On the basis of the reaction
\(\frac{4}{3}\text{Al} + \text{O}_2 \rightarrow \frac{2}{3}\text{Al}_2\text{O}_3, \quad \Delta_r G = -827\ \text{kJ}\) per mol of O2,
what is the minimum emf required to carry out the electrolysis of Al2O3? (1 F = 96500 C mol−1)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2.14 V
The Gibbs energy of a cell reaction and the emf are linked by \( \Delta_rG=-nFE_{cell} \). Electrolysis of \( \ce{Al2O3} \) is the reverse of the given reaction, so it needs \( \Delta_rG=+827\,\mathrm{kJ} \) per mol of \( \ce{O2} \) released. The applied emf must supply at least this much energy.
\( \displaystyle \frac{2}{3}\,\ce{Al2O3 ->} \frac{4}{3}\,\ce{Al + O2} \qquad \Delta_rG=+827\,\mathrm{kJ} \)
Electrons transferred (n): the reaction has \( \frac{4}{3} \) mol of \( \ce{Al^3+} \), and each takes 3 electrons, so \( n=\frac{4}{3}\times3=4 \). (The same count from the oxide side: \( \ce{2O^2- -> O2 + 4e^-} \).)
\( \displaystyle E=\frac{\Delta_rG}{nF} \)
\( \displaystyle E=\frac{827000\,\mathrm{J\,mol^{-1}}}{4\times96500\,\mathrm{C\,mol^{-1}}} \)
\( \displaystyle E=2.14\,\mathrm{V} \)
So the minimum emf needed for the electrolysis is 2.14 V.