Q48 / 120 · NEET 2012 (Mains) · Chemistry
The Gibbs' energy for the decomposition of $\mathrm{Al}_2 \mathrm{O}_3$ at $500^{\circ} \mathrm{C}$ is as follows
$$
\begin{gathered}
\frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3 \longrightarrow \frac{4}{3} \mathrm{Al}+\mathrm{O}_2 ; \\
\Delta_r G=+960 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{gathered}
$$
The potential difference needed for the electrolytic reduction of aluminium oxide $\left(\mathrm{Al}_2 \mathrm{O}_3\right)$ at $500^{\circ} \mathrm{C}$ is at least
- A$4.5 \mathrm{~V}$
- B$3.0 \mathrm{~V}$
- Correct answerC$2.5 \mathrm{~V}$
- D$5.0 \mathrm{~V}$
Solution
The half cell reaction are at anode
$\left.2 \mathrm{O}^{2-}+4 e^{-} \longrightarrow \mathrm{O}_2\right] \times 3$
at cathode
$\left.\mathrm{Al}^{3+} \longrightarrow \mathrm{Al}+3 e^{-}\right] 4$
Net reaction
$4 \mathrm{Al}+6 \mathrm{O}^{2-} \longrightarrow 3 \mathrm{O}_2+4 \mathrm{Al}$
$\begin{aligned}
& \text {or } \frac{4}{3} \mathrm{Al}+2 \mathrm{O}^{2-} \longrightarrow \mathrm{O}_2+\frac{4}{3} \mathrm{Al} \\
& \therefore \quad n=\frac{12}{3}=4 \\
& \Delta G^{\circ}=-n F E^{\circ} \\
& \text {Here, } \Delta G^{\circ}=+960 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& =960 \times 1000 \mathrm{~J} \mathrm{~mol}^{-1} \\
& n=4 \\
& F=96500 \text {coulomb } \\
& \therefore \quad 960 \times 1000=-4 \times 96500 \times E^{\circ} \\
& E^{\circ}=-\frac{960000}{4 \times 96500} \\
& =-2.48 \mathrm{~V} \\
& \text {Potential difference } \approx 2.5 \mathrm{~V}
\end{aligned}$
Hence, the correct answer is option: 'C'.