Question

On the basis of the information available from the reaction: $\frac 43 Al+O_2→\frac 23 Al_2O_2$, ∆G = -827 KJ mol-1 of O2, the minimum e.m.f. required to carry out electrolysis of Al2O3 is (F = 96500 C mol–1)

• A. 2.14 V
• B. 4.28 V
• C. 6.42 V
• D. 8.56 V

Answer 1

Answer: (A)

2.14 V

Answer 2

∆G = -nFE
Where:
∆G is the Gibbs free energy change (-827 kJ/mol).
n is the number of moles of electrons transferred.
F is the Faraday constant (96500 C/mol).
E is the cell potential.
$\frac 43 Al+O_2→\frac 23 Al_2O_2$
It's clear that $\frac 43$ moles of electrons are transferred for every 1 mole of O2 consumed. So, n = $\frac 43$
Now, plug these values into the equation:
-827 kJ = -nFE
-827 kJ = -($\frac 43$) . (96500 C/mol) . E
E = $\frac {(-827\ kJ) }{[-(\frac 43) . (96500 \ C/mol)] }$E ≈ 2.14 V

So, the minimum e.m.f. required to carry out the electrolysis of Al2O3 is approximately 2.14 V.
Therefore, the correct option is (A): 2.14 V.