Electrolytic silver deposition


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> A current is applied to two electrolytic cells in a series. In the first, silver is deposited; in the second, a zinc electrode is consumed. How much $\ce{Ag}$ is plated out if $\pu{7.5 g}$ of $\ce{Zn}$ dissolves? 
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What I ended up doing was


$$\frac{\pu{7.5 g}~\ce{Zn}}{\pu{65.382 g mol}~\ce{Zn}} \times \frac{\pu{2 mol}~\ce{Ag}}{\pu{1 mol}~\ce{Zn}} \times \pu{107 g mol-1}~\ce{Ag}$$


but I'm not even sure I'm doing this right.




I'd suggest to start with writing down the processes occurring during the electrolysis:


\begin{align}
&\text{Cathode (–):} &\ce{2 Ag+ + 2 e- &-> Ag^0} \\
&\text{Anode (+):} &\ce{Zn^0 &-> Zn^2+ + 2e} \\
\hline
&\text{Net} &\ce{2 Ag+ +Zn^0 &-> 2 Ag^0 + Zn^2+}
\end{align}


Applying the unified equation for [Faraday's laws of electrolysis](https://en.wikipedia.org/wiki/Faraday%27s_laws_of_electrolysis):


$$m = \frac{QM}{zF}$$


for each cell, and taking into account that in series circuits current is the same for every element ($I = \text{const}$):


$$\frac{Q}{F} = \frac{z\_i m\_i}{M\_i} = \text{const}$$


$$\frac{z(\ce{Ag}) \cdot m(\ce{Ag})}{M(\ce{Ag})} = \frac{z(\ce{Zn}) \cdot m(\ce{Zn})}{M(\ce{Zn})}$$


The valency numbers of ions are 1 and 2 for silver and zinc, correspondingly. The last thing to do is to express and calculate $m(\ce{Ag})$.



> 
>  $$m(\ce{Ag}) = \frac{z(\ce{Zn}) \cdot M(\ce{Ag})}{z(\ce{Ag}) \cdot M(\ce{Zn})} \cdot m(\ce{Zn}) = \frac{2 \cdot \pu{107.87 g mol-1}}{1 \cdot \pu{65.38 g mol-1}} \cdot \pu{7.5 g} = \pu{24.7 g}$$
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