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The Gibbs energy for the decomposition of $\mathrm{Al}_2 \mathrm{O}_3$ at $500^{\circ} \mathrm{C}$ is as follows :
$$
\frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3 \rightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_2, \Delta_{\mathrm{r}} \mathrm{G}=+966 \mathrm{~kJ} \mathrm{~mol}^{-1}
$$
The potential difference needed for electrolytic reduction of $\mathrm{Al}_2 \mathrm{O}_3$ at $500^{\circ} \mathrm{C}$ is at least
Options:
$$
\frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3 \rightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_2, \Delta_{\mathrm{r}} \mathrm{G}=+966 \mathrm{~kJ} \mathrm{~mol}^{-1}
$$
The potential difference needed for electrolytic reduction of $\mathrm{Al}_2 \mathrm{O}_3$ at $500^{\circ} \mathrm{C}$ is at least
Solution:
1943 Upvotes
Verified Answer
The correct answer is:
$2.5 \mathrm{~V}$
$2.5 \mathrm{~V}$
$\Delta G=-n F E \quad \Rightarrow E=\frac{-\Delta G}{n F}$
$E=-\frac{966 \times 10^3}{4 \times 96500}$
$=-2.5 \mathrm{~V}$
$\therefore$ The potential difference needed for the reduction $=2.5 \mathrm{~V}$
$E=-\frac{966 \times 10^3}{4 \times 96500}$
$=-2.5 \mathrm{~V}$
$\therefore$ The potential difference needed for the reduction $=2.5 \mathrm{~V}$
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