The Gibbs energy for the decomposition of $$Al_2O_3$$ at 500°C is as follows :
$$\frac{2}{3}Al_2O_3 \rightarrow \frac{4}{3}Al + O_2$$, $$\Delta_r G = +940$$ kJ mol$$^{-1}$$
The potential difference needed for the electrolytic reduction of aluminium oxide at 500°C should be at least :

The given reaction is $$\frac{2}{3}Al_{2}O_{3} \rightarrow \frac{4}{3}Al + O_{2}$$ with $$\Delta_{r}G = +940 \text{ kJ mol}^{-1}$$. To find the minimum electrolytic potential, we use the relation $$\Delta_{r}G = nFE$$, where $$n$$ is the number of moles of electrons transferred and $$F = 96485 \text{ C mol}^{-1}$$ is Faraday's constant.

In this reaction, aluminium is reduced from the $$+3$$ to the $$0$$ oxidation state. For $$\frac{4}{3}$$ moles of Al, the total electron transfer is $$\frac{4}{3} \times 3 = 4$$ moles of electrons. Equivalently, oxygen goes from $$-2$$ to $$0$$: for $$1 \text{ mol } O_{2}$$ (from 2 oxide ions), $$2 \times 2 = 4$$ electrons are released, confirming $$n = 4$$.

Substituting into the formula: $$E = \frac{\Delta_{r}G}{nF} = \frac{940 \times 10^{3}}{4 \times 96485} = \frac{940000}{385940} \approx 2.43 \text{ V}$$.

The minimum potential difference needed for the electrolytic reduction of aluminium oxide at 500°C is therefore at least approximately $$2.43 \text{ V}$$. Among the given options, the closest value that meets or exceeds this threshold is $$3.0 \text{ V}$$.