The Gibbs' energy for the decomposition of Al_2 O_3 at 500° C is as follows … The potential difference needed for the electrolytic reduction of…

The half cell reaction are at anode $\left.2 \mathrm{O}^{2-}+4 e^{-} \longrightarrow \mathrm{O}_2\right] \times 3$2O2−+4e−⟶O2​]×3 at cathode $\left.\mathrm{Al}^{3+} \longrightarrow \mathrm{Al}+3 e^{-}\right] 4$Al3+⟶Al+3e−]4 Net reaction $4 \mathrm{Al}+6 \mathrm{O}^{2-} \longrightarrow 3 \mathrm{O}_2+4 \mathrm{Al}$4Al+6O2−⟶3O2​+4Al $$\begin{aligned} & \text {or } \frac{4}{3} \mathrm{Al}+2 \mathrm{O}^{2-} \longrightarrow \mathrm{O}_2+\frac{4}{3} \mathrm{Al} \\ & \therefore \quad n=\frac{12}{3}=4 \\ & \Delta G^{\circ}=-n F E^{\circ} \\ & \text {Here, } \Delta G^{\circ}=+960 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & =960 \times 1000 \mathrm{~J} \mathrm{~mol}^{-1} \\ & n=4 \\ & F=96500 \text {coulomb } \\ & \therefore \quad 960 \times 1000=-4 \times 96500 \times E^{\circ} \\ & E^{\circ}=-\frac{960000}{4 \times 96500} \\ & =-2.48 \mathrm{~V} \\ & \text {Potential difference } \approx 2.5 \mathrm{~V} \end{aligned}$$​or 34​Al+2O2−⟶O2​+34​Al∴n=312​=4ΔG∘=−nFE∘Here, ΔG∘=+960 kJ mol−1=960×1000 J mol−1n=4F=96500coulomb ∴960×1000=−4×96500×E∘E∘=−4×96500960000​=−2.48 VPotential difference ≈2.5 V​