Given ${\mathrm{E}}_{\mathrm{Cr}+3/\mathrm{Cr}}^0=-0.72\mathrm{V};{\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^0=-0.42\mathrm{V}$ The potential for the cell, $\mathrm{Cr}\mid {\mathrm{Cr}}^{3+}\left(0.1\mathrm{M}\parallel {\mathrm{Fe}}^{2+}(0.01\mathrm{M}\mid \right.$ Fe is

$$\begin{gathered} {{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cathode}}^{\mathrm{o}}- {\mathrm{E}}_{\mathrm{anode}}^{\mathrm{o}} \\ {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}=(-0.42)-(-0.72)=+0.30\mathrm{V} \end{gathered}$$

For the given galvanic cell the reaction is written as 

$$\begin{gathered} 2\mathrm{Cr} +\hspace{0.17em}3{\mathrm{Fe}}_{(0.01\mathrm{M})}^{2+}\leftrightharpoons 2{\mathrm{Cr}}_{(0.1\mathrm{M})}^{3+}+3\mathrm{Fe} \\ \mathrm{The} \mathrm{reaction} \mathrm{Quitient} \left(\mathrm{Qc}\right)=\frac{{\left[{\mathrm{Cr}}^{3+}\right]}^2}{{\left[{\mathrm{Fe}}^{2+}\right]}^3}=\frac{{[0.1]}^2}{{[0.01]}^3}={10}^4 \\ \mathrm{The} \mathrm{Nersnt} \mathrm{equation} \mathrm{is} \mathrm{written} \mathrm{as} \\ {\mathrm{E}}_{\mathrm{cell}={\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}}-\frac{0.0591}{\mathrm{n}}\log \mathrm{Qc} \\ {\mathrm{E}}_{\mathrm{cell}}=0.30-\frac{0.0591}{6}\log {10}^4 \\ \mathrm{on} \mathrm{solving} \\ {\mathrm{E}}_{\mathrm{cell} = }+0.261\mathrm{V}\hspace{0.17em} \end{gathered}$$