0 1 2 NaN Prince George's Community College General Chemistry for Engineering CHM 2000 NaN Unit I: Atoms Unit II: Molecules Unit III: States of Matter Unit IV: Reactions Unit V: Kinetics & Equilibrium Unit VI: Thermo & Electrochemistry Unit VII: Nuclear Chemistry Unit I: Atoms Unit II: Molecules Unit III: States of Matter Unit IV: Reactions Unit V: Kinetics & Equilibrium Unit VI: Thermo & Electrochemistry Unit VII: Nuclear Chemistry Unit I: Atoms Unit II: Molecules Unit III: States of Matter Unit IV: Reactions Unit V: Kinetics & Equilibrium Unit VI: Thermo & Electrochemistry Unit VII: Nuclear Chemistry Learning Objective To know how to predict the relative strengths of various oxidants and reductants. We can use the procedure described in Section 17.2 to measure the standard potentials for a wide variety of chemical substances, some of which are listed in Table 17.3.1 . ( Standard Reduction Potentials at 25°C" contains a more extensive listing.) These data allow us to compare the oxidative and reductive strengths of a variety of substances. The half-reaction for the standard hydrogen electrode (SHE) lies more than halfway down the list in Table 17.3.1 . All reactants that lie above the SHE in the table are stronger oxidants than H + , and all those that lie below the SHE are weaker. The strongest oxidant in the table is F 2 , with a standard electrode potential of 2.87 V. This high value is consistent with the high electronegativity of fluorine and tells us that fluorine has a stronger tendency to accept electrons (it is a stronger oxidant) than any other element. Table 17.3.1 Standard Potentials for Selected Reduction Half-Reactions at 25°C Half-Reaction E° (V) F2(g) + 2e−→ 2F−(aq) 2.87 H2O2(aq) + 2H+(aq) + 2e− → 2H2O(l) 1.78 Ce4+(aq) + e− → Ce3+(aq) 1.72 PbO2(s) + HSO4−(aq) + 3H+(aq) + 2e− → PbSO4(s) + 2H2O(l) 1.69 Cl2(g) + 2e− → 2Cl−(aq) 1.36 Cr2O72−(aq) + 14H+(aq) + 6e− → 2Cr3+(aq) + 7H2O(l) 1.23 O2(g) + 4H+(aq) + 4e− → 2H2O(l) 1.23 MnO2(s) + 4H+(aq) + 2e− → Mn2+(aq) + 2H2O(l) 1.22 Br2(aq) + 2e− → 2Br−(aq) 1.09 NO3−(aq) + 3H+(aq) + 2e− → HNO2(aq) + H2O(l) 0.93 Ag+(aq) + e− → Ag(s) 0.80 Fe3+(aq) + e− → Fe2+(aq) 0.77 H2SeO3(aq) + 4H+ + 4e− → Se(s) + 3H2O(l) 0.74 O2(g) + 2H+(aq) + 2e− → H2O2(aq) 0.70 MnO4−(aq) + 2H2O(l) + 3e− → MnO2(s) + 4OH−(aq) 0.60 MnO42−(aq) + 2H2O(l) + 2e− → MnO2(s) + 4OH−(aq) 0.60 I2(s) + 2e− → 2I−(aq) 0.54 H2SO3(aq) + 4H+(aq) + 4e− → S(s) + 3H2O(l) 0.45 O2(g) + 2H2O(l) + 4e− → 4OH−(aq) 0.40 Cu2+(aq) + 2e− → Cu(s) 0.34 AgCl(s) + e− → Ag(s) + Cl−(aq) 0.22 Cu2+(aq) + e− → Cu+(aq) 0.15 Sn4+(aq) + 2e− → Sn2+(aq) 0.15 2H+(aq) + 2e− → H2(g) 0.00 Sn2+(aq) + 2e− → Sn(s) −0.14 2SO42−(aq) + 4H+(aq) + 2e− → S2O62−(aq) + 2H2O(l) −0.22 Ni2+(aq) + 2e− → Ni(s) −0.26 PbSO4(s) + 2e− → Pb(s) + SO42−(aq) −0.36 Cd2+(aq) + 2e− → Cd(s) −0.40 Cr3+(aq) + e− → Cr2+(aq) −0.41 Fe2+(aq) + 2e− → Fe(s) −0.45 Ag2S(s) + 2e− → 2Ag(s) + S2−(aq) −0.69 Zn2+(aq) + 2e− → Zn(s) −0.76 Al3+(aq) + 3e− → Al(s) −1.662 Be2+(aq) + 2e− → Be(s) −1.85 Li+(aq) + e− → Li(s) −3.04 Similarly, all species in Table 17.3.1 that lie below H 2 are stronger reductants than H 2 , and those that lie above H 2 are weaker. The strongest reductant in the table is thus metallic lithium, with a standard electrode potential of −3.04 V. This fact might be surprising because cesium, not lithium, is the least electronegative element. The apparent anomaly can be explained by the fact that electrode potentials are measured in aqueous solution, where intermolecular interactions are important, whereas ionization potentials and electron affinities are measured in the gas phase. Due to its small size, the Li + ion is stabilized in aqueous solution by strong electrostatic interactions with the negative dipole end of water molecules. These interactions result in a significantly greater Δ H hydration for Li + compared with Cs + . Lithium metal is therefore the strongest reductant (most easily oxidized) of the alkali metals in aqueous solution. Note the Pattern Species in Table 17.3.1 that lie below H 2 are stronger reductants (more easily oxidized) than H 2 . Species that lie above H 2 are stronger oxidants. Because the half-reactions shown in Table 17.3.1 are arranged in order of their E ° values, we can use the table to quickly predict the relative strengths of various oxidants and reductants. Any species on the left side of a half-reaction will spontaneously oxidize any species on the right side of another half-reaction that lies below it in the table. Conversely, any species on the right side of a half-reaction will spontaneously reduce any species on the left side of another half-reaction that lies above it in the table. We can use these generalizations to predict the spontaneity of a wide variety of redox reactions ( E ° cell > 0), as illustrated in Example 5. Example 5 The black tarnish that forms on silver objects is primarily Ag 2 S. The half-reaction for reversing the tarnishing process is as follows: \( AgS\left (s \right ) 2e^{-}\rightarrow Ag\left (s \right )+ S^{2-}\left (aq \right ) \;\;\; E^{o}=-0.69\; V \) Referring to Table 17.3.1 , predict which species—H 2 O 2 (aq), Zn(s), I − (aq), Sn 2+ (aq)—can reduce Ag 2 S to Ag under standard conditions. Of these species—H 2 O 2 (aq), Zn(s), I − (aq), Sn 2+ (aq), identify which is the strongest reducing agent in aqueous solution and thus the best candidate for a commercial product. From the data in Table 17.3.1 , suggest an alternative reducing agent that is readily available, inexpensive, and possibly more effective at removing tarnish. Given: reduction half-reaction, standard electrode potential, and list of possible reductants Asked for: reductants for Ag 2 S, strongest reductant, and potential reducing agent for removing tarnish Strategy: A From their positions in Table 17.3.1 , decide which species can reduce Ag 2 S. Determine which species is the strongest reductant. B Use Table 17.3.1 to identify a reductant for Ag 2 S that is a common household product. Solution: We can solve the problem in one of two ways: (1) compare the relative positions of the four possible reductants with that of the Ag 2 S/Ag couple in Table 17.3.1 or (2) compare E ° for each species with E ° for the Ag 2 S/Ag couple (−0.69 V). A The species in Table 17.3.1 are arranged from top to bottom in order of increasing reducing strength. Of the four species given in the problem, I − (aq), Sn 2+ (aq), and H 2 O 2 (aq) lie above Ag 2 S, and one [Zn(s)] lies below it. We can therefore conclude that Zn(s) can reduce Ag 2 S(s) under standard conditions, whereas I − (aq), Sn 2+ (aq), and H 2 O 2 (aq) cannot. Sn 2+ (aq) and H 2 O 2 (aq) appear twice in the table: on the left side (oxidant) in one half-reaction and on the right side (reductant) in another. The strongest reductant is Zn(s), the species on the right side of the half-reaction that lies closer to the bottom of Table 17.3.1 than the half-reactions involving I − (aq), Sn 2+ (aq), and H 2 O 2 (aq). (Commercial products that use a piece of zinc are often marketed as a “miracle product” for removing tarnish from silver. All that is required is to add warm water and salt for electrical conductivity.) B Of the reductants that lie below Zn(s) in Table 17.3.1 , and therefore are stronger reductants, only one is commonly available in household products: Al(s), which is sold as aluminum foil for wrapping foods. Exercise Refer to Table 17.3.1 to predict which species—Sn 4+ (aq), Cl − (aq), Ag + (aq), Cr 3+ (aq), and/or H 2 O 2 (aq)—can oxidize MnO 2 (s) to MNO 4 − under standard conditions. which species—Sn 4+ (aq), Cl − (aq), Ag + (aq), Cr 3+ (aq), and/or H 2 O 2 (aq)—is the strongest oxidizing agent in aqueous solution. Answer: Ag + (aq); H 2 O 2 (aq) H 2 O 2 (aq) Example 17.3.1 Use the data in Table 17.3.1 to determine whether each reaction is likely to occur spontaneously under standard conditions: Sn(s) + Be 2+ (aq) → Sn 2+ (aq) + Be(s) MnO 2 (s) + H 2 O 2 (aq) + 2H + (aq) → O 2 (g) + Mn 2+ (aq) + 2H 2 O(l) Given: redox reaction and list of standard electrode potentials ( Table 17.3.1 ) Asked for: reaction spontaneity Strategy: A Identify the half-reactions in each equation. Using Table 17.3.1 , determine the standard potentials for the half-reactions in the appropriate direction. B Use Equation 17.2.2 to calculate the standard cell potential for the overall reaction. From this value, determine whether the overall reaction is spontaneous. Solution: A Metallic tin is oxidized to Sn 2+ (aq), and Be 2+ (aq) is reduced to elemental beryllium. We can find the standard electrode potentials for the latter (reduction) half-reaction (−1.85 V) and for the former (oxidation) half-reaction (−0.14 V) directly from Table 17.3.1 . B Adding the two half-reactions gives the overall reaction: \( \begin{matrix} cathode: & MnO_{2}\left ( s \right )+4H^{+}\left ( aq \right )+ 2e^{-}\rightarrow Mn^{2+}\left ( aq \right ) +2H_{2}O\left ( l \right ) & E_{cathode}^{o}=1.22\; V \\ anode: & H_{2}O_{2}\left ( aq \right )\rightarrow O_{2}\left( g \right ) + 2H^{+}\left ( aq \right ) + 2e^{-}& E_{anode}^{o}=1.22\; V \\ overall: & MnO_{2}\left ( s \right )+2H^{+}\left ( aq \right )+ H_{2}O_{2}\left ( aq \right )\rightarrow & E_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o}\\ & O_{2}\left( g \right )+ Mn^{2+}\left ( aq \right ) +2H_{2}O\left ( l \right ) & -0.53 \; V \end{matrix} \) The standard cell potential is quite negative, so the reaction will not occur spontaneously as written. That is, metallic tin cannot be used to reduce Be 2+ to beryllium metal under standard conditions. Instead, the reverse process, the reduction of stannous ions (Sn 2+ ) by metallic beryllium, which has a positive value of E ° cell , will occur spontaneously. A MnO 2 is the oxidant (Mn 4+ is reduced to Mn 2+ ), while H 2 O 2 is the reductant (O 2− is oxidized to O 2 ). We can obtain the standard electrode potentials for the reduction and oxidation half-reactions directly from Standard Potentials for Selected Reduction Half-Reactions at 25°C . B The two half-reactions and their corresponding potentials are as follows: \( \begin{matrix} cathode: & MnO_{2}\left ( s \right )+4H^{+}\left ( aq \right )+ 2e^{-}\rightarrow Mn^{2+}\left ( aq \right ) +2H_{2}O\left ( l \right ) & E_{cathode}^{o}=1.22\; V \\ anode: & H_{2}O_{2}\left ( aq \right )\rightarrow O_{2}\left( g \right ) + 2H^{+}\left ( aq \right ) + 2e^{-}& E_{anode}^{o}=1.22\; V \\ overall: & MnO_{2}\left ( s \right )+2H^{+}\left ( aq \right )+ H_{2}O_{2}\left ( aq \right )\rightarrow & E_{cell}^{o}=E_{cathode}^{o}-E_{anode}^{o}\\ & O_{2}\left( g \right )+ Mn^{2+}\left ( aq \right ) +2H_{2}O\left ( l \right ) & -0.53 \; V \end{matrix} \) The standard potential for the reaction is positive, indicating that under standard conditions, it will occur spontaneously as written. Hydrogen peroxide will reduce MnO 2 , and oxygen gas will evolve from the solution. Exercise Use the data in Table 17.3.2 to determine whether each reaction is likely to occur spontaneously under standard conditions: 2Ce 4+ (aq) + 2Cl − (aq) → 2Ce 3+ (aq) + Cl 2 (g) 4MnO 2 (s) + 3O 2 (g) + 4OH − (aq) → 4MnO 4 − (aq) + 2H 2 O Answer: spontaneous ( E ° cell = 0.36 V) nonspontaneous ( E ° cell = −0.20 V) Although the sign of E ° cell tells us whether a particular redox reaction will occur spontaneously under standard conditions, it does not tell us to what extent the reaction proceeds, and it does not tell us what will happen under nonstandard conditions. To answer these questions requires a more quantitative understanding of the relationship between electrochemical cell potential and chemical thermodynamics, as described in Section 17.4 . Summary The oxidative and reductive strengths of a variety of substances can be compared using standard electrode potentials. Apparent anomalies can be explained by the fact that electrode potentials are measured in aqueous solution, which allows for strong intermolecular electrostatic interactions, and not in the gas phase. Key Takeaway The relative strengths of various oxidants and reductants can be predicted using E ° values. Conceptual Problems The order of electrode potentials cannot always be predicted by ionization potentials and electron affinities. Why? Do you expect sodium metal to have a higher or a lower electrode potential than predicted from its ionization potential? What is its approximate electrode potential? Without referring to tabulated data, of Br 2 /Br − , Ca 2+ /Ca, O 2 /OH − , and Al 3+ /Al, which would you expect to have the least negative electrode potential and which the most negative? Why? Because of the sulfur-containing amino acids present in egg whites, eating eggs with a silver fork will tarnish the fork. As a chemist, you have all kinds of interesting cleaning products in your cabinet, including a 1 M solution of oxalic acid (H 2 C 2 O 4 ). Would you choose this solution to clean the fork that you have tarnished from eating scrambled eggs? The electrode potential for the reaction Cu 2+ (aq) + 2e − → Cu(s) is 0.34 V under standard conditions. Is the potential for the oxidation of 0.5 mol of Cu equal to −0.34/2 V? Explain your answer. Answer No; E ° = −0.691 V for Ag 2 S(s) + 2e − → Ag(s) + S 2− (aq), which is too negative for Ag 2 S to be spontaneously reduced by oxalic acid [ E ° = 0.49 V for 2CO 2 (g) + 2H + (aq) + 2e − → H 2 C 2 O 4 (aq)] Contributors Anonymous Modified by Joshua Halpern ( Howard University ), Scott Sinex, and Scott Johnson (PGCC)