Chemistry - Electrochemistry Question with Solution | TestHub

$\begin{array}{rl}& \mathrm{M}\mid {\mathrm{M}}^{+2}\Vert \mathrm{X}/{\mathrm{X}}^{2-}\\ & {\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}={\mathrm{E}}_{\mathrm{M}/{\mathrm{M}}^{+2}}^{\mathrm{o}}+{\mathrm{E}}_{\mathrm{X}/{\mathrm{X}}^{-2}}^{\mathrm{o}}\\ & =-0.46+0.34=-0.12\mathrm{V}\end{array}$
As ${\mathrm{E}}_{\text{cell }}^{\mathrm{o}}$ is negative so anode becomes cathode and cathode become anode. Spontaneous reaction will be ${\mathrm{M}}^{+2}+{\mathrm{X}}^{2-}⟶\mathrm{M}+\mathrm{X}$