To find the standard potential of M^3+ / M electrode, the following cell is constituted: Pt / M / M^3+(0.001 mol L^-1) / Ag^+(0.01. mol L^-1) / Ag…

To find the standard potential of $\displaystyle \mathrm{M}^{3+} / \mathrm{M}$ electrode, the following cell is constituted : $\displaystyle \mathrm{Pt} / \mathrm{M} / \mathrm{M}^{3+}\left(0.001 \mathrm{~mol} \mathrm{~L}^{-1}\right) / \mathrm{Ag}^{+}(0.01$ $\displaystyle \left.\mathrm{mol} \mathrm{L}^{-1}\right) / \mathrm{Ag}$The emf of the cell is found to be $\displaystyle 0.421$ volt at $\displaystyle 298$ K. The standard potential of half reaction $\displaystyle \mathrm{M}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{M}$ at $\displaystyle 298$ K will be : (Given $\displaystyle \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\ominus}$ at $\displaystyle 298 \mathrm{~K}=0.80$ Volt)