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The Gibbs energy for the decomposition of $\mathrm{Al}_2 \mathrm{O}_3$ at $500 \mathrm{~C}$ is as follows :
$\frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3 \rightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_2, \Delta_r G=+940 \mathrm{~kJ} \mathrm{~mol}^{-1}$
The potential difference needed for the electrolytic reduction of aluminium oxide at $500^{\circ} \mathrm{C}$ should be at least :
Options:
$\frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3 \rightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_2, \Delta_r G=+940 \mathrm{~kJ} \mathrm{~mol}^{-1}$
The potential difference needed for the electrolytic reduction of aluminium oxide at $500^{\circ} \mathrm{C}$ should be at least :
Solution:
1297 Upvotes
Verified Answer
The correct answer is:
$3.0 \mathrm{~V}$
$3.0 \mathrm{~V}$
In the reaction
$\frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3 \longrightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_2$
For the oxidation half-reaction
$\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}$
no. of electron transfered $(n)=3$
$\begin{aligned}
& \Delta \mathrm{G}^{\circ}=-n \mathrm{FE}^{\circ} \\
& 940=3 \times 96500 \times \mathrm{E}^{\circ}
\end{aligned}$
$\begin{aligned}
\mathrm{E}^{\circ} & =\frac{940 \times 10^3 \mathrm{~J}}{3 \times 96500} \\
& =3.24 \approx 3 \mathrm{~V}
\end{aligned}$
$\frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3 \longrightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_2$
For the oxidation half-reaction
$\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}$
no. of electron transfered $(n)=3$
$\begin{aligned}
& \Delta \mathrm{G}^{\circ}=-n \mathrm{FE}^{\circ} \\
& 940=3 \times 96500 \times \mathrm{E}^{\circ}
\end{aligned}$
$\begin{aligned}
\mathrm{E}^{\circ} & =\frac{940 \times 10^3 \mathrm{~J}}{3 \times 96500} \\
& =3.24 \approx 3 \mathrm{~V}
\end{aligned}$
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