Answer
Step 3: Plug answers in the first equation...
$$
\begin{aligned}
W_{\text {max }} & =-n F E \\
& =-(1.26)(96485)(0.672 \mathrm{~V}) \\
& =-8.19 \times 10^{4} \mathrm{~J}
\end{aligned}
$$
Step 4: Convert work energy per mole...
$$
\begin{aligned}
& =\left(-8.19 \times 10^{4} \mathrm{~J}\right) /(0.631 \mathrm{~mol} \text { of } \mathrm{Sn}) \\
& =-129675 \mathrm{~J} / \mathrm{mol} \text { of } \mathrm{Sn}=-130 \mathrm{KJ} / \mathrm{mol} \text { of } \mathrm{Sn}
\end{aligned}
$$
Work Step by Step
$W_{\text {max }}=-n F E$
$n=$ mol of $e^{-}$
$F=96485$
Step 1: $\operatorname{Sn}(s) \rightarrow \operatorname{Sn}^{+2}(a q)+2 e^{-} \quad E_{\text {red }}=-0.136 \mathrm{~V}$
$$
I_{2}(s)+2 e^{-} \rightarrow 2 I^{-}(a q) \quad E_{\text {red }}=+0.536 \mathrm{~V}
$$
$E_{\text {Cell }}=E_{\text {red }}($ cathode $/$ reduced $)-E_{\text {red }}($ Anode $/$ Oxidized $)=0.536 \mathrm{~V}-(-0.136 \mathrm{~V})=0.672 \mathrm{~V}$
Step 2: find $\mathrm{n}$ given that $75 \mathrm{~g}$ of $\mathrm{Sn}$ has been consumed.
$$
\begin{aligned}
& \frac{75 \mathrm{~g}}{118.71 \mathrm{~g} / \mathrm{mol}}=\frac{\text { mass }}{\text { molar mass }}=\text { moles of } \mathrm{Sn}=0.631 \\
& 0.631 \mathrm{~mol} \text { of } \mathrm{S} n \times \frac{2 \text { mol of } e^{-}}{1 \text { mol of } \mathrm{S} n}=1.26=n
\end{aligned}
$$
Step 3: Plug answers in the first equation...
$$
\begin{aligned}
W_{\text {max }} & =-n F E \\
& =-(1.26)(96485)(0.672 \mathrm{~V}) \\
& =-8.19 \times 10^{4} \mathrm{~J}
\end{aligned}
$$
Step 4: Convert work energy per mole...
$$
\begin{aligned}
& =\left(-8.19 \times 10^{4} \mathrm{~J}\right) /(0.631 \mathrm{~mol} \text { of } \mathrm{Sn}) \\
& =-129675 \mathrm{~J} / \mathrm{mol} \text { of } \mathrm{Sn}=-130 \mathrm{KJ} / \mathrm{mol} \text { of } \mathrm{Sn}
\end{aligned}
$$
