Answer
Step 3: Plug answers in the first equation...
$$
\begin{aligned}
W_{\text {max }} & =-n F E \\
& =-(1.53)(96485)(1.1 \mathrm{~V}) \\
& =-1.62 \times 10^{5} \mathrm{~J}
\end{aligned}
$$
Step 4: Convert work energy per mole...
$$
\begin{aligned}
& =\left(-1.62 \times 10^{5} \mathrm{~J}\right) /(0.765 \mathrm{~mol} \text { of } \mathrm{Cu}) \\
& =-212267 \mathrm{~J} / \mathrm{mol} \text { of } \mathrm{Sn}=-212 \mathrm{KJ} / \mathrm{mol} \text { of } \mathrm{Cu}
\end{aligned}
$$
Work Step by Step
$W_{\text {max }}=-n F E$
$n=$ mol of $e^{-}$
$F=96485$
Step 1: $Z n(s) \rightarrow Z n^{+2}(a q)+2 e^{-} \quad E_{\text {red }}=-0.763 \mathrm{~V}$
$$
\mathrm{Cu}^{+2}(s)+2 e^{-} \rightarrow \mathrm{Cu}(s) \quad E_{\text {red }}=+0.337 \mathrm{~V}
$$
$E_{\text {Cell }}=E_{\text {red }}($ cathode $/$ reduced $)-E_{\text {red }}($ Anode/Oxidized $)=0.337 \mathrm{~V}-(-0.763 \mathrm{~V})=1.1 \mathrm{~V}$
Step 2: find $\mathrm{n}$ given that $50.0 \mathrm{~g}$ of $\mathrm{Cu}$ has been consumed.
$$
\begin{aligned}
& \frac{50 \mathrm{~g}}{65.39 \mathrm{~g} / \mathrm{mol}}=\frac{\text { mass }}{\text { molarmass }}=\text { moles of } \mathrm{Cu}=0.765 \\
& 0.765 \mathrm{~mol} \text { of } \mathrm{Cu} \times \frac{2 \mathrm{~mol} \text { of } \mathrm{e}^{-}}{1 \text { mol of } \mathrm{Cu}}=1.53=n
\end{aligned}
$$
Step 3: Plug answers in the first equation...
$$
\begin{aligned}
W_{\text {max }} & =-n F E \\
& =-(1.53)(96485)(1.1 \mathrm{~V}) \\
& =-1.62 \times 10^{5} \mathrm{~J}
\end{aligned}
$$
Step 4: Convert work energy per mole...
$$
\begin{aligned}
& =\left(-1.62 \times 10^{5} \mathrm{~J}\right) /(0.765 \mathrm{~mol} \text { of } \mathrm{Cu}) \\
& =-212267 \mathrm{~J} / \mathrm{mol} \text { of } \mathrm{Sn}=-212 \mathrm{KJ} / \mathrm{mol} \text { of } \mathrm{Cu}
\end{aligned}
$$
