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]

Question

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

Answer 1

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:

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

Answer 2

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