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

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

Applying the unified equation for Faraday's 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 $m(\ce{Ag})$.


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