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The Gibbs energy for the decomposition of $Al_2O_3$ at $500^\circ C$ is as follows: $\frac{2}{3}Al_2O_3 \rightarrow \frac{4}{3}Al + O_2$; $\Delta_rG = +960$ kJ $mol^{-1}$. The potential difference needed for the electrolytic reduction of aluminium oxide ($Al_2O_3$) at $500^\circ C$ is at least
A
5.0 V
B
4.5 V
C
3.0 V
D
2.5 V
Detailed Solution
$\frac{2}{3}Al_2O_3 \rightarrow \frac{4}{3}Al + O_2$
$\frac{4}{3}$ mol of $Al^{3+}$ is reduced to Al, so the number of electrons transferred is $n = \frac{4}{3}\times3 = 4$.
$\Delta G = -nFE \Rightarrow 960\times10^3 = -4\times96500\times E$
$E = -\frac{960\times10^3}{4\times96500} = -2.49$ V
So a potential difference of at least about 2.5 V is needed.
$\frac{4}{3}$ mol of $Al^{3+}$ is reduced to Al, so the number of electrons transferred is $n = \frac{4}{3}\times3 = 4$.
$\Delta G = -nFE \Rightarrow 960\times10^3 = -4\times96500\times E$
$E = -\frac{960\times10^3}{4\times96500} = -2.49$ V
So a potential difference of at least about 2.5 V is needed.