Given $E^{\circ }\left({\mathrm{Cr}}^{3+}\mid \mathrm{Cr}\right)=-0.72\mathrm{V}$ and $E^{\circ }\left({\mathrm{Fe}}^{2+}\mid \mathrm{Fe}\right)=-0.42\mathrm{V}$ The potential of the cell $\mathrm{Cr}\left|{\mathrm{Cr}}^{3+}\left(0.1\mathrm{M}\right)\parallel {\mathrm{Fe}}^{2+}\left(0.01\mathrm{M}\right)\right|\mathrm{Fe}$ is

The cell potential is given by $E_{\text{cell }}=E_{\mathrm{R}}-E_{\mathrm{L}}$ where 

Right half-cell  ${\mathrm{Fe}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Fe}$

 $$\begin{gathered} E_{\mathrm{R}}=E_{\mathrm{R}}^{\circ }-\frac{RT}{2F}\ln \frac{1}{\left[{\mathrm{Fe}}^{2+}\right]/c^{\circ }} \\ =-0.42\mathrm{V}+\left(\frac{0.059\mathrm{V}}{2}\right)\log \left(0.01\right) \\ =-0.42\mathrm{V}-0.06\mathrm{V}=-0.48\mathrm{V} \end{gathered}$$ 

Left half-cell  ${\mathrm{Cr}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Cr}$

$$\begin{gathered} E_{\mathrm{L}}=E_{\mathrm{L}}^{\circ }-\frac{RT}{3F}\ln \frac{1}{\left[{\mathrm{Cr}}^{3+}\right]/c^{\circ }}=-0.72+\left(\frac{0.059\mathrm{V}}{3}\right)\log \left(0.1\right) \\ =-0.72\mathrm{V}-0.02\mathrm{V}=-0.74\mathrm{V} \end{gathered}$$

Hence, $E_{\mathrm{cell}}=-0.48\mathrm{V}-(-0.74\mathrm{V})=0.26\mathrm{V}$