4/3 Al+O2 gives out 2/3 Al2O3. Find the minimum emf required to carry out the electrolysis of Al2O3.
delta G=-827kJ per mol of O2
We know that, ΔG = -nF E
Here, in the given reaction, 4/3 Al + O2 -> 2/3 Al2O3
The balanced equation becomes by multiplying both side with 3,
4 Al + 3O2 -> 2Al2O3
So, the reaction produces, Al -> Al+3 + 3e-
Hence, the no. of electrons exchanged, n = 3
F = 96.5 KJ/mol and ΔG = -827KJ/mol. Volt
So, E = -ΔG /nF = 827 / 3 x 96.5 = 2.85 Volt.