On the basis of the information available from the reaction
\(\frac{4}{3}Al + O_2 \longrightarrow \frac{2}{3}Al_2O_3,\ \Delta G = -827\text{ kJ mol}^{-1}\)
The minimum e.m.f. required to carry the electrolysis of \(Al_2O_3\) is \((F = 96500 \text{ C mol}^{-1})\)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2.14 V
The Gibbs energy change of a redox reaction is related to the e.m.f. by
\( \Delta G = -nFE \)
where n is the number of electrons transferred. Electrolysis of \( \ce{Al2O3} \) is the reverse of the given reaction, so it needs +827 kJ, and the external e.m.f. must be at least the value of E that corresponds to this ΔG.
Electrons transferred: in the given reaction 4/3 mol Al goes from 0 to +3.
\( n = \frac{4}{3} \times 3 = 4 \)
(the same count from oxygen: 1 mol \( \ce{O2} \) → \( \ce{2O^{2-}} \) takes 4 electrons).
\( \displaystyle E = \frac{\Delta G}{nF} = \frac{827 \times 10^{3}\ \mathrm{J\,mol^{-1}}}{4 \times 96500\ \mathrm{C\,mol^{-1}}} \)
\( E = 2.14\ \mathrm{V} \)
So the minimum e.m.f. required is 2.14 V.