Chemistry - Electrochemistry Question with Solution | TestHub

The cell reaction is written as : $\begin{array}{rl}& {\mathrm{E}}^{\circ }{\mathrm{Co}}^{2+}/{\mathrm{Co}}^{3+}=-1.81\mathrm{V}(\mathrm{SOP})\\ & \therefore {\mathrm{E}}^{\circ }{\mathrm{Co}}^{3+}/{\mathrm{Co}}^{2+}=+1.81\mathrm{V}\text{ (SRP) }\\ & {\mathrm{E}}_{\mathrm{A}//{\mathrm{Al}}^{3+}}^{\circ }=+1.66\mathrm{V}(\mathrm{SOP})\\ & \therefore {\mathrm{E}}_{{\mathrm{Al}}^{3+}/\mathrm{Al}}^{\circ }=-1.66\mathrm{V}(\mathrm{SRP})\\ & \end{array}$ Here ${\mathrm{E}}^{\circ }$ is given in standard Reduction Potential $\therefore {\mathrm{Co}}^{3+}/\mathrm{Co}$ will undergo reduction acting as cathode and ${\mathrm{Al}}^{3+}/\mathrm{Al}$ will undergo oxidation acting as anode. $\begin{array}{rl}\therefore {\mathrm{E}}_{\text{cell }}^{\circ }& ={\mathrm{E}}_{\text{cathode (Red.) }}^{\circ }-{\mathrm{E}}_{\text{Anode (oxide) }}^{\circ }\\ {\mathrm{E}}_{\text{cell }}^{\circ }& =1.81-(-1.66)\\ {\mathrm{E}}_{\text{cell }}^{\circ }& =+3.47\mathrm{V}\end{array}$