Given \(E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}=+0.76 \mathrm{~V}\) and \(E_{\mathrm{I}_{2} / \mathrm{I}^{-}}^{\circ}=+0.55 \mathrm{~V}\). The equilibrium constant for the reaction taking place in galvanic cell consisting of above two electrodes is \(\left[\frac{2.303 R T}{F}=0.06\right]\)

\(E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\circ}=0.76 \mathrm{~V}\) and \(E_{\mathrm{I}_{2} / \mathrm{I}^{-}}^{\circ}=0.55 \mathrm{~V}\) \(K=\) ?
Also \(\frac{2.303 R T}{F}=0.06\)
Here, \(\mathrm{I}_{2} / \mathrm{I}^{-}\)will act as anode and \(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\) will act as cathode.
\(2 \mathrm{Fe}^{3+}+2 \mathrm{I}^{-} \longrightarrow 2 \mathrm{Fe}^{2+}+\mathrm{I}_{2}\)
Here, \(n=2\)
\(\begin{aligned}
E_{\text {cell }}^{\circ} &=E_{\text {cathode }}-E_{\text {anode }}=0.76-0.55=+0.21 \\
\Delta G^{\circ} &=-n F E_{\text {cell }}^{\circ}=-2(96500) \times 0.21 \\
&=-40530 \mathrm{~J}
\end{aligned}\)
Also, \(\Delta G=-2.303 R T \log K_{c}\)
\(\begin{aligned} \text { So, } \log K_{c} &=\frac{\Delta G}{-\left(\frac{2303 R T}{F}\right) \times F} \\ &=\frac{+40530}{-(0.06) \times 96500}=\frac{40530}{5790} \\ \log K_{c} &=7 \\ K_{c} &=1 \times 10^{7} \end{aligned}\)