Two half cell reactions are given below : CO^ 3+ +e^ - arrow Co^ 2+...

The cell reaction is written as : $$\begin{aligned} & \mathrm{E}^{\circ} \mathrm{Co}^{2+} / \mathrm{Co}^{3+}=-1.81 \mathrm{~V}(\mathrm{SOP}) \\ & \therefore \quad \mathrm{E}^{\circ} \mathrm{Co}^{3+} / \mathrm{Co}^{2+}=+1.81 \mathrm{~V} \text{ (SRP) } \\ & \mathrm{E}_{\mathrm{A} / / \mathrm{Al}^{3+}}^{\circ}=+1.66 \mathrm{~V}(\mathrm{SOP}) \\ & \therefore \mathrm{E}_{\mathrm{Al}^{3+} / \mathrm{Al}}^{\circ}=-1.66 \mathrm{~V}(\mathrm{SRP}) \\ & \end{aligned}$$ 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. $$\begin{aligned} \therefore \quad \mathrm{E}_{\text{cell }}^{\circ} & =\mathrm{E}_{\text{cathode (Red.) }}^{\circ}-\mathrm{E}_{\text{Anode (oxide) }}^{\circ} \\ \mathrm{E}_{\text{cell }}^{\circ} & =1.81-(-1.66) \\ \mathrm{E}_{\text{cell }}^{\circ} & =+3.47 \mathrm{~V} \end{aligned}$$