For the electrolysis of $Al_2O_3$, the reaction is the reverse of the given formation reaction: $\frac{2}{3}Al_2O_3(s) \rightarrow \frac{4}{3}Al(s) + O_2(g)$ Therefore, the Gibbs free energy change for the electrolysis is $\Delta G = +827 \text{ kJ mol}^{-1} = 827000 \text{ J mol}^{-1}$.

The number of electrons involved ($n$) in this reaction can be calculated from either the reduction of Al or oxidation of O: Number of moles of Al formed = $\frac{4}{3}$ moles. Since $Al^{3+} + 3e^- \rightarrow Al$, the total moles of electrons ($n$) = $\frac{4}{3} \times 3 = 4$ moles of electrons.

Using the relation between Gibbs free energy and cell potential: $\Delta G = nFE$ $827000 = 4 \times 96500 \times E$ $827000 = 386000 \times E$ $E = \frac{827000}{386000} = 2.142 \text{ V} \approx 2.14 \text{ V}$