If $E^{\circ }\left({\mathrm{Cr}}^{2+},{\mathrm{Cr}}^{3+}\mid \mathrm{Pt}\right)=-0.42\mathrm{V}$ and $E^{\circ }\left({\mathrm{Cr}}^{2+}\mid \mathrm{Cr}\right)=-0.90\mathrm{V}$ then $E^{\circ }$ for the half-cell ${\mathrm{Cr}}^{3+}\mid \mathrm{Cr}$ will be

Add $\begin{array}{l}{\mathrm{Cr}}^{3+}+{\mathrm{e}}^{-}\to {\mathrm{Cr}}^{2+}\mathrm{\Delta }{G}_1^{\circ }=-\left(1\right)F(-0.42\mathrm{V})\\ {\mathrm{Cr}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cr}\mathrm{\Delta }{G}_2=-\left(2\right)F(-0.90\mathrm{V})\\ \overline{{\mathrm{Cr}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Cr}}\overline{\mathrm{\Delta }{G}_3^{\circ }=-F(-0.42-2\times 0.90)\mathrm{V}}\end{array}$  

Also $\mathrm{\Delta }{G}_3^0=-3F{E_{{\mathrm{Cr}}^{3+}}}^{\circ }\mathrm{Cr}$ Hence $3E_{{\mathrm{Cr}}^{3+}\mid \mathrm{Cr}}^{\circ }=-2.22\mathrm{V}\text{ or }{E}^{\circ }=-0.74\mathrm{V}$