Answer
$$
\begin{aligned}
& K=e^{\frac{(3 e^{-})(96485)(1.29)}{(8.314)(298)}} \\
& K=e^{150.711} \\
& K=2.83 \times 10^{65}
\end{aligned}
$$
Work Step by Step
Step 1: $3 C e^{+4}(a q)+3 e^{-} \rightarrow 3 C e^{+3}(a q) \quad E_{\text {red }}=+1.61 \mathrm{~V}$
$$
\mathrm{Bi}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{BiO}^{+}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})+3 e^{-} \mathrm{E}_{\mathrm{red}}=+0.32 \mathrm{~V}
$$
Step 2: $E_{\text {red }}($ cathode/reduced $)-E_{\text {red }}($ Anode/Oxidized $)$
$$
1.61 \mathrm{~V}-0.32 \mathrm{~V}=1.29 \mathrm{~V}=\mathrm{E}_{\text {Cell }}
$$
Step 3: Convert the emf calculated to the Equilibrium Constant via the formula...
$$
\begin{aligned}
& E_{\text {Cell }}=\frac{R T}{n F} \ln (K) \\
& n F E=R T \ln (K) \\
& \frac{n F E}{R T}=\ln (K) \\
& K=e^{\frac{n F E}{R T}} \\
& n=3 e^{-}, F=96485 \mathrm{C} \mathrm{mol}^{-}, E=1.29 \mathrm{~V}, R=8.314 \mathrm{~J} \mathrm{~mol}^{-} \mathrm{K}^{-}, T=298 \mathrm{~K}
\end{aligned}
$$
Step 4: Input all the known values into its appropriate variable...
$$
\begin{aligned}
& K=e^{\frac{(3 e)(96485)(1.29)}{(8.314)(298)}} \\
& K=e^{150.711} \\
& K=2.83 \times 10^{65}
\end{aligned}
$$
