi></math></p><p>[Given : The standard enthalpy change for the reaction is <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mn>165</mn><mo> </mo><mi>kJ</mi><msup><mrow><mo> </mo><mi>mol</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math>]. The temperature in <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">K</mi></math> at which the reaction attains equilibrium is (Nearest Integer)</p></div>

answer:
<div class="solution"><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mn>4</mn><mi>Fe</mi><mfenced><mi mathvariant="normal">s</mi></mfenced><mo>+</mo><mn>3</mn><msub><mi mathvariant="normal">O</mi><mn>2</mn></msub><mfenced><mi mathvariant="normal">g</mi></mfenced><mo>→</mo><mn>3</mn><msub><mi>Fe</mi><mn>2</mn></msub><msub><mi mathvariant="normal">O</mi><mn>3</mn></msub><mfenced><mi mathvariant="normal">s</mi></mfenced><mo>  </mo><msub><mi>ΔS</mi><mi mathvariant="normal">r</mi></msub><mo>=</mo><mo>-</mo><mn>550</mn><mo> </mo><mi mathvariant="normal">J</mi><mo>/</mo><mi mathvariant="normal">K</mi></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>ΔS</mi><mi mathvariant="normal">r</mi></msub><mo>=</mo><mo>-</mo><mn>165</mn><mo> </mo><mi>kJ</mi><mo>/</mo><mi>mole</mi></math></p><p>At equilibrium <math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ΔG</mi><mo>=</mo><mn>0</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">T</mi><mo>=</mo><mfenced><mfrac><mi>ΔH</mi><mi>ΔS</mi></mfrac></mfenced></math> at equilibrium</p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>=</mo><mfrac><mrow><mo>-</mo><mn>165</mn><mo>×</mo><msup><mn>10</mn><mn>3</mn></msup></mrow><mrow><mo>-</mo><mn>550</mn></mrow></mfrac><mo>=</mo><mn>300</mn><mo> </mo><mi mathvariant="normal">K</mi></math></p></div>


title: the-standard-free-energy-change-in-j-for-the-reaction-3-fe-2-aq-2-cr-s-2-cr-3-aq-3-fe-s-given
slug: the-standard-free-energy-change-in-j-for-the-reaction-3-fe-2-aq-2-cr-s-2-cr-3-aq-3-fe-s-given-58901
tags: ['Chemistry', 'Electrochemistry', 'KVPY', 'KVPY 2014 (SB/SX)']
source_db: MarksBatch2_P3.db
question:
<div class="question">The standard free energy change (in $J)$ for the reaction $3 \mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{Cr}(\mathrm{s})=2 \mathrm{Cr}^{3+}(\mathrm{aq})+3 \mathrm{Fe}(\mathrm{s})$ given</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">57,900</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$-57,900$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$-173,700$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">173,700</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$-173,700$</span> </div>

answer:
<div class="solution">$\mathrm{E}_{\mathrm{Cell}}^{\circ}=\mathrm{SRP}$ cathode - SRP anode $=-0.44-(-0.74)=0.3 \mathrm{~V}$<br/>$\underset{Y C}{\Delta} \mathrm{G}=-\mathrm{nF} \mathrm{E}_{\mathrm{Cel|}}^{\circ}\left(\mathrm{n}=6\right.$ ele $\left.^{-}\right)$</div>


tags: ['Chemistry', 'Thermodynamics (C)', 'VITEEE', 'VITEEE 2008']
source_db: MarksBatch2_P3.db
title:
the-standard-free-energy-change-of-a-reaction-is-g-115-kj-at-298-k-calculate-the-equilibrium-constant-k-p-in-lo-g-k-p-r-8314-jk-1-mol-1

slug:
the-standard-free-energy-change-of-a-reaction-is-g-115-kj-at-298-k-calculate-the-equilibrium-constant-k-p-in-lo-g-k-p-r-8314-jk-1-mol-1-90945

question:
<div class="question">The standard free energy change of a reaction is $\Delta \mathrm{G}^{\circ}=-115 \mathrm{~kJ}$ at $298 \mathrm{~K}$. Calculate the equilibrium constant $\mathrm{k}_{\mathrm{p}}$ in $\log \mathrm{k}_{\mathrm{p}}$ $\left(\mathrm{R}=8.314 \mathrm{Jk}^{-1} \mathrm{~mol}^{-1}\right)$</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$20.16$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$2.303$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$2.016$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$13.83$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$20.16$</span> </div>

answer:
<div class="solution">$\begin{aligned} \Delta \mathrm{G}^{\circ} &amp;=-115 \mathrm{~kJ} \text { at } 298 \mathrm{~K} . \\ \mathrm{Now}, \Delta \mathrm{G}^{\circ} &amp;=-2.303 \mathrm{RT} \log \mathrm{k}_{\mathrm{p}} \\ \mathrm{R} &amp;=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \&amp; \mathrm{~T}=298 \mathrm{~K} . \\ \Delta \mathrm{G}^{\circ} &amp;=-2.303 \times 8.314 \times 298 \times \log \mathrm{k}_{\mathrm{p}} \\ \log \mathrm{k}_{\mathrm{P}} &amp;=\frac{-115 \times 10^{3}}{-2.303 \times 8.314 \times 298}=20.155 \\ &amp;=20.16 \end{aligned}$</div>


tags: ['Chemistry', 'Chemical Equilibrium', 'JEE Main', 'JEE Main 2015 (04 Apr)']
source_db: MarksBatch2_P3.db
title:
the-standard-gibbs-energy-change-at-300-k-for-the-reaction-2-a-b-c-is-24942-j-at-a-given-time-the-composition-of-the-reaction-mixture-is-a-1-2-b-2-and

slug:
the-standard-gibbs-energy-change-at-300-k-for-the-reaction-2-a-b-c-is-24942-j-at-a-given-time-the-composition-of-the-reaction-mixture-is-a-1-2-b-2-and-74582

question:
<div class="question">The standard Gibbs energy change at 300 K for the reaction <math><mn>2</mn><mi mathvariant="normal">A</mi><mo>⇌</mo><mi mathvariant="normal">B</mi><mo>+</mo><mi mathvariant="normal">C</mi></math> is 2494.2 J. At a given time, the composition of the reaction mixture is <math><mfenced close="]" open="[" separators="|"><mrow><mi mathvariant="normal">A</mi></mrow></mfenced><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mo>,</mo><mi mathvariant="normal"> </mi><mfenced close="]" open="[" separators="|"><mrow><mi mathvariant="normal">B</mi></mrow></mfenced><mo>=</mo><mn>2</mn></math> and <math><mfenced close="]" open="[" separators="|"><mrow><mi mathvariant="normal">C</mi></mrow></mfenced><mo>=</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac></math> . The reaction proceeds in the: <math><mo>[</mo><mi mathvariant="normal">R</mi><mo>=</mo><mn>8.314</mn><mi mathvariant="normal"> </mi><mrow><mrow><mi mathvariant="normal">J</mi></mrow><mo>/</mo><mrow><mi mathvariant="normal">K</mi><mo>-</mo><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">l</mi><mi mathvariant="normal"> </mi><mo>,</mo><mi mathvariant="normal"> </mi><mi mathvariant="normal"> </mi><mi mathvariant="normal">e</mi><mo>=</mo><mn>2.718</mn><mo>] </mo></mrow></mrow></math> {Given antilog (-0.44)=0.36}</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">Reverse direction because  <math><mi mathvariant="normal">Q</mi><mo>&lt;</mo><msub><mrow><mi mathvariant="normal">K</mi></mrow><mrow><mi mathvariant="normal">C</mi></mrow></msub></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">Forward direction because  <math><mi mathvariant="normal">Q</mi><mo>&gt;</mo><msub><mrow><mi mathvariant="normal">K</mi></mrow><mrow><mi mathvariant="normal">C</mi></mrow></msub></math></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">Reverse direction because <math><mi mathvariant="normal">Q</mi><mo>&gt;</mo><msub><mrow><mi mathvariant="normal">K</mi></mrow><mrow><mi mathvariant="normal">C</mi></mrow></msub></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">Forward direction because  <math><mi mathvariant="normal">Q</mi><mo>&lt;</mo><msub><mrow><mi mathvariant="normal">K</mi></mrow><mrow><mi mathvariant="normal">C</mi></mrow></msub></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">Reverse direction because <math><mi mathvariant="normal">Q</mi><mo>&gt;</mo><msub><mrow><mi mathvariant="normal">K</mi></mrow><mrow><mi mathvariant="normal">C</mi></mrow></msub></math></span> </div>

answer:
<div class="solution"><math> <mrow> <mo>Δ</mo> <msup> <mtext>G</mtext> <mo>∘</mo> </msup> <mo>=</mo> <mo>-</mo> <msub> <mtext>RT ln K</mtext> <mtext>C</mtext> </msub> </mrow> </math><br/><br/><math> <mrow> <mtext>2494.2</mtext> <mo>=</mo> <mo>-</mo> <mtext>8.314</mtext> <mo>×</mo> <msub> <mtext>300 ln K</mtext> <mtext>C</mtext> </msub> </mrow> </math><br/><br/><math> <mrow> <mtext>2494.2</mtext> <mo>=</mo> <mo>-</mo> <mtext>8.314</mtext> <mo>×</mo> <mn>300</mn> <mo>×</mo> <msub> <mtext>2.303 log K</mtext> <mtext>C</mtext> </msub> </mrow> </math><br/><br/><math> <mrow> <mo>-</mo> <mfrac> <mrow> <mtext>2494.2</mtext> </mrow> <mrow> <mtext>2.303</mtext> <mo>×</mo> <mn>300</mn> <mo>×</mo> <mtext>8.314</mtext> </mrow> </mfrac> <mo>=</mo> <mo>-</mo> <mtext>0.44</mtext> <mo>=</mo> <msub> <mtext>log K</mtext> <mtext>C</mtext> </msub> </mrow> </math><br/><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mtext>log K</mtext><mtext>C</mtext></msub><mo>=</mo><mo>-</mo><mtext>0.44</mtext></math><br/><br/><math> <mrow> <msub> <mtext>K</mtext> <mtext>C</mtext> </msub> <mo>=</mo> <mtext>0.36</mtext> </mrow> </math><br/><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Q</mtext><mo>=</mo><mfrac><mrow><mo>[</mo><mi mathvariant="normal">B</mi><mo>]</mo><mo>[</mo><mi mathvariant="normal">C</mi><mo>]</mo></mrow><mrow><mo>[</mo><mi mathvariant="normal">A</mi><msup><mo>]</mo><mn>2</mn></msup></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>2</mn><mo>×</mo><mfrac><mn>1</mn><mn>2</mn></mfrac></mrow><msup><mfenced><mfrac><mn>1</mn><mn>2</mn></mfrac></mfenced><mn>2</mn></msup></mfrac><mo>=</mo><mn>4</mn></math><br/><br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>Q</mtext><mo>&gt;</mo><msub><mtext>K</mtext><mtext>C</mtext></msub></math>, reaction will shift in reverse direction.</div>


tags: ['Chemistry', 'Electrochemistry', 'AP EAMCET', 'AP EAMCET 2022 (06 Jul Shift 2)']
source_db: MarksBatch2_P3.db
title:
the-standard-gibbs-energy-change-for-daniel-cell-reaction-is-zn-s-cu-2-a-q-zn-2-a-q-cu-s-e-cell-11-v

slug:
the-standard-gibbs-energy-change-for-daniel-cell-reaction-is-zn-s-cu-2-a-q-zn-2-a-q-cu-s-e-cell-11-v-83862

question:
<div class="question">The standard Gibbs' energy change for Daniel cell reaction is<br/>$$<br/>\begin{aligned}<br/>&amp; \mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s) \\<br/>&amp; E_{\text {cell }}^{\circ}=1.1 \mathrm{~V} \text {. }<br/>\end{aligned}<br/>$$</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$-212.3 \mathrm{~kJ}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$106.15 \mathrm{~kJ}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$+212.3 \mathrm{~kJ}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$100 \mathrm{~kJ}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$-212.3 \mathrm{~kJ}$</span> </div>

answer:
<div class="solution">Given,<br/>$$<br/>\begin{aligned}<br/>n, E_{\text {cell }}^{\circ} &amp; =1.1 \mathrm{~V} \\<br/>n &amp; =2 \\<br/>F &amp; =96500 \mathrm{C} / \mathrm{mol} \\<br/>\Delta G &amp; =-n F E^{\circ} \\<br/>&amp; =-2 \times 96500 \times 1.1 \\<br/>&amp; =-212300 \mathrm{~J} \\<br/>&amp; =-212.3 \mathrm{~kJ}<br/>\end{aligned}<br/>$$</div>


tags: ['Chemistry', 'Electrochemistry', 'JEE Main', 'JEE Main 2019 (09 Apr Shift 1)']
source_db: MarksBatch2_P3.db
title:
the-standard-gibbs-energy-for-the-given-cell-reaction-in-k-j-m-o-l-1-at-298-k-is-z-n-s-c-u-2-a-q-z-n-2-a-q-c-u-s-e-0-2-v-a-t-298-k-f-a-r-a-d-a-y-s-c-o

slug:
the-standard-gibbs-energy-for-the-given-cell-reaction-in-k-j-m-o-l-1-at-298-k-is-z-n-s-c-u-2-a-q-z-n-2-a-q-c-u-s-e-0-2-v-a-t-298-k-f-a-r-a-d-a-y-s-c-o-23464

question:
<div class="question">The standard Gibbs energy for the given cell reaction in <math><mi mathvariant="normal">k</mi><mi mathvariant="normal">J</mi><mi mathvariant="normal"> </mi><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><msup><mrow><mi mathvariant="normal">l</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math> at <math><mn>298</mn><mi mathvariant="normal"> </mi><mi mathvariant="normal">K</mi></math> is:<br/><math><mi mathvariant="normal">Z</mi><mi mathvariant="normal">n</mi><mfenced separators="|"><mrow><mi mathvariant="normal">s</mi></mrow></mfenced><mo>+</mo><mi mathvariant="normal">C</mi><msup><mrow><mi mathvariant="normal">u</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mfenced separators="|"><mrow><mi mathvariant="normal">a</mi><mi mathvariant="normal">q</mi></mrow></mfenced><mo>⟶</mo><mi mathvariant="normal">Z</mi><msup><mrow><mi mathvariant="normal">n</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mfenced separators="|"><mrow><mi mathvariant="normal">a</mi><mi mathvariant="normal">q</mi></mrow></mfenced><mo>+</mo><mi mathvariant="normal">C</mi><mi mathvariant="normal">u</mi><mo>(</mo><mi mathvariant="normal">s</mi><mo>)</mo></math> ,<br/><math><msup><mrow><mi mathvariant="normal">E</mi></mrow><mrow><mn>0</mn></mrow></msup><mo>=</mo><mn>2</mn><mi mathvariant="normal"> </mi><mi mathvariant="normal">V</mi><mi mathvariant="normal"> </mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">t</mi><mi mathvariant="normal"> </mi><mn>298</mn><mi mathvariant="normal"> </mi><mi mathvariant="normal">K</mi></math><br/><math><mo>(</mo><mi mathvariant="normal">F</mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">r</mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">d</mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">y</mi><mi mathvariant="normal">'</mi><mi mathvariant="normal">s</mi><mi mathvariant="normal"> </mi><mi mathvariant="normal">c</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal">s</mi><mi mathvariant="normal">t</mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal">t</mi><mi mathvariant="normal"> </mi><mo>,</mo><mi mathvariant="normal">F</mi><mo>=</mo><mn>96000</mn><mi mathvariant="normal"> </mi><mi mathvariant="normal">C</mi><mi mathvariant="normal"> </mi><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><msup><mrow><mi mathvariant="normal">l</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup><mo>)</mo></math></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math><mo>-</mo><mn>192</mn></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math><mn>192</mn></math></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math><mn>384</mn></math></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><math><mo>-</mo><mn>384</mn></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math><mo>-</mo><mn>384</mn></math></span> </div>

answer:
<div class="solution">For the reaction,<br/><math><mi mathvariant="normal">C</mi><msup><mrow><mi mathvariant="normal">u</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mfenced separators="|"><mrow><mi mathvariant="normal">a</mi><mi mathvariant="normal">q</mi></mrow></mfenced><mo>+</mo><mi mathvariant="normal">Z</mi><mi mathvariant="normal">n</mi><mfenced separators="|"><mrow><mi mathvariant="normal">s</mi></mrow></mfenced><mo>⟶</mo><mi mathvariant="normal">Z</mi><msup><mrow><mi mathvariant="normal">n</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mfenced separators="|"><mrow><mi mathvariant="normal">a</mi><mi mathvariant="normal">q</mi></mrow></mfenced><mo>+</mo><mi mathvariant="normal">C</mi><mi mathvariant="normal">u</mi><mfenced separators="|"><mrow><mi mathvariant="normal">s</mi></mrow></mfenced></math><br/><math><mi mathvariant="normal">n</mi><mo>=</mo><mn>2</mn></math>, two electrons are exchanged<br/><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∆</mo><msup><mi mathvariant="normal">G</mi><mi mathvariant="normal">o</mi></msup><mo>=</mo><mo>-</mo><mi mathvariant="normal">n</mi><mi mathvariant="normal">F</mi><msup><mi mathvariant="normal">E</mi><mi mathvariant="normal">o</mi></msup></math><br/><math><mo>=</mo><mo>-</mo><mn>2</mn><mo>×</mo><mn>96000</mn><mo>×</mo><mn>2</mn><mo>=</mo><mo>-</mo><mn>384000</mn></math><br/><math><mo>=</mo><mo>-</mo><mn>384</mn><mi mathvariant="normal"> </mi><mi mathvariant="normal">k</mi><mi mathvariant="normal">J</mi></math></div>


tags: ['Chemistry', 'Thermodynamics (C)', 'AP EAMCET', 'AP EAMCET 2022 (07 Jul Shift 2)']
source_db: MarksBatch2_P3.db
title:
the-standard-gibbs-energy-g-for-the-following-reaction-is-a-s-b-2-a-q-a-2-a-q-b-s-k-c-1-0-12-at-2-5-c-k-c-equilibrium-constant

slug:
the-standard-gibbs-energy-g-for-the-following-reaction-is-a-s-b-2-a-q-a-2-a-q-b-s-k-c-1-0-12-at-2-5-c-k-c-equilibrium-constant-48101

question:
<div class="question">The standard Gibb's energy $\left(\Delta G^{\circ}\right)$ for the following reaction is<br/>$\begin{aligned} &amp; A(s)+B^{2+}(a q) \rightleftharpoons A^{2+}(a q)+B(s), \\ &amp; K_C=10^{12} \text { at } 25^{\circ} \mathrm{C}\left(K_C=\text { equilibrium constant }\right)\end{aligned}$</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$-150 \mathrm{~kJ}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$-96.80 \mathrm{~kJ}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$-68.47 \mathrm{~kJ}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$-100 \mathrm{~kJ}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$-68.47 \mathrm{~kJ}$</span> </div>

answer:
<div class="solution">The standard Gibb's free energy change $\left(\Delta G^{\circ}\right)$ at standard conditions is given by the formula $\Delta G^{\circ}=-R T \ln K_{\text {eq }}$<br/>where, $R$ is gas constant $=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, $T=(25+273) \mathrm{K}=298 \mathrm{~K}$ and $K_{\mathrm{eq}}$ is the equilibrium constant<br/>$$<br/>\begin{aligned}<br/>\Delta G^{\circ} &amp; =-8.314 \times 298 \times 2.303 \log 10^{12} \\<br/>&amp; =-68,47018 \mathrm{~J}=-68.47 \mathrm{~kJ}<br/>\end{aligned}<br/>$$</div>


tags: ['Chemistry', 'Electrochemistry', 'KVPY', 'KVPY 2011 (SB/SX)']
source_db: MarksBatch2_P3.db
title:
the-standard-gibbs-free-energy-change-g-in-kj-mol-1-in-a-daniel-cell-e-cell-11-v-when-2-moles-of-zn-s-is-oxidized-at-298-k-is-closest-to

slug:
the-standard-gibbs-free-energy-change-g-in-kj-mol-1-in-a-daniel-cell-e-cell-11-v-when-2-moles-of-zn-s-is-oxidized-at-298-k-is-closest-to-44469

question:
<div class="question">The standard Gibbs free energy change $\left(\Delta \mathrm{G}^{\circ}\right.$ in $\mathrm{kJ} \mathrm{mol}^{-1}$ ), in a Daniel cell $\left(\mathrm{E}_{\text {cell }}^{\circ}=1.1 \mathrm{~V}\right.$ ), when 2 moles of $\mathrm{Zn}(\mathrm{s})$ is oxidized at $298 \mathrm{~K}$, is closest to</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$-212.3$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$-106.2$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$-424.6$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$-53.1$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$-424.6$</span> </div>

answer:
<div class="solution">$\mathrm{Zn}+\mathrm{Cu}^{+2} \rightarrow \mathrm{Zn}^{+2}+\mathrm{Cu}$ <br/>$2 \mathrm{Zn}+2 \mathrm{Cu}^{+2} \rightarrow 2 \mathrm{Zn}^{+2}+2 \mathrm{Cu}$<br/>For 2 moles of $\mathrm{Zn}, \mathrm{n}=4$<br/>$\Delta \mathrm{G}^{\circ}=-\mathrm{nFE}_{\mathrm{Cell}}^{\circ}=-4 \times 96500 \times 1.1=-424.6 \mathrm{~kJ}$</div>


title: the-standard-gibbs-free-energy-change-g-is-related-to-equilibrium-constant-k-p-as-1
slug: the-standard-gibbs-free-energy-change-g-is-related-to-equilibrium-constant-k-p-as-1-13488
tags: ['Chemistry', 'Thermodynamics (C)', 'JEE Main']
source_db: MarksBatch2_P3.db
question:
<div class="question">The standard Gibbs free energy change $\Delta G^{\prime}$ is related to equilibrium constant $K_p$ as</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$K_p=-R T \ln \Delta G^{\circ}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$K_p=\left(\frac{e}{R T}\right)^{\Delta G^e}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$K_p=-\frac{\Delta G^o}{R T}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$K_p=e^{-\frac{\Delta G^T}{R T}}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$K_p=e^{-\frac{\Delta G^T}{R T}}$</span> </div>

answer:
<div class="solution">$K_p=e^{-\frac{\Delta C^T}{R T}}$<br/>For equilibrium condition $\Delta \mathrm{G}=0$<br/>So,<br/>$0=\Delta G^{\mathrm{a}}+R T \ln K_p$<br/>$\ln K_{\mathrm{p}}=-\Delta G^{\circ} / R T$<br/>Now<br/>$K_p=e^{-\frac{\Delta G^T}{R T}}$</div>


title: the-standard-gibbs-free-energy-change-g-is-related-to-equilibrium-constant-k-p-as
slug: the-standard-gibbs-free-energy-change-g-is-related-to-equilibrium-constant-k-p-as-77009
tags: ['Chemistry', 'Electrochemistry', 'VITEEE', 'VITEEE 2010']
source_db: MarksBatch2_P3.db
question:
<div class="question">The standard Gibb's free energy change, $\Delta \mathrm{G}^{\circ}$ is related to equilibrium constant, $\mathrm{K}_{\mathrm{p}}$ as</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$K_{p}=-R T \ln \Delta G^{\circ}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$K_{p}=\left[\frac{e}{R T}\right]^{\Delta G^{\circ}}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\mathrm{K}_{\mathrm{p}}=-\frac{\Delta \mathrm{G}}{\mathrm{RT}}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$K_{p}=e^{-\Delta G^{0} / R T}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$K_{p}=e^{-\Delta G^{0} / R T}$</span> </div>

answer:
<div class="solution">$\Delta G^{\circ}$ and $K_{p}$ are related as<br/>$$<br/>\begin{array}{l}<br/>\Delta G^{\circ}=-R T \ln K_{p} \\<br/>\Rightarrow \ln K_{p}=\frac{\Delta G^{\circ}}{R T} \Rightarrow K_{p}=e^{-\Delta G^{\circ} / R T}<br/>\end{array}<br/>$$</div>


tags: ['Chemistry', 'Thermodynamics (C)', 'WBJEE', 'WBJEE 2013']
source_db: MarksBatch2_P3.db
title:
the-standard-gibbs-free-energy-change-g-o-at-2-5-c-for-the-dissociation-of-n-2-o-4-g-to-no-2-g-is-given-equilibrium-constant-015-r-8314-jk-mol

slug:
the-standard-gibbs-free-energy-change-g-o-at-2-5-c-for-the-dissociation-of-n-2-o-4-g-to-no-2-g-is-given-equilibrium-constant-015-r-8314-jk-mol-45174

question:
<div class="question">The standard Gibbs free energy change $\left(\Delta G^{o}\right)$ at $25^{\circ} \mathrm{C}$ for the dissociation of $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})$ to $\mathrm{NO}_{2}(g)$ is (given, equilibrium constant $=0.15, R=8.314 \mathrm{JK} / \mathrm{mol})$</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$1.1 \mathrm{KJ}$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$4.7 \mathrm{KJ}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$8.1 \mathrm{KJ}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$38.2 \mathrm{KJ}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$4.7 \mathrm{KJ}$</span> </div>

answer:
<div class="solution">$$
<br/>\begin{array}{l}
<br/>=-2.303 \times 8.314 \times 298 \times \log 0.15 \\
<br/>=-2.303 \times 8.314 \times 298 \times(-0.82) \\
<br/>=4678.7 \mathrm{J}=4.67 \mathrm{kJ}
<br/>\end{array}
<br/>$$</div>


tags: ['Chemistry', 'Ionic Equilibrium', 'AIIMS', 'AIIMS 2010']
source_db: MarksBatch2_P3.db
title:
the-standard-halfcell-reduction-potential-for-ag-ag-is-07991-v-at-2-5-c-given-the-experimental-value-k-s-p-156-1-0-10-for-agcl-calculate-the-standard-

slug:
the-standard-halfcell-reduction-potential-for-ag-ag-is-07991-v-at-2-5-c-given-the-experimental-value-k-s-p-156-1-0-10-for-agcl-calculate-the-standard-12848

question:
<div class="question">The standard half-cell reduction potential for $\mathrm{Ag}^{+} \mid \mathrm{Ag}$ is $0.7991 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. Given the experimental value $K_{s p}=1.56 \times 10^{-10}$ for $\mathrm{AgCl}$, calculate the standard half-cell reduction potential for the $\mathrm{Ag} \mid \mathrm{AgCl}$ electrode.</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$0.2192 \mathrm{~V}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$-0.2192 \mathrm{~V}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$-1.2192 \mathrm{~V}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$1.2192 \mathrm{~V}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$0.2192 \mathrm{~V}$</span> </div>

answer:
<div class="solution">For the desired reaction,<br/>$$<br/>\mathrm{AgCl}+1 e^{-} \rightarrow \mathrm{Ag}+\mathrm{Cl}^{-}, \quad E^{\circ}=\frac{-\Delta_r G^{\circ}}{(1) F}<br/>$$<br/>The needed $\Delta_r G^{\circ}$ can be obtained by adding the values of $\Delta_r G^{\circ}$ for the reactions<br/>$$<br/>\begin{aligned}<br/>&amp; \mathrm{Ag}^{+}+1 e^{-} \rightarrow \mathrm{Ag} ; \Delta G^{\circ}=-n F E^{\circ} \\<br/>&amp; \mathrm{AgCl} \rightarrow \mathrm{Ag}^{+}+\mathrm{Cl}^{-} ; \Delta G^{\circ}=-R T \ln K_{s p}, \text { giving } \\<br/>&amp; \Delta_r G^{\circ}=-(1 \mathrm{~mol})\left(9.648 \times 10^4 \mathrm{~J} \mathrm{~mol}{ }^{-1} \mathrm{~V}^{-1}\right)(0.7991 \mathrm{~V}) \\<br/>&amp; \quad-\left(8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)(298 \mathrm{~K}) \ln \left(1.56 \times 10^{-10}\right) \\<br/>&amp; =-77.10 \mathrm{~kJ}+55.95 \mathrm{~kJ}=-21.15 \mathrm{~kJ}<br/>\end{aligned}<br/>$$<br/>The potential is $E^{\circ}=\frac{-(-21.15)}{(1)(96.485)}=+0.2192 \mathrm{~V}$.</div>


tags: ['Chemistry', 'Thermodynamics (C)', 'JEE Main', 'JEE Main 2020 (07 Jan Shift 2)']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">-192</span> </div>
source_db: MarksBatch2_P3.db
title:
the-standard-heat-of-formation-f-h-298-0-of-ethane-in-k-j-m-o-l-if-the-heat-of-combustion-of-ethane-hydrogen-and-graphite-are-1560-3935-and-286-k-j-m--1

slug:
the-standard-heat-of-formation-f-h-298-0-of-ethane-in-k-j-m-o-l-if-the-heat-of-combustion-of-ethane-hydrogen-and-graphite-are-1560-3935-and-286-k-j-m-1-98447

question:
<div class="question">The standard heat of formation <math><mfenced separators="|"><mrow><msub><mrow><mo>∆</mo></mrow><mrow><mi>f</mi></mrow></msub><msubsup><mrow><mi>H</mi></mrow><mrow><mn>298</mn></mrow><mrow><mn>0</mn></mrow></msubsup></mrow></mfenced></math> of ethane (in <math><mrow><mrow><mi mathvariant="normal">k</mi><mi mathvariant="normal">J</mi></mrow><mo>/</mo><mrow><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">l</mi></mrow></mrow></math> ), if the heat of combustion of ethane, hydrogen and graphite are <math><mo>-</mo><mn>1560</mn><mo>,</mo><mo>-</mo><mn>393.5</mn></math> and <math><mo>-</mo><mn>286 </mn><mrow><mrow><mi mathvariant="normal">k</mi><mi mathvariant="normal">J</mi></mrow><mo>/</mo><mrow><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">l</mi></mrow></mrow></math> , respectively is ________</div>

answer:
<div class="solution"><math> <mrow> <mn>2</mn><mtext>C</mtext><mo stretchy="false">(</mo><mtext> graphite </mtext><mo stretchy="false">)</mo><mo>+</mo><mn>3</mn><msub> <mtext>H</mtext> <mn>2</mn> </msub> <mo stretchy="false">(</mo><mtext>g</mtext><mo stretchy="false">)</mo><mo>â</mo><msub> <mtext>C</mtext> <mn>2</mn> </msub> <msub> <mtext>H</mtext> <mn>6</mn> </msub> <mo stretchy="false">(</mo><mtext>g</mtext><mo stretchy="false">)</mo></mrow> </math><br/><br/><math> <mrow> <msub> <mi>Î</mi> <mtext>f</mtext> </msub> <mtext>H</mtext><mfenced> <mrow> <msub> <mtext>C</mtext> <mn>2</mn> </msub> <msub> <mtext>H</mtext> <mn>6</mn> </msub> </mrow> </mfenced><mo>=</mo><mn>2</mn><mi>Î</mi><msub> <mtext>H</mtext> <mrow> <mtext>comb </mtext></mrow> </msub> <mfenced> <mrow> <msub> <mtext>C</mtext> <mrow> <mtext>gaphite </mtext></mrow> </msub> </mrow> </mfenced><mo>+</mo><mn>3</mn><mi>Î</mi><msub> <mtext>H</mtext> <mrow> <mtext>comb </mtext></mrow> </msub> <mfenced> <mrow> <msub> <mtext>H</mtext> <mn>2</mn> </msub> </mrow> </mfenced></mrow> </math><br/><br/><math> <mrow> <mo>â</mo><mi>Î</mi><msub> <mtext>H</mtext> <mrow> <mtext>comb </mtext></mrow> </msub> <mfenced> <mrow> <msub> <mtext>C</mtext> <mn>2</mn> </msub> <msub> <mtext>H</mtext> <mn>6</mn> </msub> </mrow> </mfenced></mrow> </math><br/><br/><math> <mrow> <mo>=</mo><mo>â</mo><mo stretchy="false">(</mo><mn>286</mn><mo>Ã</mo><mn>2</mn><mo stretchy="false">)</mo><mo>â</mo><mo stretchy="false">(</mo><mn>393.5</mn><mo>Ã</mo><mn>3</mn><mo stretchy="false">)</mo><mo>â</mo><mo stretchy="false">(</mo><mo>â</mo><mn>1560</mn><mo stretchy="false">)</mo></mrow> </math><br/><br/><math> <mrow> <mo>=</mo><mo>â</mo><mn>572</mn><mo>â</mo><mn>1180.5</mn><mo>+</mo><mn>1560</mn><mo>=</mo><mo>â</mo><mn>192.5 </mn><mtext>kJ</mtext><mo>/</mo><mtext>mole</mtext></mrow> </math></div>


tags: ['Chemistry', 'Thermodynamics (C)', 'NEET']
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">192.5</span> </div>
source_db: MarksBatch2_P3.db
title:
the-standard-heat-of-formation-f-h-298-0-of-ethane-in-k-j-m-o-l-if-the-heat-of-combustion-of-ethane-hydrogen-and-graphite-are-1560-3935-and-286-k-j-m-

slug:
the-standard-heat-of-formation-f-h-298-0-of-ethane-in-k-j-m-o-l-if-the-heat-of-combustion-of-ethane-hydrogen-and-graphite-are-1560-3935-and-286-k-j-m-58256

question:
<div class="question">The standard heat of formation <math><mo>(</mo><msub><mrow><mi mathvariant="normal">Î</mi></mrow><mrow><mi mathvariant="normal">f</mi></mrow></msub><msubsup><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>298</mn></mrow><mrow><mn>0</mn></mrow></msubsup><mo>)</mo></math> of ethane (in <math><mi mathvariant="normal">k</mi><mi mathvariant="normal">J</mi><mo>/</mo><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">l</mi></math> ), if the heat of combustion of ethane, hydrogen and graphite are â1560, â393.5 and â286 <math><mi mathvariant="normal">k</mi><mi mathvariant="normal">J</mi><mo>/</mo><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">l</mi></math> , respectively is .</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">192.5</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">â 192.5</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">96.25</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">+96.25</span> </li> </ul>

answer:
<div class="solution"><math><mtext>2C</mtext><mo>(</mo><mtext>graphite</mtext><mo>)</mo><mo>+</mo><mtext>3</mtext><msub><mrow><mtext>H</mtext></mrow><mrow><mn>2</mn></mrow></msub><mo>(</mo><mi mathvariant="normal">g</mi><mo>)</mo><mo>â</mo><mtext>1</mtext><msub><mrow><mtext>C</mtext></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>6</mn></mrow></msub><mo>+</mo><mi mathvariant="normal">Î</mi><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mi mathvariant="normal">f</mi></mrow></msub><mo>(</mo><msub><mrow><mi mathvariant="normal">C</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>6</mn></mrow></msub><mo>)</mo></math> <br/>(A) <math><msub><mrow><mi mathvariant="normal">C</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>6</mn></mrow></msub><mo>+</mo><mfrac><mrow><mn>7</mn></mrow><mrow><mn>2</mn></mrow></mfrac><msub><mrow><mi mathvariant="normal">O</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>â</mo><mtext>2C</mtext><msub><mrow><mtext>O</mtext></mrow><mrow><mn>2</mn></mrow></msub><mo>+</mo><mtext>3</mtext><msub><mrow><mtext>H</mtext></mrow><mrow><mn>2</mn></mrow></msub><mi mathvariant="normal">O</mi></math> <br/><math><mi mathvariant="normal">Î</mi><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mi mathvariant="normal">C</mi></mrow></msub><mo>=</mo><mo>â</mo><mtext>1560</mtext></math> <br/>(B) <math><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>+</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><msub><mrow><mi mathvariant="normal">O</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>â</mo><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mi mathvariant="normal">O</mi></math> <br/><math><mi mathvariant="normal">Î</mi><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mi mathvariant="normal">C</mi></mrow></msub><mo>=</mo><mo>â</mo><mtext>393</mtext><mo>.</mo><mn>5</mn></math> <br/>(C) C(graphite) <math><mo>+</mo><msub><mrow><mi mathvariant="normal">O</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>â</mo><mtext>C</mtext><msub><mrow><mtext>O</mtext></mrow><mrow><mn>2</mn></mrow></msub></math> <br/><math><mi mathvariant="normal">Î</mi><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mi mathvariant="normal">C</mi></mrow></msub><mo>=</mo><mo>â</mo><mtext>286kJ/mol</mtext></math> <br/><math><mi mathvariant="normal">Î</mi><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mi mathvariant="normal">f</mi></mrow></msub><mo>(</mo><msub><mrow><mi mathvariant="normal">C</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>6</mn></mrow></msub><mo>)</mo><mo>=</mo><mn>3</mn><mo>(</mo><mo>â</mo><mtext>393</mtext><mo>.</mo><mn>5</mn><mo>)</mo><mo>+</mo><mn>2</mn><mo>(</mo><mo>â</mo><mtext>286</mtext><mo>)</mo><mo>+</mo><mtext>156</mtext><mn>0</mn></math> <br/><math><mo>=</mo><mo>â</mo><mtext>118</mtext><mn>0.5</mn><mo>+</mo><mo>(</mo><mo>â</mo><mtext>572</mtext><mo>)</mo><mo>+</mo><mtext>156</mtext><mn>0</mn></math> <br/><math><mo>=</mo><mtext>192</mtext><mo>.</mo><mn>5</mn></math> <br/><math><mn>2</mn><mi mathvariant="normal">C</mi><mfenced separators="|"><mrow><mi mathvariant="normal">g</mi><mi mathvariant="normal">r</mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">p</mi><mi mathvariant="normal">h</mi><mi mathvariant="normal">i</mi><mi mathvariant="normal">t</mi><mi mathvariant="normal">e</mi></mrow></mfenced><mo>+</mo><mn>3</mn><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>â</mo><msub><mrow><mi mathvariant="normal">C</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>6</mn></mrow></msub></math> (1) eq<sup>n</sup> <br/><math><msub><mrow><mi mathvariant="normal">C</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>6</mn></mrow></msub><mo>+</mo><mfrac><mrow><mn>7</mn></mrow><mrow><mn>2</mn></mrow></mfrac><msub><mrow><mi mathvariant="normal">O</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>â</mo><mn>2</mn><mi mathvariant="normal">C</mi><msub><mrow><mi mathvariant="normal">O</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>+</mo><mn>3</mn><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mi mathvariant="normal">O</mi></math> (2) eq<sup>n</sup> <br/><math><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>+</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><msub><mrow><mi mathvariant="normal">O</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>â</mo><mn>2</mn><mi mathvariant="normal">C</mi><msub><mrow><mi mathvariant="normal">O</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>+</mo><mn>3</mn><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mi mathvariant="normal">O</mi></math> (3) eq<sup>n</sup> <br/><math><mi mathvariant="normal">C</mi><mfenced separators="|"><mrow><mi mathvariant="normal">g</mi><mi mathvariant="normal">r</mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">p</mi><mi mathvariant="normal">h</mi><mi mathvariant="normal">i</mi><mi mathvariant="normal">t</mi><mi mathvariant="normal">e</mi></mrow></mfenced><mo>+</mo><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mi mathvariant="normal">O</mi><mo>â</mo><mi mathvariant="normal">C</mi><msub><mrow><mi mathvariant="normal">O</mi></mrow><mrow><mn>2</mn></mrow></msub></math> (4) eq<sup>n</sup> <br/>To get equation (1) eq<sup>n</sup> (2) + 2 Ã eq<sup>n</sup> (3) 2 Ã eq<sup>n</sup> <br/><math><mo>â</mo><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mi mathvariant="normal">f</mi></mrow></msub></math> of ethane <math><mo>=</mo><mn>1560</mn><mo>+</mo><mn>2</mn><mo>Ã</mo><mfenced separators="|"><mrow><mo>-</mo><mn>286</mn></mrow></mfenced><mo>+</mo><mn>3</mn><mo>(</mo><mo>-</mo><mn>393.5</mn><mo>)</mo></math> <br/><math><mo>=</mo><mn>1560</mn><mo>-</mo><mn>572</mn><mo>-</mo><mn>1180.5</mn><mo>=</mo><mn>192.5</mn></math></div>


tags: ['Chemistry', 'Thermodynamics (C)', 'NEET']
source_db: MarksBatch2_P3.db
title:
the-standard-molar-enthalpies-of-formation-of-cycohexane-l-and-benzene-l-at-25-c-are-156-and-49-kj-m-o-l-1-respectively-the-standard-enthalpy-of-hydro

slug:
the-standard-molar-enthalpies-of-formation-of-cycohexane-l-and-benzene-l-at-25-c-are-156-and-49-kj-m-o-l-1-respectively-the-standard-enthalpy-of-hydro-32885

question:
<div class="question">The standard molar enthalpies of formation of cycohexane (l) and benzene (l) at <math><mn>25</mn><mo>Â°</mo><mi mathvariant="normal">C</mi></math> are -156 and +49 KJ <math><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><msup><mrow><mi mathvariant="normal">l</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math> respectively. The standard enthalpy of hydrogenation of cyclo hexene (l) at <math><mn>25</mn><mo>Â°</mo><mi mathvariant="normal">C</mi></math> is -116 KJ <math><mi mathvariant="normal">m</mi><mi mathvariant="normal">o</mi><msup><mrow><mi mathvariant="normal">l</mi></mrow><mrow><mo>-</mo><mn>1</mn></mrow></msup></math> . Use these data to estimate the magnitude of the resonance energy of benzene.</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><math> <mrow> <mtext>152</mtext><mtext> </mtext><mtext>kJ</mtext><mtext> </mtext><msup> <mrow> <mtext>mol</mtext></mrow> <mrow> <mo>−</mo><mn>1</mn></mrow> </msup> </mrow></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math> <mrow> <mn>1.52</mn><mtext> </mtext><mtext>kJ</mtext><mtext> </mtext><msup> <mrow> <mtext>mol</mtext></mrow> <mrow> <mo>−</mo><mn>1</mn></mrow> </msup> </mrow></math></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math> <mrow> <mn>15.2</mn><mtext> </mtext><mtext>kJ</mtext><mtext> </mtext><msup> <mrow> <mtext>mol</mtext></mrow> <mrow> <mo>−</mo><mn>1</mn></mrow> </msup> </mrow></math></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">None of these</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math> <mrow> <mtext>152</mtext><mtext> </mtext><mtext>kJ</mtext><mtext> </mtext><msup> <mrow> <mtext>mol</mtext></mrow> <mrow> <mo>−</mo><mn>1</mn></mrow> </msup> </mrow></math></span> </div>

answer:
<div class="solution">Enthalpy of hydrogenation of these double bond in benzene<br/><math> <mrow> <mo>=</mo><mn>3</mn><mo>Ã</mo><mo>â</mo><mn>119</mn><mo>=</mo><mo>â</mo><mn>357</mn><mtext>â</mtext><mtext>kJ</mtext><mtext>â</mtext><msup> <mrow> <mtext>mol</mtext></mrow> <mrow> <mo>â</mo><mn>1</mn></mrow> </msup> </mrow> </math><br/><math> <mrow> <mtext>ΔH(reaction)</mtext><mtext> </mtext><mtext>=</mtext><mtext> </mtext><mtext>ΣH(products)</mtext><mo>−</mo><mtext>ΔΣH(reactants)</mtext></mrow></math><br/><math> <mrow> <mo>=</mo><mo>â</mo><mn>156</mn><mo>â</mo><mfenced> <mrow> <mn>49</mn><mo>+</mo><mn>0</mn></mrow> </mfenced><mo>=</mo><mo>â</mo><mn>205</mn><mtext>â</mtext><mtext>kJ</mtext><mtext>â</mtext><msup> <mrow> <mtext>mol</mtext></mrow> <mrow> <mo>â</mo><mn>1</mn></mrow> </msup> </mrow> </math><br/>Resonance energy = Calculated heat of hydrogenation â Actual heat of hydrogenation<br/><math> <mrow> <mo>=</mo><mn>357</mn><mo>â</mo><mn>205</mn><mo>=</mo><mn>152</mn><mtext>â</mtext><mtext>kJ</mtext><mtext>â</mtext><msup> <mrow> <mtext>mol</mtext></mrow> <mrow> <mo>â</mo><mn>1</mn></mrow> </msup> <mo>.</mo></mrow> </math></div>


tags: ['Chemistry', 'Thermodynamics (C)', 'AP EAMCET', 'AP EAMCET 2022 (07 Jul Shift 2)']
source_db: MarksBatch2_P3.db
title:
the-standard-molar-enthalpy-of-vaporisation-vap-h-of-a-b-and-c-liquids-is-233-41-and-29-kj-mol-1-respectively-the-correct-order-of-diopledipole-attrac

slug:
the-standard-molar-enthalpy-of-vaporisation-vap-h-of-a-b-and-c-liquids-is-233-41-and-29-kj-mol-1-respectively-the-correct-order-of-diopledipole-attrac-60438

question:
<div class="question">The standard molar enthalpy of vaporisation $\left(\Delta_{\text {vap }} H^{\circ}\right)$ of $A, B$ and $C$ liquids is $23.3,41$ and $29 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The correct order of diople-dipole attractive forces in these liquids is</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$B&gt;C&gt;A$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$B&gt;A&gt;C$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$A&gt;C&gt;B$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$A&gt;B&gt;C$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$B&gt;C&gt;A$</span> </div>

answer:
<div class="solution">( The enthalpy of vaporisation is the amount of heat that must be added to a liquid substance to transform it into a gas. Dipole-Dipole interactions are intermolecular force of attraction between polar molecules. Stronger the dipole-dipole interactions, higher will be the heat energy required to break these interactions and vapourise the liquid. Since, the liquids $A, B$ and $C$ have molar enthalpy of vaporisation as $23.3,41$ and $29 \mathrm{~kJ} \mathrm{~mol}^{-1}$, the order of dipole-dipole attractive interactions will be $B&gt;C&gt;A$.</div>


tags: ['Chemistry', 'Thermodynamics (C)', 'NEET']
source_db: MarksBatch2_P3.db
title:
the-standard-molar-entropy-of-h-2-o-l-is-70-j-k-1-mol-1-will-the-standard-molar-entropy-of-h-2-os-be-more-or-less-than-70-j-k-1-mol-1

slug:
the-standard-molar-entropy-of-h-2-o-l-is-70-j-k-1-mol-1-will-the-standard-molar-entropy-of-h-2-os-be-more-or-less-than-70-j-k-1-mol-1-10948

question:
<div class="question">The standard molar entropy of H<sub>2</sub>O (l) is 70 J K<sup>â1</sup>mol<sup>â1</sup>. Will the standard molar entropy of H<sub>2</sub>O(s) be more, or less than 70 J K<sup>â1</sup>mol<sup>â1</sup>?</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">Less, because H<sub>2</sub>O(l) is more ordered than ice.</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">More, because ice is more ordered than H<sub>2</sub>O(l).</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">Less, because ice is more ordered than H<sub>2</sub>O(l).</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">More, because H<sub>2</sub>O(l) is more ordered than ice.</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">Less, because ice is more ordered than H<sub>2</sub>O(l).</span> </div>

answer:
<div class="solution">Correct Option is : (C)<br/>Less, because ice is more ordered than H<sub>2</sub>O(l).</div>


tags: ['Chemistry', 'Thermodynamics (C)', 'JEE Main']
source_db: MarksBatch2_P3.db
title:
the-standard-molar-heat-of-formation-of-ethane-c-o-2-and-water-are-respectively-211-941-and-683-kcal-the-standard-molar-heat-of-combustion-of-ethane-w

slug:
the-standard-molar-heat-of-formation-of-ethane-c-o-2-and-water-are-respectively-211-941-and-683-kcal-the-standard-molar-heat-of-combustion-of-ethane-w-25179

question:
<div class="question">The standard molar heat of formation of ethane, <math><mi mathvariant="normal">C</mi><msub><mrow><mi mathvariant="normal">O</mi></mrow><mrow><mn>2</mn></mrow></msub></math> and water are respectively <math><mo>-</mo><mn>21.1,</mn><mo>-</mo><mn>94.1</mn></math> and <math> <mrow> <mo>â</mo><mtext>â</mtext><mn>68.3</mn><mtext>â</mtext><mtext>kcal</mtext><mtext>.</mtext></mrow> </math> The standard molar heat of combustion of ethane will be:</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><math><mo>-</mo><mn>372</mn></math> kcal/mol</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math> <mrow> <mtext>162 kcal/mol</mtext></mrow></math></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math><mo>-</mo><mn>240</mn></math> kcal/mol</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math> <mrow> <mtext>183</mtext><mtext>.5 kcal/mol</mtext></mrow></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math><mo>-</mo><mn>372</mn></math> kcal/mol</span> </div>

answer:
<div class="solution">Given, <math> <mrow> <mtext>2C</mtext><mtext> </mtext><mtext>+</mtext><mtext> </mtext><msub> <mrow> <mtext>3H</mtext></mrow> <mtext>2</mtext> </msub> <mo>→</mo><msub> <mtext>C</mtext> <mtext>2</mtext> </msub> <msub> <mtext>H</mtext> <mtext>6</mtext> </msub> <mtext>; </mtext><mtext> </mtext><mtext>ΔH</mtext><mtext> </mtext><mtext>=</mtext><mtext> </mtext><mo>−</mo><mtext>21</mtext><mtext>.1</mtext></mrow></math><br/><math> <mrow> <mtext>C</mtext><mtext>â</mtext><mtext>+</mtext><mtext>â</mtext><msub> <mtext>O</mtext> <mtext>2</mtext> </msub> <mo>â</mo><msub> <mrow> <mtext>CO</mtext></mrow> <mtext>2</mtext> </msub> <mo>;</mo><mo> </mo><mtext>ÎH</mtext><mo>=</mo><mo>â</mo><mn>94.1</mn></mrow> </math><br/><math> <mrow> <msub> <mtext>H</mtext> <mtext>2</mtext> </msub> <mo>+</mo><mfrac> <mn>1</mn> <mn>2</mn> </mfrac> <msub> <mtext>O</mtext> <mtext>2</mtext> </msub> <mo>â</mo><msub> <mtext>H</mtext> <mtext>2</mtext> </msub> <mtext>O</mtext><mo>;</mo><mo> </mo><mtext>â</mtext><mtext>ÎH</mtext><mo>=</mo><mo>â</mo><mn>68.3</mn></mrow> </math><br/>Eqs. <math> <mrow> <mn>2</mn><mo>Ã</mo><mtext>(ii)</mtext><mo>+</mo><mn>3</mn><mo>Ã</mo><mtext>(iii)</mtext><mo>â</mo><mtext>(i)</mtext></mrow> </math><br/><math><msub><mrow><mi mathvariant="normal">C</mi></mrow><mrow><mn>2</mn></mrow></msub><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>6</mn></mrow></msub><mo>+</mo><mfrac><mrow><mn>7</mn></mrow><mrow><mn>2</mn></mrow></mfrac><msub><mrow><mi mathvariant="normal">O</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>â</mo><mn>2</mn><mi mathvariant="normal">C</mi><msub><mrow><mi mathvariant="normal">O</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>+</mo><mn>3</mn><msub><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mn>2</mn></mrow></msub><mi mathvariant="normal">O</mi></math><br/><math> <mrow> <mtext>Îx</mtext><mo>=</mo><mn>2</mn><mfenced> <mrow> <mo>â</mo><mn>94.1</mn></mrow> </mfenced><mo>+</mo><mn>3</mn><mfenced> <mrow> <mo>â</mo><mn>68.3</mn></mrow> </mfenced><mo>â</mo><mfenced> <mrow> <mo>â</mo><mn>21.1</mn></mrow> </mfenced></mrow> </math><br/><math> <mrow> <mo>=</mo><mo>â</mo><mn>372</mn><mo> </mo><mtext>kcal/mol</mtext></mrow> </math></div>


tags: ['Chemistry', 'Thermodynamics (C)', 'VITEEE', 'VITEEE 2014']
source_db: MarksBatch2_P3.db
title:
the-standard-molar-heat-of-formation-of-ethane-co-2-and-water-1-are-respectively-211-941-and-683-kcal-the-standard-molar-heat-of-combustion-of-ethane-

slug:
the-standard-molar-heat-of-formation-of-ethane-co-2-and-water-1-are-respectively-211-941-and-683-kcal-the-standard-molar-heat-of-combustion-of-ethane-82454

question:
<div class="question">The standard molar heat of formation of ethane, $\mathrm{CO}_{2}$ and water (1) are respectively $-21.1,-94.1$ and $-68.3 \mathrm{kcal}$. The standard molar heat of combustion of ethane will be</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$-372 \mathrm{kcal}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$162 \mathrm{kcal}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$-240 \mathrm{kcal}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$183.5 \mathrm{kcal}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$-372 \mathrm{kcal}$</span> </div>

answer:
<div class="solution">Given,<br/>(a) $2 \mathrm{C}+3 \mathrm{H}_{2} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6} ; \Delta \mathrm{H}=-21.1 \mathrm{kcal}$<br/>(b) $\mathrm{C}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2} ; \Delta \mathrm{H}=-94.1 \mathrm{kcal}$<br/>(c) $\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} ; \Delta \mathrm{H}=-68.3 \mathrm{kcal}$<br/>Now, eqs. $2 \times(\mathrm{b})+3 \times(\mathrm{c})-(\mathrm{a})$<br/>$\mathrm{C}_{2} \mathrm{H}_{6}+\frac{3}{2} \mathrm{O}_{2} \longrightarrow 2 \mathrm{CO}_{2}+3 \mathrm{H}_{2} \mathrm{O}$<br/>Heat of combustion of Ethane $\Delta \mathrm{x}=2(-94.1)$<br/>$\begin{array}{l}<br/>+3(-68.3)-(-21.1) \\<br/>=(-188.2)+(-204.9)-21.1 \\<br/>=-372 \mathrm{kcal}<br/>\end{array}$</div>


tags: ['Chemistry', 'Electrochemistry', 'AIIMS', 'AIIMS 2008']
answer: <div class="solution">$E_{\text {cell }}^{\circ}=0.76-0.41=0.35 \mathrm{~V}$</div>
source_db: MarksBatch2_P3.db
title:
the-standard-oxidation-potential-ecirc-for-the-half-cell-reaction-are-beginarrayll-mathrmzn-rightarrow-mathrmzn22-e-ecirc076-mathrmv-mathrmfe-rightarr

slug:
the-standard-oxidation-potential-ecirc-for-the-half-cell-reaction-are-beginarrayll-mathrmzn-rightarrow-mathrmzn22-e-ecirc076-mathrmv-mathrmfe-rightarr-47085

question:
<div class="question">The standard oxidation potential \(E^{\circ}\) for the half cell reaction are<br/>\[<br/>\begin{array}{ll}<br/>\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+2 e^{-} ; &amp; E^{\circ}=+0.76 \mathrm{~V} \\<br/>\mathrm{Fe} \rightarrow \mathrm{Fe}^{2+}+2 e^{-} ; &amp; E^{\circ}=+0.41 \mathrm{~V}<br/>\end{array}<br/>\]<br/>\(E M F\) of the cell reaction is<br/>\[<br/>\mathrm{Zn}+\mathrm{Fe}^{2+} \rightarrow \mathrm{Zn}^{2+}+\mathrm{Fe}<br/>\]</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">\(-0.35 \mathrm{~V}\)</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">\(+0.35 \mathrm{~V}\)</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">\(0.17 \mathrm{~V}\)</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">\(1.17 \mathrm{~V}\).</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">\(+0.35 \mathrm{~V}\)</span> </div>


tags: ['Chemistry', 'Electrochemistry', 'JEE Main']
source_db: MarksBatch2_P3.db
title:
the-standard-oxidation-potential-of-mathrmni-mathrmni2-electrode-is-0236-mathrmv-if-this-is-combined-with-a-hydrogen-electrode-in-acid-solution-at-wha

slug:
the-standard-oxidation-potential-of-mathrmni-mathrmni2-electrode-is-0236-mathrmv-if-this-is-combined-with-a-hydrogen-electrode-in-acid-solution-at-wha-97780

question:
<div class="question">The standard oxidation potential of \( \mathrm{Ni} / \mathrm{Ni}^{2+} \) electrode is \( 0.236 \mathrm{~V} \). If this is combined with a hydrogen electrode in acid solution, at what \( \mathrm{pH} \) of the solution will the measured emf be zero at \( 25{ }^{\circ} \mathrm{C} ? \) Assume \( \left[\mathrm{Ni}^{2+}\right]=1 \mathrm{M} \) and \( \mathrm{p}_{\mathrm{H}_{2}}=1 \) atm</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">\( \mathrm{pH}=4 \)</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">\( \mathrm{pH}=2 \)</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">\( \mathrm{pH}=3 \)</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">\( \mathrm{pH}=1 \)</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">\( \mathrm{pH}=4 \)</span> </div>

answer:
<div class="solution"><math> <mrow> <msubsup> <mtext>E</mtext> <mrow> <msup> <mtext>Ni</mtext> <mrow> <mn>2</mn> <mo>+</mo> </mrow> </msup> <mo>/</mo> <mtext>Ni</mtext> <mfenced close=")" open="(" separators=","> <mrow> <mtext>s</mtext> </mrow> </mfenced> </mrow> <mo>∘</mo> </msubsup> <mo>=</mo> <mo>-</mo> <mtext>0.236 V</mtext> </mrow> </math><br/>The cell is represented as:<br/><math> <mrow> <mtext>Ni</mtext> <mfenced close=")" open="(" separators=","> <mrow> <mtext>s</mtext> </mrow> </mfenced> <mfenced close="|" open="|" separators=","> <mrow> <msubsup> <mtext>Ni</mtext> <mfenced close=")" open="(" separators=","> <mrow> <mtext>aq</mtext> </mrow> </mfenced> <mrow> <mn>2</mn> <mo>+</mo> </mrow> </msubsup> <mfenced close=")" open="(" separators=","> <mrow> <mtext>1 M</mtext> </mrow> </mfenced> </mrow> </mfenced> <mfenced close="|" open="|" separators=","> <mrow> <msubsup> <mtext>H</mtext> <mfenced close=")" open="(" separators=","> <mrow> <mtext>aq</mtext> </mrow> </mfenced> <mo>+</mo> </msubsup> <mtext>, pH</mtext> <mo>=</mo> <mtext>??</mtext> </mrow> </mfenced> <mtext>Pt</mtext> <mfenced close=")" open="(" separators=","> <mrow> <mtext>s</mtext> </mrow> </mfenced> <msub> <mo>, H</mo> <mn>2</mn> </msub> <mfenced close=")" open="(" separators=","> <mrow> <mtext>g</mtext> </mrow> </mfenced> <mo>1 atm</mo> </mrow> </math><br/><math> <mrow> <mtable> <mtr> <mtd> <mtext>Ni</mtext> <mfenced close=")" open="(" separators=","> <mrow> <mtext>s</mtext> </mrow> </mfenced> </mtd> <mtd> <mo>→</mo> </mtd> <mtd> <msup> <mtext>Ni</mtext> <mrow> <mn>2</mn> <mo>+</mo> </mrow> </msup> <mo>+</mo> <mn>2</mn> <msup> <mtext>e</mtext> <mo>-</mo> </msup> </mtd> </mtr> <mtr> <mtd> <mn>2</mn> <mrow> <msup> <mtext>H</mtext> <mo>+</mo> </msup> <mo>+</mo> <mn>2</mn> <msup> <mtext>e</mtext> <mo>-</mo> </msup> </mrow> </mtd> <mtd> <mo>→</mo> </mtd> <mtd> <msub> <mtext>H</mtext> <mn>2</mn> </msub> <mfenced close=")" open="(" separators=","> <mrow> <mtext>g</mtext> </mrow> </mfenced> <mo>1 atm</mo> </mtd> </mtr> </mtable> </mrow> </math><br/>--------------------------------------------------------------------<br/><math> <mrow> <mtable> <mtr> <mtd> <mtext>Ni</mtext> <mfenced close=")" open="(" separators=","> <mrow> <mtext>s</mtext> </mrow> </mfenced> <mo>+</mo> <mn>2</mn> <msubsup> <mtext>H</mtext> <mfenced close=")" open="(" separators=","> <mrow> <mtext>aq</mtext> </mrow> </mfenced> <mo>+</mo> </msubsup> </mtd> <mtd> <mover> <mo>→</mo> <mrow> <mtext>n</mtext> <mo>=</mo> <mn>2</mn> </mrow> </mover> </mtd> <mtd> <msubsup> <mtext>Ni</mtext> <mfenced close=")" open="(" separators=","> <mrow> <mtext>aq</mtext> </mrow> </mfenced> <mrow> <mn>2</mn> <mo>+</mo> </mrow> </msubsup> <mo>+</mo> <msub> <mtext>H</mtext> <mn>2</mn> </msub> <mfenced close=")" open="(" separators=","> <mrow> <mtext>g</mtext> </mrow> </mfenced> <mo>(1 atm)</mo> </mtd> </mtr> </mtable> </mrow> </math><br/><math> <mrow> <msub> <mtext>E</mtext> <mtext>cell</mtext> </msub> <mo>=</mo> <mn>0</mn> <mo>=</mo> <msubsup> <mtext>E</mtext> <mtext>cell</mtext> <mo>∘</mo> </msubsup> <mo>-</mo> <mfrac> <mrow> <mtext>0.059</mtext> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mtext>log</mtext> <mfrac> <mrow> <mfenced close="]" open="[" separators=","> <mrow> <msup> <mtext>Ni</mtext> <mrow> <mn>2</mn> <mo>+</mo> </mrow> </msup> </mrow> </mfenced> </mrow> <mrow> <msup> <mrow> <mfenced close="]" open="[" separators=","> <mrow> <msup> <mtext>H</mtext> <mo>+</mo> </msup> </mrow> </mfenced> </mrow> <mn>2</mn> </msup> </mrow> </mfrac> </mrow> </math><br/><math> <mrow> <mn>0</mn> <mo>=</mo> <mfenced close="]" open="[" separators=","> <mrow> <mn>0</mn> <mo>-</mo> <mfenced close=")" open="(" separators=","> <mrow> <mo>-</mo> <mtext>0.236</mtext> </mrow> </mfenced> </mrow> </mfenced> <mo>-</mo> <mfrac> <mrow> <mtext>0.059</mtext> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mtext>log</mtext> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <msup> <mrow> <mfenced close="]" open="[" separators=","> <mrow> <msup> <mtext>H</mtext> <mo>+</mo> </msup> </mrow> </mfenced> </mrow> <mn>2</mn> </msup> </mrow> </mfrac> </mrow> </math><br/><math> <mrow> <mfrac> <mrow> <mtext>0.236</mtext> <mo>×</mo> <mn>2</mn> </mrow> <mrow> <mtext>0.059</mtext> </mrow> </mfrac> <mo>=</mo> <mtext>log</mtext> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <msup> <mrow> <mfenced close="]" open="[" separators=","> <mrow> <msup> <mtext>H</mtext> <mo>+</mo> </msup> </mrow> </mfenced> </mrow> <mn>2</mn> </msup> </mrow> </mfrac> </mrow> </math><br/><math> <mrow> <mo>=</mo> <mfrac> <mrow> <mtext>0.472</mtext> </mrow> <mrow> <mtext>0.059</mtext> </mrow> </mfrac> <mo>=</mo> <mn>8</mn> <mo>=</mo> <mtext>log</mtext> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <msup> <mrow> <mfenced close="]" open="[" separators=","> <mrow> <msup> <mtext>H</mtext> <mo>+</mo> </msup> </mrow> </mfenced> </mrow> <mn>2</mn> </msup> </mrow> </mfrac> </mrow> </math><br/><math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <msup> <mrow> <mfenced close="]" open="[" separators=","> <mrow> <msup> <mtext>H</mtext> <mo>+</mo> </msup> </mrow> </mfenced> </mrow> <mn>2</mn> </msup> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> <msup> <mn>0</mn> <mn>8</mn> </msup> <mo>⇒</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mfenced close="]" open="[" separators=","> <mrow> <msup> <mtext>H</mtext> <mo>+</mo> </msup> </mrow> </mfenced> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> <msup> <mn>0</mn> <mrow> <mo>+</mo> <mn>4</mn> </mrow> </msup> </mrow> </math><br/><math> <mrow> <mfenced close="]" open="[" separators=","> <mrow> <msup> <mtext>H</mtext> <mo>+</mo> </msup> </mrow> </mfenced> <mo>=</mo> <mn>1</mn> <msup> <mn>0</mn> <mrow> <mo>-</mo> <mn>4</mn> </mrow> </msup> <mo>  ∴  </mo> <mtext>pH</mtext> <mo>=</mo> <mn>4</mn> </mrow> </math></div>


tags: ['Chemistry', 'Electrochemistry', 'JEE Main']
source_db: MarksBatch2_P3.db
title:
the-standard-oxidation-potential-of-zinc-and-silver-in-water-at-298-k-are-zn-s-zn-2-2-e-e-076-v-ag-s-ag-2-2-e-e-080-v-which-of-the-following-reactions

slug:
the-standard-oxidation-potential-of-zinc-and-silver-in-water-at-298-k-are-zn-s-zn-2-2-e-e-076-v-ag-s-ag-2-2-e-e-080-v-which-of-the-following-reactions-84400

question:
<div class="question">The standard oxidation potential of zinc and silver in water at $298 \mathrm{~K}$ are<br/>$\mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} ; E=0.76 \mathrm{~V}$<br/>$\mathrm{Ag}(\mathrm{s}) \rightarrow \mathrm{Ag}^{2+}+2 \mathrm{e}^{-} ; E=-0.80 \mathrm{~V}$<br/>Which of the following reactions actually take place</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$\mathrm{Zn}(\mathrm{s})+2 \mathrm{Ag}^{+}(a q) \rightarrow Z \mathrm{n}^{++}(a q)+2 \mathrm{Ag}(\mathrm{s})$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$Z n^{++}(a q)+2 A g(s) \rightarrow 2 A g^{+}(a q)+Z n(s)$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\mathrm{Zn}(\mathrm{s})+\mathrm{Ag}(\mathrm{s}) \rightarrow \mathrm{Zn}^{++}(a q)+\mathrm{Ag}^{+}(a q)$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$Z n^{++}(a q)+A g^{+}(a q) \rightarrow Z n(s)+A g(s)$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\mathrm{Zn}(\mathrm{s})+2 \mathrm{Ag}^{+}(a q) \rightarrow Z \mathrm{n}^{++}(a q)+2 \mathrm{Ag}(\mathrm{s})$</span> </div>

answer:
<div class="solution"><img src="https://cdn-question-pool.getmarks.app/modules/qc/Mhsq9PYaXwfIeyxLcyvfTm4fCkY7uzPQOmj7IxlgNlo.original.fullsize.png"/><br/><br/>In this reaction zinc act as a anode and Ag  act as a cathode.</div>


title: the-standard-petal-of-the-papilionaceous-corolla-is-also-called
slug: the-standard-petal-of-the-papilionaceous-corolla-is-also-called-89164
question: <div class="question">The standard petal of the papilionaceous corolla is also called:</div>
tags: ['Biology', 'Morphology of Flowering Plants', 'NEET', 'NEET 2016 (Phase 1)']
source_db: MarksBatch2_P3.db
options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">Carina</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">Pappus</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">Vexillum</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">Corona</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">Vexillum</span> </div>

answer:
<div class="solution"><p><img alt="" src="https://cdn-question-pool.getmarks.app/pyq/neet/Biology_q24_ans.png" style="width: 173px; height: 97px;"/></p><p>Papilionaceous corolla.</p><br/>A single, large, upper petal is known as the banner also vexillum or standard petal of a papilionaceous flower.</div>


tags: ['Chemistry', 'Electrochemistry', 'JEE Main']
source_db: MarksBatch2_P3.db
title:
the-standard-potential-of-the-cell-formed-by-combining-the-cl-2-cl-aq-halfcell-with-the-standard-hydrogen-electrode-is-136v-and-e-o-t-p-12-1-0-3-vk-1-

slug:
the-standard-potential-of-the-cell-formed-by-combining-the-cl-2-cl-aq-halfcell-with-the-standard-hydrogen-electrode-is-136v-and-e-o-t-p-12-1-0-3-vk-1-12781

question:
<div class="question">The standard potential of the cell formed by combining the Cl<sub>2</sub>/Cl<sup>â</sup> (aq) half-cell with the standard hydrogen electrode is +1.36V and <math> <mrow> <msub> <mrow> <mfenced close=")" open="(" separators=","> <mrow> <mfrac> <mrow> <mo>â</mo> <msup> <mtext>E</mtext> <mtext>o</mtext> </msup> </mrow> <mrow> <mo>â</mo> <mtext>T</mtext> </mrow> </mfrac> </mrow> </mfenced> </mrow> <mtext>P</mtext> </msub> <mo>=</mo> <mo>-</mo> <mtext>1.2</mtext> <mo>Ã</mo> <mn>1</mn> <msup> <mn>0</mn> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </msup> <msup> <mo>VK</mo> <mn>-1</mn> </msup> </mrow> </math>. What is the value of <math> <mrow> <mtext>Î</mtext> <msubsup> <mtext>S</mtext> <mtext>reaction</mtext> <mtext>o</mtext> </msubsup> </mrow> </math> for reaction <math> <mrow> <msub> <mtext>H</mtext> <mrow> <mn>2</mn> <mfenced close=")" open="(" separators=","> <mrow> <mtext>g</mtext> </mrow> </mfenced> </mrow> </msub> <mo>+</mo> <msub> <mtext>Cl</mtext> <mrow> <mn>2</mn> <mfenced close=")" open="(" separators=","> <mrow> <mtext>g</mtext> </mrow> </mfenced> </mrow> </msub> <mo>â</mo> <mn>2</mn> <msubsup> <mtext>H</mtext> <mfenced close=")" open="(" separators=","> <mrow> <mtext>aq</mtext> </mrow> </mfenced> <mo>+</mo> </msubsup> <mo>+</mo> <mn>2</mn> <msubsup> <mtext>Cl</mtext> <mfenced close=")" open="(" separators=","> <mrow> <mtext>aq</mtext> </mrow> </mfenced> <mo>-</mo> </msubsup> </mrow> </math></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><math><mrow> <mo>-</mo> <mn>1</mn> <mo>×</mo> <mn>1</mn> <msup> <mn>0</mn> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </msup> <msup> <mo>JK</mo> <mn>-1</mn> </msup></mrow></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math><mrow> <mn>2.3</mn> <mo>×</mo> <mn>1</mn> <msup> <mn>0</mn> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </msup> <msup> <mo>JK</mo> <mn>-1</mn> </msup></mrow></math></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><math> <mrow> <mo>-</mo> <mn>2.3</mn> <mo>×</mo> <mn>1</mn> <msup> <mn>0</mn> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </msup> <msup> <mo>JK</mo> <mn>-1</mn> </msup> </mrow> </math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math><mrow> <mn>1</mn> <msup> <mo>JK</mo> <mn>-1</mn> </msup></mrow></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math> <mrow> <mo>-</mo> <mn>2.3</mn> <mo>×</mo> <mn>1</mn> <msup> <mn>0</mn> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </msup> <msup> <mo>JK</mo> <mn>-1</mn> </msup> </mrow> </math></span> </div>

answer:
<div class="solution"><math> <mrow> <mtext>Δ</mtext> <msubsup> <mtext>S</mtext> <mtext>reaction</mtext> <mtext>o</mtext> </msubsup> <mo>=</mo> <mo>nF</mo> <msub> <mrow> <mfenced close=")" open="(" separators=","> <mrow> <mfrac> <mrow> <mo>∂</mo> <msup> <mtext>E</mtext> <mtext>o</mtext> </msup> </mrow> <mrow> <mo>∂</mo> <mtext>T</mtext> </mrow> </mfrac> </mrow> </mfenced> </mrow> <mtext>P</mtext> </msub> </mrow> </math><br/><br/><math> <mrow> <mo>=</mo> <mn>2</mn> <mo>×</mo> <mn>9</mn> <mn>6</mn> <mn>4</mn> <mn>8</mn> <mn>0</mn> <msup> <mo>C mol</mo> <mn>-1</mn> </msup> <mo>×</mo> <mfenced close=")" open="(" separators=","> <mrow> <mo>-</mo> <mtext>1.2</mtext> <mo>×</mo> <mn>1</mn> <msup> <mn>0</mn> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </msup> <msup> <mo>VK</mo> <mn>-1</mn> </msup> </mrow> </mfenced> </mrow> </math><br/><br/><math> <mrow> <mo>=</mo> <mo>-</mo> <mn>2.3</mn> <mo>×</mo> <mn>1</mn> <msup> <mn>0</mn> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </msup> <msup> <mo>JK</mo> <mn>-1</mn> </msup> </mrow></math></div>


tags: ['Chemistry', 'Electrochemistry', 'JEE Main', 'JEE Main 2012 (19 May Online)']
source_db: MarksBatch2_P3.db
title:
the-standard-potentials-of-ag-ag-hg-2-2-2-hg-cu-2-cu-and-mg-2-mg-electrodes-are-080-079-034-and-237-v-respectively-an-aqueous-solution-which-contains-

slug:
the-standard-potentials-of-ag-ag-hg-2-2-2-hg-cu-2-cu-and-mg-2-mg-electrodes-are-080-079-034-and-237-v-respectively-an-aqueous-solution-which-contains-95355

question:
<div class="question">The standard potentials of $\mathrm{Ag}^{+} / \mathrm{Ag}, \mathrm{Hg}_2{ }^{2+} / 2 \mathrm{Hg}$, $\mathrm{Cu}^{2+} / \mathrm{Cu}$ and $\mathrm{Mg}^{2+} / \mathrm{Mg}$ electrodes are $0.80,0.79$, $0.34$ and $-2.37 \mathrm{~V}$, respectively. An aqueous solution which contains one mole per litre of the salts of each of the four metals is electrolyzed. With increasing voltage, the correct sequence of deposition of the metals at the cathode is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\mathrm{Ag}, \mathrm{Hg}, \mathrm{Cu}, \mathrm{Mg}$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\mathrm{Cu}, \mathrm{Hg}$, Ag only<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>Ag, Hg, Cu only<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\mathrm{Mg}, \mathrm{Cu}, \mathrm{Hg}, \mathrm{Ag}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><br/>Ag, Hg, Cu only<br/></span> </div>

answer:
<div class="solution">More the reduction potential, more is the deposition of metals at cathode. Cation having $\mathrm{E}^0$ value less than $-0.83 \mathrm{~V}$ (reduction potential of $\mathrm{H}_2 \mathrm{O}$ ) will not deposit from aqueous solution. Hence correct order of A deposition of the metal at the cathode is</div>


tags: ['Chemistry', 'Thermodynamics (C)', 'JEE Main', 'JEE Main 2019 (11 Jan Shift 2)']
source_db: MarksBatch2_P3.db
title:
the-standard-reaction-gibbs-energy-for-a-chemical-reaction-at-an-absolute-temperature-t-is-given-by-g-a-bt-where-a-and-b-are-nonzero-constants-which-o

slug:
the-standard-reaction-gibbs-energy-for-a-chemical-reaction-at-an-absolute-temperature-t-is-given-by-g-a-bt-where-a-and-b-are-nonzero-constants-which-o-87042

question:
<div class="question">The standard reaction Gibbs energy for a chemical reaction at an absolute temperature $\mathrm{T}$ is given by $\Delta \mathrm{G}^{\circ}=\mathrm{A}-\mathrm{BT}$ where A and $B$ are non-zero constants. Which of the following is true about this reaction?</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">Exothermic if $B  &lt; 0$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">Endothermic if $\mathrm{A}&gt;0$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">Endothermic if $\mathrm{A}  &lt; 0$ and $\mathrm{B}&gt;0$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">Exothermic if $\mathrm{A}&gt;0$ and $\mathrm{B}  &lt; 0$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">Endothermic if $\mathrm{A}&gt;0$</span> </div>

answer:
<div class="solution">$\Delta G^{\circ}=A-B T$
<br/> <br/>A and B are non-zero constants
<br/> <br/>$\therefore \Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}=A-B T$
<br/> <br/>$\therefore$ Reaction will be endothermic if $\Delta \mathrm{G}^{\circ}&gt;0 .$
<br/> <br/>Hence, $\mathrm{A}&gt;0$ and $\mathrm{B}  &lt; 0$</div>


tags: ['Chemistry', 'Biomolecules', 'KCET', 'KCET 2017']
source_db: MarksBatch2_P3.db
title:
the-standard-reduction-potential-at-298-k-for-the-following-half-cell-reaction-z-n-2-a-q-2-e-z-n-s-e-0762-v-c-r-3-a-q-3-e-c-r-s-e-0740-v-2-h-a-q-2-e-h

slug:
the-standard-reduction-potential-at-298-k-for-the-following-half-cell-reaction-z-n-2-a-q-2-e-z-n-s-e-0762-v-c-r-3-a-q-3-e-c-r-s-e-0740-v-2-h-a-q-2-e-h-12051

question:
<div class="question">The standard reduction potential at $298 \mathrm{~K}$ for the following half cell reaction<br/>$Z n^{2+}(a q)+2 e^{-} \rightarrow Z n(s) ; \quad E^{\circ}=-0.762 V$<br/>$C r^{3+}(a q)+3 e^{-} \rightarrow C r(s) ; \quad E^{\circ}=0.740 V$<br/>$2 H^{+}(a q)+2 e^{-} \rightarrow H_{2}(g) ; \quad E^{\circ}=0.0 V$<br/>$F_{2}(g)+2 e^{-} \rightarrow 2 F^{-}(a q) ; \quad E^{\circ}=2.87 V$<br/>Which of the following is strongest reducing agent ?</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">\( Z n(s) \)</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">\( C r(s) \)</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">\( H_{2}(g) \)</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">\( F_{2}(g) \)</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">\( Z n(s) \)</span> </div>

answer:
<div class="solution">The more negative the \( E^{\circ} \) value. The more readily the metal loses electrons. The more negative the value the stronger<br/>is the reducing agent. Therefore, strongest reducing agent is \( \mathrm{Zn} \)</div>


tags: ['Chemistry', 'Electrochemistry', 'NEET']
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">1.10 V</span> </div>
source_db: MarksBatch2_P3.db
title:
the-standard-reduction-potential-e-for-the-half-reactions-are-as-zn-2-2-e-zn-e-076-v-cu-2-2-e-cu-e-034-v-the-standard-cell-voltage-for-the-cell-reacti

slug:
the-standard-reduction-potential-e-for-the-half-reactions-are-as-zn-2-2-e-zn-e-076-v-cu-2-2-e-cu-e-034-v-the-standard-cell-voltage-for-the-cell-reacti-63900

question:
<div class="question">The standard reduction potential <math><msup><mrow><mtext>E</mtext></mrow><mrow><mo>Â°</mo></mrow></msup></math> for the half reactions are as:<br/> <br/> <math> <mrow> <msup> <mrow> <mtext>Zn</mtext></mrow> <mrow> <mn>2</mn><mo>+</mo></mrow> </msup> <mtext> </mtext><mo>+</mo><mtext> </mtext><mn>2</mn><msup> <mtext>e</mtext> <mo>â</mo> </msup> <mtext> </mtext><mo>â</mo><mtext>Zn</mtext><mo>,</mo><msup> <mrow> <mtext>      E</mtext></mrow> <mo>Â°</mo> </msup> <mo>=</mo><mo>â</mo><mn>0.76</mn><mtext> V</mtext></mrow> </math><br/> <br/> <math> <mrow> <msup> <mrow> <mtext>Cu</mtext></mrow> <mrow> <mn>2</mn><mo>+</mo></mrow> </msup> <mtext> </mtext><mo>+</mo><mtext> </mtext><mn>2</mn><msup> <mtext>e</mtext> <mo>â</mo> </msup> <mtext> </mtext><mo>â</mo><mtext>Cu</mtext><mo>,</mo><msup> <mrow> <mtext>      E</mtext></mrow> <mo>Â°</mo> </msup> <mtext> </mtext><mo>=</mo><mtext> </mtext><mn>0.34</mn><mtext> V</mtext></mrow> </math><br/> <br/>
The standard cell voltage for the cell reaction is?<br/> <br/> <math><mtext mathvariant="normal">Z</mtext><mtext mathvariant="normal">n</mtext><mtext mathvariant="normal"> </mtext><mo>+</mo><mtext mathvariant="normal"> </mtext><msup><mrow><mtext mathvariant="normal">C</mtext><mtext mathvariant="normal">u</mtext></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mtext mathvariant="normal"> </mtext><mo>â</mo><mtext mathvariant="normal"> </mtext><msup><mrow><mtext mathvariant="normal">Z</mtext><mtext mathvariant="normal">n</mtext></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mo>+</mo><mtext mathvariant="normal"> </mtext><mtext mathvariant="normal">C</mtext><mtext mathvariant="normal">u</mtext></math></div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">0.42 V</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">-0.42 V</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">-1.1 V</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">1.10 V</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

answer:
<div class="solution"><math> <msubsup> <mtext>E</mtext> <mrow> <mtext>cell</mtext></mrow> <mtext>o</mtext> </msubsup> <msubsup> <mtext>= E</mtext> <mrow> <mtext>Cathode</mtext></mrow> <mtext>o</mtext> </msubsup> <msubsup> <mtext>- E</mtext> <mrow> <mtext>Anode</mtext></mrow> <mtext>o</mtext> </msubsup> </math><br/><br/><math> <msubsup> <mtext>E</mtext> <mrow> <mtext>cell</mtext></mrow> <mtext>o</mtext> </msubsup> <msubsup> <mtext>= E</mtext> <mrow> <msup> <mrow> <mtext>Cu</mtext></mrow> <mrow> <mtext>+2</mtext></mrow> </msup> <mtext>/Cu</mtext></mrow> <mtext>o</mtext> </msubsup> <msubsup> <mtext>- E</mtext> <mrow> <msup> <mrow> <mtext>Zn</mtext></mrow> <mrow> <mtext>+2</mtext></mrow> </msup> <mtext>/Zn</mtext></mrow> <mtext>o</mtext> </msubsup> <mtext> </mtext> </math><br/><br/><math><mtable columnalign="left"><mtr><mtd><mo>=</mo><mtext>â</mtext><mn>0.34</mn><mo>â</mo><mo stretchy="false">(</mo><mo>â</mo><mn>0.76</mn><mo stretchy="false">)</mo> </mtd> </mtr> <mtr> <mtd> <mtext>â</mtext><mtext>â</mtext><mtext>â</mtext><mtext>â</mtext><mtext>â</mtext><mtext>â</mtext><mtext>â</mtext><mtext>â</mtext><mtext>â</mtext><mo>=</mo><mtext>â</mtext><mn>1.10 </mn><mtext>V</mtext> </mtd> </mtr> </mtable> </math></div>


tags: ['Chemistry', 'Electrochemistry', 'NEET']
source_db: MarksBatch2_P3.db
title:
the-standard-reduction-potential-for-c-u-2-c-u-is-034-v-the-reduction-potential-at-ph-14-for-the-above-couple-k-s-p-c-u-o-h-2-1-1-0-19-is

slug:
the-standard-reduction-potential-for-c-u-2-c-u-is-034-v-the-reduction-potential-at-ph-14-for-the-above-couple-k-s-p-c-u-o-h-2-1-1-0-19-is-90748

question:
<div class="question">The standard reduction potential for <math><mi mathvariant="normal">C</mi><msup><mrow><mi mathvariant="normal">u</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mo>/</mo><mi mathvariant="normal">C</mi><mi mathvariant="normal">u</mi></math> is 0.34 V. The reduction potential at pH = 14 for the above couple <math><mo>(</mo><msub><mrow><mi mathvariant="normal">K</mi></mrow><mrow><mi mathvariant="normal">s</mi><mi mathvariant="normal">p</mi></mrow></msub><mi mathvariant="normal">C</mi><mi mathvariant="normal">u</mi><mo>(</mo><mi mathvariant="normal">O</mi><mi mathvariant="normal">H</mi><msub><mrow><mo>)</mo></mrow><mrow><mn>2</mn></mrow></msub><mo>=</mo><mn>1</mn><mo>Ã</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mn>19</mn><mo>-</mo></mrow></msup><mo>)</mo></math> is</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">-0.22 V</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">-0.34 V</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">+0.22 V</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">+0.34 V</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">-0.22 V</span> </div>

answer:
<div class="solution"><math><mi mathvariant="normal">p</mi><mi mathvariant="normal">H</mi><mo>=</mo><mn>14</mn></math><br/><math><mi mathvariant="normal">p</mi><mi mathvariant="normal">O</mi><mi mathvariant="normal">H</mi><mo>=</mo><mn>0</mn></math><br/><math> <mrow> <mfenced close="]" open="["> <mrow> <msup> <mrow> <mtext>OH</mtext></mrow> <mo>−</mo> </msup> </mrow> </mfenced><mo>=</mo><msup> <mrow> <mn>10</mn></mrow> <mn>0</mn> </msup> <mo>=</mo><mn>1</mn></mrow></math><br/><math><mi mathvariant="normal">C</mi><msup><mrow><mi mathvariant="normal">u</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup><mo>=</mo><mfrac><mrow><msub><mrow><mi mathvariant="normal">K</mi></mrow><mrow><mi mathvariant="normal">s</mi><mi mathvariant="normal">p</mi></mrow></msub></mrow><mrow><mo>[</mo><mi mathvariant="normal">O</mi><msup><mrow><mi mathvariant="normal">H</mi></mrow><mrow><mo>-</mo></mrow></msup><msup><mrow><mo>]</mo></mrow><mrow><mn>2</mn></mrow></msup></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>1</mn><mo>Ã</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mo>-</mo><mn>19</mn></mrow></msup></mrow><mrow><mn>1</mn></mrow></mfrac><mo>=</mo><mn>1</mn><mo>Ã</mo><mn>1</mn><msup><mrow><mn>0</mn></mrow><mrow><mo>-</mo><mn>19</mn></mrow></msup></math><br/><math> <mrow> <msub> <mtext>E</mtext> <mrow> <msup> <mrow> <mtext>Cu</mtext></mrow> <mrow> <mtext>2+</mtext></mrow> </msup> <mtext>/Cu</mtext></mrow> </msub> <mtext>â</mtext><mtext>=</mtext><mtext>â</mtext><msup> <mtext>E</mtext> <mtext>o</mtext> </msup> <msub> <mrow></mrow> <mrow> <msup> <mrow> <mtext>Cu</mtext></mrow> <mrow> <mtext>2+</mtext></mrow> </msup> <mtext>/Cu</mtext></mrow> </msub> <mo>â</mo><mfrac> <mrow> <mtext>0</mtext><mtext>.059</mtext></mrow> <mtext>2</mtext> </mfrac> <mtext>â</mtext><mtext>log</mtext><mtext>â</mtext><mfrac> <mtext>1</mtext> <mrow> <msup> <mrow> <mtext>1Ã10</mtext></mrow> <mrow> <mo>â</mo><mtext>19</mtext></mrow> </msup> </mrow> </mfrac> </mrow> </math><br/><math> <mrow> <mo>=</mo><mn>0.34</mn><mtext>â</mtext><mtext>V</mtext><mo>â</mo><mfrac> <mrow> <mn>0.059</mn></mrow> <mn>2</mn> </mfrac> <mo>Ã</mo><mn>19</mn><mo>=</mo><mo>â</mo><mn>0.22</mn><mtext>â</mtext><mtext>V</mtext></mrow> </math></div>


tags: ['Chemistry', 'Electrochemistry', 'VITEEE', 'VITEEE 2017']
source_db: MarksBatch2_P3.db
title:
the-standard-reduction-potential-for-cu-2-cu-is-034-calculate-the-reduction-potential-at-ph-14-for-the-above-couple-k-sp-cu-oh-2-1-1-0-19

slug:
the-standard-reduction-potential-for-cu-2-cu-is-034-calculate-the-reduction-potential-at-ph-14-for-the-above-couple-k-sp-cu-oh-2-1-1-0-19-69416

question:
<div class="question">The standard reduction potential for $\mathrm{Cu}^{2+} / \mathrm{Cu}$ is $+0.34$. Calculate the reduction potential at $\mathrm{pH}$
<br/>$=14$ for the above couple. $\left(\mathrm{K}_{\text {sp }} \mathrm{Cu}(\mathrm{OH})_{2}\right.$ $\left.=1 \times 10^{-19}\right)$</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$-0.22 \mathrm{~V}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$+0.22 \mathrm{~V}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$-0.44 \mathrm{~V}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$+0.44 \mathrm{~V}$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$-0.22 \mathrm{~V}$</span> </div>

answer:
<div class="solution">When $\mathrm{pH}=14\left[\mathrm{H}^{+}\right]=10^{-14}$ and $\left[\mathrm{OH}^{-}\right]=1 \mathrm{M}$
<br/>$\begin{array}{l}
<br/>\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}=10^{-19} \\
<br/>\therefore \quad\left[\mathrm{Cu}^{2+}\right]=\frac{10^{-19}}{\left[\mathrm{OH}^{-}\right]^{2}}=10^{-19}
<br/>\end{array}$
<br/>The half cell reaction
<br/>$\begin{array}{l}
<br/>\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} \\
<br/>\mathrm{E}=\mathrm{E}^{\circ}-\frac{0.059}{2} \log \frac{1}{\left[\mathrm{Cu}^{2+}\right]} \\
<br/>=0.34-\frac{0.059}{2} \log \frac{1}{10^{-19}}=-0.22 \mathrm{~V}
<br/>\end{array}$</div>


tags: ['Chemistry', 'Electrochemistry', 'JEE Main']
source_db: MarksBatch2_P3.db
title:
the-standard-reduction-potential-for-cu-2-cu-is-034-v-calculate-the-reduction-potential-at-ph-14-for-the-above-couple-k-sp-of-cu-oh-2-1-10-19

slug:
the-standard-reduction-potential-for-cu-2-cu-is-034-v-calculate-the-reduction-potential-at-ph-14-for-the-above-couple-k-sp-of-cu-oh-2-1-10-19-97340

question:
<div class="question">The standard reduction potential for <math> <mrow> <msup> <mrow> <mtext>Cu</mtext></mrow> <mrow> <mn>2</mn><mo>+</mo></mrow> </msup> <mtext>/Cu</mtext></mrow> </math> is <math><mo>+</mo><mi> </mi><mn>0.34</mn><mi> </mi><mtext>V</mtext><mo>.</mo></math> Calculate the reduction potential at <math> <mrow> <mtext>pH</mtext><mo>=</mo><mn>14</mn></mrow> </math> for the above couple. <math> <mrow> <mfenced> <mrow> <msub> <mtext>K</mtext> <mrow> <mtext>sp</mtext></mrow> </msub> <mtext> of</mtext> <mtext> Cu</mtext><msub> <mrow> <mfenced> <mrow> <mtext>OH</mtext></mrow> </mfenced></mrow> <mn>2</mn> </msub> <mo>=</mo><mn>1</mn><mo>Ã</mo><msup> <mrow> <mn>10</mn></mrow> <mrow> <mo>â</mo><mn>19</mn></mrow> </msup> </mrow> </mfenced></mrow> </math></div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><math><mo>-</mo><mn>0.22</mn><mi> </mi><mtext>V</mtext></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><math><mo>+</mo><mn>0.22</mn><mi>Â </mi><mtext>V</mtext></math></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><math><mo>-</mo><mn>0.44</mn><mi>Â </mi><mtext>V</mtext></math></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><math><mo>+</mo><mn>0.44</mn><mi>Â </mi><mtext>V</mtext></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value"><math><mo>-</mo><mn>0.22</mn><mi> </mi><mtext>V</mtext></math></span> </div>

answer:
<div class="solution">When <math> <mrow> <mtext>â</mtext><mtext>p</mtext><mtext>H</mtext><mo>=</mo><mn>14, </mn><mrow><mo>[</mo> <mrow><msup><mtext>H</mtext> <mo>+</mo> </msup> </mrow> <mo>]</mo></mrow><mo>=</mo><msup> <mrow> <mn>10</mn></mrow> <mrow> <mo>â</mo><mn>14</mn></mrow> </msup> <mtext>â</mtext><mtext>and</mtext><mtext>â</mtext><mrow><mo>[</mo> <mrow> <mtext>O</mtext><msup> <mtext>H</mtext> <mo>â</mo> </msup> </mrow> <mo>]</mo></mrow><mo>=</mo><mn>1</mn><mi> M</mi></mrow> </math><br/><br/><math> <mrow> <msub> <mtext>K</mtext> <mrow> <mtext>s</mtext><mtext>p</mtext></mrow> </msub> <mo>=</mo><mrow><mo>[</mo> <mrow> <mtext>C</mtext><msup> <mtext>u</mtext> <mrow> <mn>2</mn><mo>+</mo></mrow> </msup> </mrow> <mo>]</mo></mrow><msup> <mrow> <mrow><mo>[</mo> <mrow> <mtext>O</mtext><msup> <mtext>H</mtext> <mo>â</mo> </msup> </mrow> <mo>]</mo></mrow></mrow> <mn>2</mn> </msup> <mo>=</mo><msup> <mrow> <mn>10</mn></mrow> <mrow> <mo>â</mo><mn>19</mn></mrow> </msup> </mrow> </math><br/><br/><math> <mrow> <mo>â´</mo><mrow><mo>[</mo> <mrow> <mtext>C</mtext><msup> <mtext>u</mtext> <mrow> <mn>2</mn><mo>+</mo></mrow> </msup> </mrow> <mo>]</mo></mrow><mo>=</mo><mfrac> <mrow> <msup> <mrow> <mn>10</mn></mrow> <mrow> <mo>â</mo><mn>19</mn></mrow> </msup> </mrow> <mrow> <msup> <mrow> <mrow><mo>[</mo> <mrow> <mtext>O</mtext><msup> <mtext>H</mtext> <mo>â</mo> </msup> </mrow> <mo>]</mo></mrow></mrow> <mn>2</mn> </msup> </mrow> </mfrac> <mo>=</mo><msup> <mrow> <mn>10</mn></mrow> <mrow> <mo>â</mo><mn>19</mn></mrow> </msup> </mrow> </math><br/><br/>The half cell reaction<br/><br/><math> <mrow> <mtext>C</mtext><msup> <mtext>u</mtext> <mrow> <mn>2</mn><mo>+</mo></mrow> </msup> <mo>+</mo><mn>2</mn><msup> <mtext>e</mtext> <mo>â</mo> </msup> <mo>â</mo><mtext>C</mtext><mtext>u</mtext></mrow> </math><br/><br/><math> <mrow> <mtext>E</mtext><mo>=</mo><msup> <mtext>E</mtext> <mtext>O</mtext> </msup> <mo>â</mo><mfrac> <mrow> <mn>0.059</mn></mrow> <mn>2</mn> </mfrac> <mi>log</mi><mfrac> <mn>1</mn> <mrow> <mrow><mo>[</mo> <mrow> <mtext>C</mtext><msup> <mtext>u</mtext> <mrow> <mn>2</mn><mo>+</mo></mrow> </msup> </mrow> <mo>]</mo></mrow></mrow> </mfrac> </mrow> </math><br/><br/><math> <mrow> <mo>=</mo><mn>0.34</mn><mo>â</mo><mfrac> <mrow> <mn>0.059</mn></mrow> <mn>2</mn> </mfrac> <mi>log</mi><mfrac> <mn>1</mn> <mrow> <msup> <mrow> <mn>10</mn></mrow> <mrow> <mo>â</mo><mn>19</mn></mrow> </msup> </mrow> </mfrac> <mo>=</mo><mo>â</mo><mn>0.22 </mn><mtext>V</mtext></mrow> </math><br/> </div>


tags: ['Chemistry', 'Electrochemistry', 'MHT CET', 'MHT CET 2011']
source_db: MarksBatch2_P3.db
title:
the-standard-reduction-potential-for-fe-2-fe-and-sn-2-sn-electrodes-are-044-and-014-v-respectively-for-the-cell-reaction-fe-2-sn-fe-sn-2-the-standard-

slug:
the-standard-reduction-potential-for-fe-2-fe-and-sn-2-sn-electrodes-are-044-and-014-v-respectively-for-the-cell-reaction-fe-2-sn-fe-sn-2-the-standard-89282

question:
<div class="question">The standard reduction potential for $\mathrm{Fe}^{2+} / \mathrm{Fe}$ and $\mathrm{Sn}^{2+} / \mathrm{Sn}$ electrodes are $-0.44$ and $-0.14 \mathrm{~V}$ respectively. For the cell reaction, $\mathrm{Fe}^{2+}+\mathrm{Sn} \longrightarrow \mathrm{Fe}+\mathrm{Sn}^{2+}$ the standard emf is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$+0.30 \mathrm{~V}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$-0.58 \mathrm{~V}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$+0.58 \mathrm{~V}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$-0.30 \mathrm{~V}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$-0.30 \mathrm{~V}$</span> </div>

answer:
<div class="solution">$\mathrm{Fe}^{2+}+\mathrm{Sn} \longrightarrow \mathrm{Fe}+\mathrm{Sn}^{2+}$<br/>$$<br/>=-0.44+(0.14)<br/>$$<br/>$=-0.30 \mathrm{~V}$</div>


tags: ['Chemistry', 'Electrochemistry', 'MHT CET', 'MHT CET 2009']
source_db: MarksBatch2_P3.db
title:
the-standard-reduction-potential-for-mg-2-mg-is-237-v-and-for-cu-2-cu-is-0337-the-e-cell-for-the-following-reaction-is-mg-cu-2-mg-2-cu

slug:
the-standard-reduction-potential-for-mg-2-mg-is-237-v-and-for-cu-2-cu-is-0337-the-e-cell-for-the-following-reaction-is-mg-cu-2-mg-2-cu-63518

question:
<div class="question">The standard reduction potential for $\mathrm{Mg}^{2+} / \mathrm{Mg}$ is $-2.37 \mathrm{~V}$ and for $\mathrm{Cu}^{2+} / \mathrm{Cu}$ is $0.337$. The $E_{\text {cell }}^{\circ}$ for the following reaction is $\mathrm{Mg}+\mathrm{Cu}^{2+} \longrightarrow \mathrm{Mg}^{2+}+\mathrm{Cu}$</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$+2.03 \mathrm{~V}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$-2.03 \mathrm{~V}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$-2.7 \mathrm{~V}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$+2.7 \mathrm{~V}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$+2.7 \mathrm{~V}$</span> </div>

answer:
<div class="solution">$E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}$<br/>$$<br/>\begin{aligned}<br/>=E_{\left(\mathrm{Cu}^{2+} / \mathrm{Cu}\right)}^{\circ}-E_{\left(\mathrm{Mg}^{2+} / \mathrm{Mg}\right)}^{\circ} &amp;=0.337-(-2.37) \mathrm{V} \\<br/>&amp;=2.7 \mathrm{~V}<br/>\end{aligned}<br/>$$</div>


tags: ['Chemistry', 'Electrochemistry', 'JIPMER', 'JIPMER 2017']
source_db: MarksBatch2_P3.db
title:
the-standard-reduction-potential-for-zn-2-zn-ni-2-ni-and-fe-2-fe-are-076-023-and-044-v-respectively-the-reaction-x-y-2-x-2-y-will-be-spontaneous-when

slug:
the-standard-reduction-potential-for-zn-2-zn-ni-2-ni-and-fe-2-fe-are-076-023-and-044-v-respectively-the-reaction-x-y-2-x-2-y-will-be-spontaneous-when-28421

question:
<div class="question">The standard reduction potential for $\mathrm{Zn}^{2+} / \mathrm{Zn}, \mathrm{Ni}^{2+} / \mathrm{Ni}$ and $\mathrm{Fe}^{2+} / \mathrm{Fe}$ are -0.76 , -0.23 and -0.44 V , respectively.<br/>The reaction $X+Y^{2+} \longrightarrow X^{2+}+Y$ will be spontaneous when</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">X=Ni, Y=Fe</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">X=Ni, Y=Zn</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">X=Fe, Y=Zn</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">X=Zn, Y=Ni</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">X=Zn, Y=Ni</span> </div>

answer:
<div class="solution">The reaction is spontaneous when $\Delta G \lt 0$ and $E_{\text {cell }}^{\circ}\gt0 .$<br/>$\begin{aligned} &amp; \because \Delta G=-n F E^{\circ} \text { and } E^{\circ}\gt0 \\ &amp; \text { Also } E_{\text {cell }}^{\circ}=E_{\text {oxi }}^{\circ}+E_{\text {red }}^{\circ} \text { (or) } E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}\end{aligned}$<br/>Here,<img src="https://cdn-question-pool.getmarks.app/pyq/jipmer/PmL2n482WKSYQr5fIbukz0QpZTzlbhtzE1kvBRnDw_8.original.fullsize.png"/><br/>Thus, $E_{\text {cell }}^{\circ}$ for<br/>(a) -0.21 V , i.e. $E_{\text {cell }}^0 \lt 0$<br/>(b) -0.53 V , i.e. $E_{\text {cell }}^0 \lt 0$<br/>(c) -0.32 V , i.e. $E_{\text {cell }}^{\circ} \lt 0$<br/>(d) 0.53 V , i.e. $E^{\circ}\gt0$</div>


tags: ['Chemistry', 'Electrochemistry', 'WBJEE', 'WBJEE 2011']
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">+0.35</span> </div>
source_db: MarksBatch2_P3.db
title:
the-standard-reduction-potential-mathrmecirc-for-half-reations-are-beginarrayll-mathrmznmathrmzn2mathrmze-mathrmecirc076-mathrmv-mathrmfemathrmfe2math

slug:
the-standard-reduction-potential-mathrmecirc-for-half-reations-are-beginarrayll-mathrmznmathrmzn2mathrmze-mathrmecirc076-mathrmv-mathrmfemathrmfe2math-84324

question:
<div class="question">The standard reduction potential \(\mathrm{E}^{\circ}\) for half reations are<br/>\(\begin{array}{ll}<br/>\mathrm{Zn}=\mathrm{Zn}^{+2}+\mathrm{Ze} &amp; \mathrm{E}^{\circ}=+0.76 \mathrm{~V} \\<br/>\mathrm{Fe}=\mathrm{Fe}^{+2}+\mathrm{Ze} &amp; \mathrm{E}^{\circ}=+0.41 \mathrm{~V}<br/>\end{array}\)<br/>The EMF of hte cell reaction \(\mathrm{Fe}^{+2}+\mathrm{Zn}=\mathrm{Zn}^{+2}+\mathrm{Fe}\) is</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">\(-0.35 \mathrm{~V}\)</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">+0.35</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">\(+1.17 \mathrm{~V}\)</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">\(-1.17 \mathrm{~V}\)</span> </li> </ul>

answer:
<div class="solution">\(\begin{aligned}<br/>&amp; \text {Hints: } \mathrm{E}_{\text {cell }}=\mathrm{E}_{\operatorname{Anu}([\mathrm{Qp})}^{\circ}-\mathrm{E}_{\text {cathode }(\mathrm{op})}^{\circ} \\<br/>&amp; =0.76-0.41 \\<br/>&amp; =+0.35 \mathrm{~V}<br/>\end{aligned}\)</div>


tags: ['Chemistry', 'Electrochemistry', 'JEE Main', 'JEE Main 2023 (06 Apr Shift 2)']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">3</span> </div>
source_db: MarksBatch2_P3.db
title:
the-standard-reduction-potentials-at-295-k-for-the-following-half-cells-are-given-below-no-3-4-h-3-e-no-g-2-h-2-o-e-o-0-97-v-v-2-aq-2-e-v-s-e-o-1-19-v

slug:
the-standard-reduction-potentials-at-295-k-for-the-following-half-cells-are-given-below-no-3-4-h-3-e-no-g-2-h-2-o-e-o-0-97-v-v-2-aq-2-e-v-s-e-o-1-19-v-58117

question:
<div class="question"><p>The standard reduction potentials at <math xmlns="http://www.w3.org/1998/Math/MathML"><mn>295</mn><mo> </mo><mi mathvariant="normal">K</mi></math> for the following half cells are given below:</p><table border="1" cellpadding="1" cellspacing="1" style="width: 500px;"> <tbody> <tr> <td style="text-align: center;"><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><msub><mi>NO</mi><mn>3</mn></msub><mo>-</mo></msup><mo>+</mo><mn>4</mn><msup><mi mathvariant="normal">H</mi><mo>+</mo></msup><mo>+</mo><mn>3</mn><msup><mi mathvariant="normal">e</mi><mo>-</mo></msup><mo>→</mo><mi>NO</mi><mfenced><mi mathvariant="normal">g</mi></mfenced><mo>+</mo><mn>2</mn><msub><mi mathvariant="normal">H</mi><mn>2</mn></msub><mi mathvariant="normal">O</mi></math></td> <td style="text-align: center;"><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi mathvariant="normal">E</mi><mi mathvariant="normal">o</mi></msup><mo>=</mo><mn>0</mn><mo>.</mo><mn>97</mn><mo> </mo><mi mathvariant="normal">V</mi></math></td> </tr> <tr> <td style="text-align: center;"><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi mathvariant="normal">V</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup><mfenced><mi>aq</mi></mfenced><mo>+</mo><mn>2</mn><msup><mi mathvariant="normal">e</mi><mo>-</mo></msup><mo>→</mo><mi mathvariant="normal">V</mi><mfenced><mi mathvariant="normal">s</mi></mfenced></math></td> <td style="text-align: center;"><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi mathvariant="normal">E</mi><mi mathvariant="normal">o</mi></msup><mo>=</mo><mo>-</mo><mn>1</mn><mo>.</mo><mn>19</mn><mo> </mo><mi mathvariant="normal">V</mi></math></td> </tr> <tr> <td style="text-align: center;"><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi>Fe</mi><mrow><mn>3</mn><mo>+</mo></mrow></msup><mfenced><mi>aq</mi></mfenced><mo>+</mo><mn>3</mn><msup><mi mathvariant="normal">e</mi><mo>-</mo></msup><mo>→</mo><mi>Fe</mi><mfenced><mi mathvariant="normal">s</mi></mfenced></math></td> <td style="text-align: center;"><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi mathvariant="normal">E</mi><mi mathvariant="normal">o</mi></msup><mo>=</mo><mo>-</mo><mn>0</mn><mo>.</mo><mn>04</mn><mo> </mo><mi mathvariant="normal">V</mi></math></td> </tr> <tr> <td style="text-align: center;"><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi>Ag</mi><mo>+</mo></msup><mfenced><mi>aq</mi></mfenced><mo>+</mo><msup><mi mathvariant="normal">e</mi><mo>-</mo></msup><mo>→</mo><mi>Ag</mi><mfenced><mi mathvariant="normal">s</mi></mfenced></math></td> <td style="text-align: center;"><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi mathvariant="normal">E</mi><mi mathvariant="normal">o</mi></msup><mo>=</mo><mn>0</mn><mo>.</mo><mn>80</mn><mo> </mo><mi mathvariant="normal">V</mi></math></td> </tr> <tr> <td style="text-align: center;"><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi>Au</mi><mrow><mn>3</mn><mo>+</mo></mrow></msup><mfenced><mi>aq</mi></mfenced><mo>+</mo><mn>3</mn><msup><mi mathvariant="normal">e</mi><mo>-</mo></msup><mo>→</mo><mi>Au</mi><mfenced><mi mathvariant="normal">s</mi></mfenced></math></td> <td style="text-align: center;"><math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi mathvariant="normal">E</mi><mi mathvariant="normal">o</mi></msup><mo>=</mo><mn>1</mn><mo>.</mo><mn>40</mn><mo> </mo><mi mathvariant="normal">V</mi></math></td> </tr> </tbody></table><p>The number of metal(s) which will be oxidised by <math xmlns="http://www.w3.org/1998/Math/MathML"><msup><msub><mi>NO</mi><mn>3</mn></msub><mo>-</mo></msup></math> in aqueous solution is _____.</p></div>

answer:
<div class="solution"><blockquote><p>For feasibility, check , <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><msup><mi mathvariant="normal">E</mi><mi mathvariant="normal">o</mi></msup><mi>cell</mi></msub><mo>=</mo><msubsup><mi mathvariant="normal">E</mi><mrow><mi>cathode</mi><mo>(</mo><mi>reduction</mi><mo>)</mo></mrow><mi mathvariant="normal">o</mi></msubsup><mo>-</mo><msubsup><mi mathvariant="normal">E</mi><mrow><mi>anode</mi><mo>(</mo><mi>oxidation</mi><mo>)</mo></mrow><mi mathvariant="normal">o</mi></msubsup><mo>&gt;</mo><mn>0</mn></math></p><p>The <math xmlns="http://www.w3.org/1998/Math/MathML"><msup><mi mathvariant="normal">E</mi><mi mathvariant="normal">o</mi></msup></math> values when metal acts as anode and <math xmlns="http://www.w3.org/1998/Math/MathML"><msubsup><mi>NO</mi><mn>3</mn><mo>-</mo></msubsup></math> reaction is cathodic reaction.</p><p>For Vanadium metal, <math xmlns="http://www.w3.org/1998/Math/MathML"><msubsup><mi mathvariant="normal">E</mi><mi>cell</mi><mi mathvariant="normal">o</mi></msubsup><mo>=</mo><mn>0</mn><mo>.</mo><mn>97</mn><mo>+</mo><mn>1</mn><mo>.</mo><mn>19</mn><mo>=</mo><mn>2</mn><mo>.</mo><mn>16</mn><mi mathvariant="normal">V</mi></math></p><p>For Iron metal, <math xmlns="http://www.w3.org/1998/Math/MathML"><msubsup><mi mathvariant="normal">E</mi><mi>cell</mi><mi mathvariant="normal">o</mi></msubsup><mo>=</mo><mn>0</mn><mo>.</mo><mn>97</mn><mo>+</mo><mn>0</mn><mo>.</mo><mn>04</mn><mo>=</mo><mn>1</mn><mo>.</mo><mn>01</mn><mi mathvariant="normal">V</mi></math></p><p>For silver metal, <math xmlns="http://www.w3.org/1998/Math/MathML"><msubsup><mi mathvariant="normal">E</mi><mi>cell</mi><mi mathvariant="normal">o</mi></msubsup><mo>=</mo><mn>0</mn><mo>.</mo><mn>97</mn><mo>-</mo><mn>0</mn><mo>.</mo><mn>80</mn><mo>=</mo><mn>0</mn><mo>.</mo><mn>17</mn><mi mathvariant="normal">V</mi></math></p><p>For Gold metal, <math xmlns="http://www.w3.org/1998/Math/MathML"><msubsup><mi mathvariant="normal">E</mi><mi>cell</mi><mi mathvariant="normal">o</mi></msubsup><mo>=</mo><mn>0</mn><mo>.</mo><mn>97</mn><mo>-</mo><mn>1</mn><mo>.</mo><mn>140</mn><mo>=</mo><mo>-</mo><mn>0</mn><mo>.</mo><mn>17</mn><mi mathvariant="normal">V</mi></math></p><p>For electrodes having oxidation potential greater than <math xmlns="http://www.w3.org/1998/Math/MathML"><mo>-</mo><mn>0</mn><mo>.</mo><mn>97</mn><mo> </mo><mi mathvariant="normal">V</mi></math>, <math xmlns="http://www.w3.org/1998/Math/MathML"><msub><msup><mi mathvariant="normal">E</mi><mi mathvariant="normal">o</mi></msup><mi>cell</mi></msub><mo>&gt;</mo><mn>0</mn></math></p><p><math xmlns="http://www.w3.org/1998/Math/MathML"><mo>∴</mo><mo> </mo><mi>Ag</mi><mo>,</mo><mo> </mo><mi>Fe</mi></math> and <math xmlns="http://www.w3.org/1998/Math/MathML"><mi mathvariant="normal">V</mi></math> can be oxidised</p></blockquote></div>


tags: ['Chemistry', 'Electrochemistry', 'JEE Main']
source_db: MarksBatch2_P3.db
title:
the-standard-reduction-potentials-at-298-k-for-single-electrodes-are-given-below-electrode-electrode-potential-volt-mg-2-mg-2-34-zn-2-zn-0-76-fe-2-fe-

slug:
the-standard-reduction-potentials-at-298-k-for-single-electrodes-are-given-below-electrode-electrode-potential-volt-mg-2-mg-2-34-zn-2-zn-0-76-fe-2-fe-11625

question:
<div class="question">The standard reduction potentials at 298 K for single electrodes are given below.
<table border="1"> <tbody> <tr> <td>Electrode</td> <td>electrode potential (volt)</td> </tr> <tr> <td><math> <mrow> <msup> <mrow> <mtext>Mg</mtext></mrow> <mrow> <mtext>2+</mtext></mrow> </msup> <mtext>/Mg</mtext></mrow> </math></td> <td><math> <mrow> <mo>â</mo><mtext>2</mtext><mtext>.34</mtext></mrow> </math></td> </tr> <tr> <td><math> <mrow> <msup> <mrow> <mtext>Zn</mtext></mrow> <mrow> <mtext>2+</mtext></mrow> </msup> <mtext>/Zn</mtext></mrow> </math></td> <td><math> <mrow> <mo>â</mo><mtext>0</mtext><mtext>.76</mtext></mrow> </math></td> </tr> <tr> <td><math> <mrow> <msup> <mrow> <mtext>Fe</mtext></mrow> <mrow> <mtext>2+</mtext></mrow> </msup> <mtext>/Fe</mtext></mrow> </math></td> <td><math> <mrow> <mo>â</mo><mtext>0</mtext><mtext>.44</mtext></mrow> </math></td> </tr> </tbody> </table>
From this we can tell that</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">Zn can reduce both  <math><mi mathvariant="normal">M</mi><msup><mrow><mi mathvariant="normal">g</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup></math>  and  <math><mi mathvariant="normal">F</mi><msup><mrow><mi mathvariant="normal">e</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup></math></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">Fe can reduce both  <math><mi mathvariant="normal">M</mi><msup><mrow><mi mathvariant="normal">g</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup></math>  and  <math><mi mathvariant="normal">Z</mi><msup><mrow><mi mathvariant="normal">n</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup></math></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">Mg can reduce both  <math><mi mathvariant="normal">Z</mi><msup><mrow><mi mathvariant="normal">n</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup></math>  and  <math><mi mathvariant="normal">F</mi><msup><mrow><mi mathvariant="normal">e</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup></math></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">Mg can reduce  <math><mi mathvariant="normal">Z</mi><msup><mrow><mi mathvariant="normal">n</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup></math> but not  <math><mi mathvariant="normal">F</mi><msup><mrow><mi mathvariant="normal">e</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup></math></span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">Mg can reduce both  <math><mi mathvariant="normal">Z</mi><msup><mrow><mi mathvariant="normal">n</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup></math>  and  <math><mi mathvariant="normal">F</mi><msup><mrow><mi mathvariant="normal">e</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup></math></span> </div>

answer:
<div class="solution">Metals with lower SRP can displace the metal with higher SRP from their solutions. Mg with lower SRP (-2.34) can reduce both  <math><mi mathvariant="normal">Z</mi><msup><mrow><mi mathvariant="normal">n</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup></math>  (-0.76) and  <math><mi mathvariant="normal">F</mi><msup><mrow><mi mathvariant="normal">e</mi></mrow><mrow><mn>2</mn><mo>+</mo></mrow></msup></math>  (-0.44).</div>


tags: ['Chemistry', 'Electrochemistry', 'JEE Main']
source_db: MarksBatch2_P3.db
title:
the-standard-reduction-potentials-at-298-k-for-the-following-half-reactions-are-given-against-each-z-n-2-a-q-2-e-z-n-s-0762-cr-3-aq-3-e-cr-s-0740-2-h-

slug:
the-standard-reduction-potentials-at-298-k-for-the-following-half-reactions-are-given-against-each-z-n-2-a-q-2-e-z-n-s-0762-cr-3-aq-3-e-cr-s-0740-2-h-78088

question:
<div class="question">The standard reduction potentials at $298 \mathrm{~K}$ for the following half reactions are given against each<br/>$\begin{aligned}<br/>&amp; Z n^{2+}(a q .)+2 e \rightleftharpoons Z n(s) ; \quad-0.762 \\<br/>&amp; \mathrm{Cr}^{3+}(\mathrm{aq})+3 e \rightleftharpoons \mathrm{Cr}(\mathrm{s}) ; \quad-0.740 \\<br/>&amp; 2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e} \rightleftharpoons \mathrm{H}_2(g) ; \quad 0.00 \\<br/>&amp; \mathrm{Fe}^{3+}(a q)+e \rightleftharpoons \mathrm{Fe}^{2+}(a q) ; \quad 0.770 \\<br/>&amp;<br/>\end{aligned}$<br/><br/>Which is the strongest reducing agent</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$\mathrm{Zn}(\mathrm{s})$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\mathrm{Cr}(\mathrm{s})$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$H_2(g)$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\mathrm{Fe}^{2+}(a q)$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\mathrm{Zn}(\mathrm{s})$</span> </div>

answer:
<div class="solution">More negative is the reduction potential, higher will be the reducing property, i.e. the power to give up electrons.</div>


tags: ['Chemistry', 'Electrochemistry', 'VITEEE', 'VITEEE 2007']
source_db: MarksBatch2_P3.db
title:
the-standard-reduction-potentials-at-298-k-for-the-following-half-reactions-are-given-against-each-zn-2-aq-2-e-zn-s-cr-3-aq-3-e-cr-s-2-h-aq-2-e-h-2-g-

slug:
the-standard-reduction-potentials-at-298-k-for-the-following-half-reactions-are-given-against-each-zn-2-aq-2-e-zn-s-cr-3-aq-3-e-cr-s-2-h-aq-2-e-h-2-g-47326

question:
<div class="question">The standard reduction potentials at $298 \mathrm{~K}$ for the following half reactions are given against each<br/>$\begin{array}{ll}<br/>\mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Zn}(\mathrm{s}) &amp; \mathrm{E}_{0}=-0.762 \mathrm{~V} \\<br/>\mathrm{Cr}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Cr}(\mathrm{s}) &amp; \mathrm{E}_{0}=-0.740 \mathrm{~V} \\<br/>2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}(\mathrm{~g}) &amp; \mathrm{E}_{0}=0.00 \mathrm{~V} \\<br/>\mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq}) &amp; \mathrm{E}_{0}=+0.762 \mathrm{~V}<br/>\end{array}$</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$\mathrm{Zn}(\mathrm{s})$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\mathrm{Cr}(\mathrm{s})$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\mathrm{H}_{2}(\mathrm{~g})$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\mathrm{Fe}^{2+}(\mathrm{aq})$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\mathrm{Zn}(\mathrm{s})$</span> </div>

answer:
<div class="solution">Lower the value of reduction potential, greater will be the reducing power of element.<br/>Since $\mathrm{Zn}$ has lowest reduction potention hence $\mathrm{Zn}$ is the strongest reducing agent.</div>


tags: ['Chemistry', 'Electrochemistry', 'MHT CET', 'MHT CET 2008']
source_db: MarksBatch2_P3.db
title:
the-standard-reduction-potentials-at-298-k-for-the-following-halfcell-reactions-are-given-on-next-page-zn-2-a-q-2-e-zn-s-0762-cr-3-a-q-3-e-cr-s-074-2-

slug:
the-standard-reduction-potentials-at-298-k-for-the-following-halfcell-reactions-are-given-on-next-page-zn-2-a-q-2-e-zn-s-0762-cr-3-a-q-3-e-cr-s-074-2-32218

question:
<div class="question">The standard reduction potentials at $298 \mathrm{~K}$ for the following half-cell reactions are given on next page<br/>$\mathrm{Zn}^{2+}(a q)+2 e^{-} \rightleftharpoons \mathrm{Zn}(s)-0.762$<br/>$\mathrm{Cr}^{3+}(a q)+3 e^{-} \rightleftharpoons \mathrm{Cr}(s)-0.74$<br/>$2 \mathrm{H}^{+}(a q)+2 e^{-} \rightleftharpoons \mathrm{H}_{2}(g)+0.00$<br/>$\mathrm{Fe}^{3+}(a q)+e^{-} \rightleftharpoons \mathrm{Fe}^{2+}(a q)+0.77$<br/>Which one of the following is the strongest reducing agent?</div>

options:
<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$\mathrm{Zn}(s)$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\operatorname{Cr}(s)$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\mathrm{H}_{2}(s)$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\mathrm{Fe}^{2+}(a q)$</span> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">$\mathrm{Zn}(s)$</span> </div>

answer:
<div class="solution">The metals having higher negative value of standard reduction potential are placed above hydrogen in electrochemical series. The metals placed above hydrogen has a great tendency to donate electrons or oxidising power. The metals having great oxidising power are strongest reducing agent. Zn has higher negative value of standard reduction potential. Therefore, it is the strongest reducing agent.</div>


tags: ['Chemistry', 'Electrochemistry', 'JEE Main', 'JEE Main 2024 (09 Apr Shift 1)']
options: None
correct_option: <div class="correct-answer"> The correct answer is: <span class="option-value">3</span> </div>
source_db: MarksBatch2_P3.db
title:
the-standard-reduction-potentials-at-298-mathrmk-for-the-following-half-cells-are-given-below-beginarrayll-mathrmcr2-mathrmo7214-mathrmh6-mathrme-righ

slug:
the-standard-reduction-potentials-at-298-mathrmk-for-the-following-half-cells-are-given-below-beginarrayll-mathrmcr2-mathrmo7214-mathrmh6-mathrme-righ-19965

question:
<div class="question">The standard reduction potentials at \(298 \mathrm{~K}\) for the following half cells are given below :<br/>\(\begin{array}{ll}<br/>\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}, \mathrm{E}^{\circ}=1.33 \mathrm{~V} \\<br/>\mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \qquad \mathrm{E}^{\circ}=-0.04 \mathrm{~V} \\<br/>\mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni} \qquad \mathrm{E}^{\circ}=-0.25 \mathrm{~V} \\<br/>\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag} \quad\qquad \mathrm{E}^{\circ}=0.80 \mathrm{~V} \\<br/>\mathrm{Au}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Au} \qquad \mathrm{E}^{\circ}=1.40 \mathrm{~V}<br/>\end{array}\)<br/>Consider the given electrochemical reactions,<br/>The number of metal(s) which will be oxidized be \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\), in aqueous solution is ______</div>

answer:
<div class="solution">$\mathrm{Fe}, \mathrm{Ni}, \mathrm{Ag}$ will be oxidized due to lower S.R.P.</div>


tags: ['Chemistry', 'Electrochemistry', 'JEE Main']
source_db: MarksBatch2_P3.db
title:
the-standard-reduction-potentials-at-298-mathrmk-for-the-following-half-reactions-are-given-against-each-beginarrayll-mathrmzn2mathrmaq2-mathrme-right

slug:
the-standard-reduction-potentials-at-298-mathrmk-for-the-following-half-reactions-are-given-against-each-beginarrayll-mathrmzn2mathrmaq2-mathrme-right-18754

question:
<div class="question">The standard reduction potentials at \( 298 \mathrm{~K} \) for the following half reactions are given against each<br/>\[<br/>\begin{array}{ll}<br/>\mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{e} \rightleftharpoons \mathrm{Zn}(\mathrm{s}) &amp; -0.762 \\<br/>\mathrm{Cr}^{3+}(\mathrm{aq})+3 \mathrm{e} \rightleftharpoons \mathrm{Cr}(\mathrm{s}) &amp; -0.740 \\<br/>2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e} \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g}) &amp; 0.000 \\<br/>\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{e} \rightleftharpoons \mathrm{Fe}^{2+}(\mathrm{aq}) &amp; 0.770<br/>\end{array}<br/>\]<br/>Which is the strongest reducing agent?</div>

options:
<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">\( \mathrm{Fe}^{2+} \) (aq)</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">\( \mathrm{H}_{2}(\mathrm{~g}) \)</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">\( \operatorname{Cr}(\mathrm{s}) \)</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">\( \operatorname{Zn}(\mathrm{s}) \)</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

correct_option:
<div class="correct-answer">
The correct answer is:
<span class="option-value">\( \operatorname{Zn}(\mathrm{s}) \)</span> </div>

answer:
<div class="solution"><div><p align="left" style="text-align:left;">For an element or an Ion to act as a reducing agent, it should itself get oxidised.  Hence, the species which has the highest negative value of reduction potential will be most easily oxidised. As Zn<sup>2+</sup> has the highest negative value for reduction, it will have the highest positive value for oxidation and thus will be the best reducing agent among the given options.</p><p align="left" style="text-align:left;"><span style="position:relative;top:0.0pt"> <math xmlns="http://www.w3.org/1998/Math/MathML"><mtext>The correct order of reducing character is</mtext><mo>:</mo><mspace linebreak="newline"></mspace><msup><mi>Zn</mi><mrow><mn>2</mn><mo>+</mo></mrow></msup><mrow><mo>(</mo><mi mathvariant="normal">E</mi><mo>°</mo><mo>=</mo><mo>-</mo><mn>0</mn><mo>.762</mo><mo>)</mo></mrow><mo>&gt;</mo><msup><mi>Cr</mi><mrow><mn>3</mn><mo>+</mo></mrow></msup><mrow><mo>(</mo><mi mathvariant="normal">E</mi><mo>°</mo><mo>=</mo><mo>−</mo><mn>0</mn><mo>.740</mo><mo>)</mo></mrow><mo>&gt;</mo><msup><mi mathvariant="normal">H</mi><mo>+</mo></msup><mrow><mo>(</mo><mi mathvariant="normal">E</mi><mo>°</mo><mo>=</mo><mn>0</mn><mo>)</mo></mrow><mo>&gt;</mo><msup><mi>Fe</mi><mrow><mn>3</mn><mo>+</mo></mrow></msup><mrow><mo>(</mo><mi mathvariant="normal">E</mi><mo>°</mo><mo>=</mo><mn>0</mn><mo>.770</mo><mo>)</mo></mrow></math></span></p></div></div>


