Answer
$$
\begin{aligned}
& K=e^{\frac{(4)(9645)(0.59)}{(8.314)(298)}} \\
& K=e^{91.90} \\
& K=8.21 \times 10^{39}
\end{aligned}
$$
Work Step by Step
Step 1: $\mathrm{N}_{2} \mathrm{H}_{5}^{+}(a q) \rightarrow N_{2}(a q)+5 H^{+}+4 e^{-} \quad E_{\text {red }}=-0.23 \mathrm{~V}$
$$
4 \mathrm{Fe}\left(\mathrm{CN}_{6}\right)^{-3}(a q)+4 e^{-} \rightarrow 4 \mathrm{Fe}\left(\mathrm{CN}_{6}\right)^{-4}(a q) \quad E_{\text {red }}=+0.36 \mathrm{~V}
$$
Step 2: $E_{\text {red }}($ cathode/reduced $)-E_{\text {red }}$ (Anode/Oxidized)
$$
0.36 \mathrm{~V}-(-0.23 \mathrm{~V})=0.59 \mathrm{~V}=E_{\text {Cell }}
$$
Step 3: Convert the emf calculated to the Equilibrium Constant via the formula...
$$
\begin{aligned}
& E_{\text {Cell }}=\frac{R T}{n F} \ln (K) \\
& n F E=R T \ln (K) \\
& \frac{n F E}{R T}=\ln (K) \\
& K=e^{\frac{n F E}{R T}} \\
& n=4 e^{-}, F=96485 \mathrm{C} \mathrm{mol}^{-}, E=0.59 \mathrm{~V}, R=8.314 \mathrm{~J} \mathrm{~mol}^{-} \mathrm{K}^{-}, T=298 \mathrm{~K}
\end{aligned}
$$
Step 4: Input all the known values into its appropriate variable...
$$
\begin{aligned}
& K=e^{\frac{(4)(9645)(0.59)}{(8.314)(298)}} \\
& K=e^{91.90} \\
& K=8.21 \times 10^{39}
\end{aligned}
$$
