The Gibbs' energy for the decomposition of Al₂O₃ at 500°C is as follows: 2/3Al₂O₃ → 4/3Al + O₂; ΔrG = +960 kJmol⁻¹. The potential difference needed for the electrolytic reduction of aluminium oxide (Al₂O₃) at 500°C is at least:

The relationship between the Gibb's free energy change and the cell potential is as given below.
ΔGo = −nFEo
where n is the number of electrons transferred
In this case, the reaction is:
2/3Al₂O₃ → 4/3Al + O₂
The balanced half-reactions are:
Al³⁺ + 3e⁻ → Al (Reduction)
2O²⁻ → O₂ + 4e⁻ (Oxidation)
To balance the electrons, multiply the reduction half-reaction by 4 and the oxidation half-reaction by 3:
4Al³⁺ + 12e⁻ → 4Al
6O²⁻ → 3O₂ + 12e⁻
Adding the two half-reactions gives the overall reaction:
4Al³⁺ + 6O²⁻ → 4Al + 3O₂
This simplifies to:
2/3Al₂O₃ → 4/3Al + O₂
From the overall balanced reaction, n = 12 electrons transferred.
ΔGo = +960 kJmol⁻¹ = +960 × 10³ Jmol⁻¹
F = 96485 Cmol⁻¹
Substituting these values into the equation ΔGo = −nFEo:
+960 × 10³ Jmol⁻¹ = −12 × 96485 Cmol⁻¹ × Eo
Eo = −(960 × 10³ Jmol⁻¹) / (12 × 96485 Cmol⁻¹)
Eo = −0.828 V
The potential difference needed for the electrolytic reduction must be at least 0.828 V, however the given options do not include this value. The question may have some errors. It's important to note that the positive ΔG indicates the reaction is non-spontaneous under standard conditions, requiring an external voltage to proceed.