4/3 Al+O₂ gives out 2/3 Al₂O_3. Find the minimum emf required to carry out the electrolysis of Al₂O_3.

delta G=-827kJ per mol of O₂

Dear student!

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.