# 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]

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:

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

• “every electron that flows through cell 1 must flow through cell 2.“ Not really, different electron flow through the middle wire than through the outer wires. It is one to one though. May 18 '19 at 4:12
• Trying to keep it simple, but I get your point. However, electrons from the the far right wire could certainly end up in the middle wire.
– Tunk
May 18 '19 at 4:23
• Electrons from one wire end up in the gas phase (hydrogen) and from the other end up in the solid phase (deposited silver). Of course there could be some exchange reactions that we can’t measure, electrons being indistinguishable. May 18 '19 at 4:30
• Ah, true! Forgot about the gaseous species.
– Tunk
May 18 '19 at 4:33
• @andselisk $x$ was not specified. May 18 '19 at 13:14

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