The Gibbs free energy for the decomposition of $\mathrm{Al}_{2} \mathrm{O}_{3}$ at $500^{\circ} \mat

Question

The Gibbs free energy for the decomposition of $\mathrm{Al}_{2} \mathrm{O}_{3}$ at $500^{\circ} \mathrm{C}$ is as follows:\ $\frac{2}{3} \mathrm{Al}_{2} \mathrm{O}_{3} ightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_{2} \quad ; \Delta G=+966 \mathrm{~kJ} \mathrm{~mol}^{-1}$\ The potential difference needed for electrolytic reduction of $\mathrm{Al}_{2} \mathrm{O}_{3}$ at $500^{\circ} \mathrm{C}$ is atleast:\

Professor PC Thomas Classes · Answer

$\Delta G=-\mathrm{nFE}$ So $\mathrm{E}=\frac{-\Delta G}{n F}=\frac{-966000}{4 	imes 96500}=2.5 \mathrm{~V}$\