The given reaction for the decomposition of $Al_2O_3$ is: $\frac{2}{3}Al_2O_3 \rightarrow \frac{4}{3}Al + O_2$ In this reaction, $\frac{4}{3}$ moles of $Al^{3+}$ ions are reduced to Al atoms. The number of electrons involved ($n$) can be calculated from either the reduction of Al or oxidation of O: For Al: $n = \frac{4}{3} \times 3 = 4$ moles of electrons. For O: $2O^{2-} \rightarrow O_2 + 4e^-$, so $n = 4$. We know the relation between standard Gibbs energy and minimum external potential ($E$) required for electrolysis: $\Delta_rG = nFE$ Given $\Delta_rG = +960 \text{ kJ mol}^{-1} = 960 \times 10^3 \text{ J mol}^{-1}$ and $F \approx 96500 \text{ C mol}^{-1}$. $960 \times 10^3 = 4 \times 96500 \times E$ $E = \frac{960 \times 10^3}{4 \times 96500} = \frac{960000}{386000} = 2.487 \text{ V} \approx 2.5 \text{ V}$ Hence, the minimum potential difference needed for the electrolytic reduction is $2.5 \text{ V}$.