Given ${\mathrm{E}}_{{\mathrm{Cr}}^{3+}/\mathrm{Cr}}^{\mathrm{o}}=-0.72\mathrm{V},{\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^{\mathrm{o}}=-0.42\mathrm{V}$. The potential for the cell 

$\mathrm{Cr} \overline{) {\mathrm{Cr}}^{+3}\left(0.1\mathrm{M}\right)}\parallel {\mathrm{Fe}}^{+2}\overline{)\left(0.01\mathrm{M}\right)} \mathrm{Fe} \mathrm{is}$

As ${\mathrm{E}}_{{\mathrm{Cr}}^{3+}/\mathrm{Cr}}^{\mathrm{o}}=-0.7\mathrm{V}$ and ${\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^{\mathrm{o}}=-0.42\mathrm{V}$

$\begin{array}{l}2\mathrm{Cr}+3{\mathrm{Fe}}^{2+}\to 3\mathrm{Fe}+2{\mathrm{Cr}}^{3+}\\ {\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}-\frac{0.0591}{6}\log \frac{{\left[{\mathrm{Cr}}^{3+}\right]}^2}{{\left[{\mathrm{Fe}}^{2+}\right]}^3}\\ =(-0.42+0.72)-\frac{0.0591}{6}\log \frac{{[0.1]}^2}{{[0.01]}^3}\\ =0.30-\frac{0.0591}{6}\log \frac{{[0.1]}^2}{{[0.01]}^3}\\ =0.30-\frac{0.0591}{6}\log {10}^4\\ {\mathrm{E}}_{\mathrm{cell}}=0.2606\mathrm{V}\end{array}$