Calculate the potential of Iron electrode in which the concentration of Fe^2+ ion is 0.01 M. ( E^0 Fe^2+ / Fe =-0.45 V at 298 K ) [Given: log 10=1 ]

\[\begin{aligned} & \mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}^{2}}=\mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}^{-\frac{0.059}{2}} \log \frac{1}{\left[F e^{2+}\right]}}^{\mathrm{O}} \\ & =-0.45 \mathrm{~V}-\frac{0.059}{2} \log \frac{1}{0.01} \\ & =-0.45 \mathrm{~V}-0.059 \mathrm{~V} \\ & =-0.509 \mathrm{~V} \end{aligned} \]