The ${\mathrm{E}}_{\text{cell }}^0$of the reaction is, 0.59 V at 25⁰ C. The equilibrium constant for the reaction is

$$\begin{gathered} {\mathrm{E}}_{\text{cell }}^{\mathrm{o}}=0.59V, \\ {\mathrm{MnO}}_4^{-}+{\mathrm{Fe}}^{2+}+{\mathrm{H}}^{+}\to {\mathrm{Mn}}^{2+}+{\mathrm{Fe}}^{3+}+{\mathrm{H}}_{2}O, \\ E={\mathrm{E}}_{\text{cell }}^{\mathrm{o}}-\frac{0.0591}{5}{\mathrm{logQ}}_{\mathrm{c}} \end{gathered}$$
At equilibrium$$\begin{gathered} \mathrm{E}=0,{\mathrm{Q}}_{\mathrm{c}}={\mathrm{K}}_{\mathrm{c}}\therefore {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}=\frac{0.0591}{5}{\mathrm{logK}}_{\mathrm{c}} \\ \therefore \frac{5\times 0.59}{0.59}={\mathrm{logK}}_{\mathrm{c}}50 \\ ={\mathrm{logK}}_{\mathrm{c}}\therefore {\mathrm{K}}_{\mathrm{c}}={10}^{50} \end{gathered}$$