Answer
$$
\begin{aligned}
& \Delta G=-R T \ln (K) \\
& -89152.1=-\left(8.314 \mathrm{~J} \mathrm{~mol}^{-1} K^{-1}\right)(298 K) \ln (K) \\
& \ln (K)=35.98 \\
& K=e^{(35.98)} \\
& K=4.24 \times 10^{15}
\end{aligned}
$$
Work Step by Step
Step 1: $\mathrm{Cu}(s) \rightarrow \mathrm{Cu}^{+2}(a q)+2 e^{-} E_{\text {red }}=+0.337 \mathrm{~V}$
$$
2 \mathrm{Ag}^{+}(a q)+2 e^{-} \rightarrow 2 A g(s) \quad E_{\text {red }}=+0.799 \mathrm{~V}
$$
Step 2: $E_{\text {red }}($ cathode/reduced $)-E_{\text {red }}$ (Anode/Oxidized $)$
$$
0.799 \mathrm{~V}-0.337 \mathrm{~V}=0.462 \mathrm{~V}=E_{\text {Cell }}
$$
Step 3: Find Gibbs Free Energy...
$$
\begin{aligned}
\Delta G & =-n F E \quad \text { } \\
\Delta G & =-(2 e)(96485 \mathrm{C} / \mathrm{mol} \mathrm{e})(0.462 \mathrm{~V}) \\
\Delta G & =-89152.1 \mathrm{~J}
\end{aligned}
$$
Step 4: Convert the calculated Energy to Equilbrium Constant...
$$
\begin{aligned}
& \Delta G=-R T \ln (K) \\
& -89152.1=-\left(8.314 \mathrm{~J} \mathrm{~mol}^{-1} K^{-1}\right)(298 K) \ln (K) \\
& \ln (K)=35.98 \\
& K=e^{(35.98)} \\
& K=4.24 \times 10^{15}
\end{aligned}
$$
