On the basis of information available from the reaction 4Al + 3O2 →2Al2O3; ∆G = – 965 kJ/mol of O2.The minimum EMF required to carry out electrolysis of Al2O3 is

Given ∆G = – 965 kJ/mol = – 965000 J/mol of O2 Therefore, number of electrons, n = 2 × 2 = 4 F = 965000 C/mol Balancing both anodic and cathodic half half cell, Anode : [Al3 → Al + 3e–] × 4 Cathode : [2O2– + 4e– → O2] × 3 Net reaction : 4Al + 6O2– → 3O2 + 4Al Or 4/3Al + 2O2– → O2 + 4/3Al We know that, ∆G = – nFE –965000 = – 4 × 96500 × E E = 2.5 V