91Ó°ÊÓ

In the reaction, chromium plates out on the cathode, therefore it gets reduced, while cadmium is oxidized. Getting the${\text{E}}_{\text{cell}}^{\text{o}}$ of the solution will give the voltage required to reverse the reaction.

$\begin{array}{l}{\text{E}}_{\text{oxidation}}^{\text{o}}{\text{= E}}_{{\text{Cr}}^{\text{3+}}}^{\text{o}}\text{= - 0.74³Õ}\\ {\text{E}}_{\text{reduction}}^{\text{o}}{\text{= E}}_{{\text{Cd}}^{\text{2+}}}^{\text{o}}\text{= - 0.40³Õ}\\ {\text{E}}_{\text{cell}}^{\text{o}}{\text{= â¶Ä‰E}}_{\text{reduction}}^{\text{o}}{\text{-E}}_{\text{oxidation}}^{\text{o}}\\ {\text{= â¶Ä‰E}}_{{\text{Cd}}^{\text{3+}}}^{\text{o}}{\text{-E}}_{{\text{Cr}}^{\text{2+}}}^{\text{o}}\\ \text{= â¶Ä‰-0.40-(-0.74)}\\ \text{= â¶Ä‰â¶Ä‰0.34³Õ}\end{array}$

${\text{E}}_{\text{cell}}^{\text{o}}\text{= 0.34³Õ}$, therefore, $0.34\text{V}$is needed to be applied to the cell to reverse the reaction. If $1.5\text{V}$of electricity is applied to the voltaic cell, the reaction will be reversed. Chromium will plate out, while the Cadmium metal on the cathode will dissolve into the electrolyte solution. Hence Chromium will plate out and cadmium will dissolve into the solution.