Calculate the reduction potential at 25^oC for  ${\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}$ electrode if the concentration of ${\mathrm{Fe}}^{2+}$ ion is 10 times higher than ${\mathrm{Fe}}^{3+}$ ion $\left[ \mathrm{Given} {\mathrm{E}}_{{\mathrm{Fe}}^{3+}{\mathrm{Fe}}^{2+}}^0=0.77\mathrm{V}\right]$

The reduction of Fe³⁺ can be given as:

${\mathrm{Fe}}^{3+}+\hspace{0.17em}{\mathrm{e}}^{-}\to {\mathrm{Fe}}^{2+}$

The Nernst equation for the reduction is 

${\mathrm{E}}_{\mathrm{cell}}= {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{0.0591}{\mathrm{n}}\log \frac{\left[{\mathrm{Fe}}^{2+}\right]}{\left[{\mathrm{Fe}}^{3+}\right]}$

As the given concentration of Fe²⁺ ions are 10 times higher than Fe³⁺ ions ;

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}^{ }= 0.77-\frac{0.0591}{1}\log \frac{10\left[{\mathrm{Fe}}^{3+}\right]}{\left[{\mathrm{Fe}}^{2+}\right]} \\ {\mathrm{E}}_{\mathrm{cell}}=+0.71\mathrm{V} \end{gathered}$$