Two cells are connected in series. If x g of silver are deposited in cell 1, what volume of oxygen is given off in cell 2? [closed]








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I got the answer correct but I'm pretty sure my methodology was wrong. How would you solve such a question? 




Silver plates out, i.e. is reduced, at the cell 1 cathode according to


$$\ce{Ag+(aq) + 2 e- → 2 Ag(s)}$$


while oxygen is produced at cell 2's anode via the electrolysis of water


$$\ce{H2O(l) → 0.5 O2(g) + 2 H+ + 2 e-}$$


Since the cells are in series, we know that every electron that flows through cell 1 must flow through cell 2. This allows us to create a relationship between the moles of silver plated and the moles of oxygen evolved. Specifically,


$$\ce{H2O(l) + 2 Ag+(aq) → 2 Ag(s) + 0.5 O2(g) + 2 H+}$$


That is, for every 4 moles of silver that plate out, we evolve 1 mole of oxygen gas. The rest is algebra to give us answer A:


$$V(\ce{O2}) = \frac{\pu{22.7 L/mol}\cdot x}{4\cdot \pu{108 g/mol}}$$


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You can solve by using the Faradays second law: use the following expression:
$$\frac{\text{mass of}~\ce{Ag}}{\text{equivalent mass of}~\ce{Ag}} = \frac{\text{mass of}~\ce{O2}} {\text{equivalent mass of}~\ce{O2}}$$
$$\frac{x}{108} = \frac{\text{mass of}~\ce{O2}} {8 }$$
so :mass of $\ce{O2} =\frac{x}{108}\times{\pu{8 g}}$


and the amount of $\ce{O2} = \frac{x}{108}\times{\frac{\pu{8 g}}{\pu{32g/mol}}}=\frac{x}{108}\times{\frac{\pu{1 mol}}{4}}$
$$V(\ce{O2}) =\frac{x}{108}\times{\frac{1}{4}}\times{\pu{22.7 L}}$$


**Another method for solving the problem** :by using the balanced chemical equations calculations and the Faraday first law as the following :


1- Write the two half cell reactions ,and make the electrons gained equal to the electrons lost: 


$$\ce{4Ag+\_\mathrm{(aq)} + 4 e- → 4 Ag\_\mathrm{(s)})}$$
$$\ce{2H2O\_\mathrm{(l)} → O2\_\mathrm{(g)} +4 H+\_\mathrm{(aq)} + 4 e-}$$


2- Calculate the amount of electricity needed to depsite $\pu{x g}$ of silver($\pu{\frac{x}{108} mol}$ of siver atom) in the 1st cell,by creating a relationship between the moles of silver plated and the amount of electricity flowing in the first cell as the following :


\begin{align}
\ce{&4Ag+\_\mathrm{(aq)} &+ &4 e- &→ &4 Ag\_\mathrm{(s)})}\\
&&& \text{4 electrons}&& \text{4 silver atoms}\\
&&&\pu{4 mol} ~\text{of electrons} &&\pu{4 mol}~\text{of silver atoms}\\
&&&\pu{4 Faraday} &&\pu{4 mol}~\text{of silver atoms}\\
&&& ? &&\pu{\frac{x}{108}mol}
\end{align}
so,the amount of electricity flows through the first cell =$\pu{\frac{x}{108}mol}\times\frac{{\pu{4 Faraday}}}{\pu{4 mol}}=\pu{\frac{x}{108}Faraday} $


Because the cells are in series, the amount of electricity flows through the first cell 
equal the amount of electricity flows through the second cell=$\pu{\frac{x}{108}Faraday} $.


3- Calculate the amount of $\ce{O2}$ evolved in the second cell
by creating a relationship between the amount of $\ce{O2}$ evolved and the amount of electricity flowed through the second cell as the following :
\begin{align}
\ce{2H2O\_\mathrm{(l)} → &O2\_\mathrm{(g)} +4 H+\_\mathrm{(aq)} + &4 e-}\\
&\pu{1 mol} &\pu{4 faraday}\\
&? &\pu{\frac{x}{108}Faraday}
\end{align}
so,the amount of $\ce{O2}$ evolved =$\pu{\frac{x}{108}Faraday}\times\frac{\pu{1 mol }}{\pu{4 Faraday}}=\frac{x}{108}\times\frac{\pu{1 mol}}{4}$


Thus,$$V(\ce{O2}) = \frac{x}{108}\times\frac{1\times{\pu{22.7L}}}{4}$$


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