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

JEE Main · 2010 · 11 Apr 2010 · Offline · Redox Reactions and Electrochemistry

The Gibbs energy for the decomposition of

Al_{2} O_{3}

50 0^{\circ} C

is as follows:

\frac{2}{3} Al_{2} O_{3} \to \frac{4}{3} Al + O_{2}, Δ_{r} G = + 966 kJ mol^{- 1}

The potential difference needed for electrolytic reduction of

Al_{2} O_{3}

50 0^{\circ} C

is at least

Answer: (C)

Electrical work can drive a non-spontaneous reaction only if it at least matches the free-energy rise, through

Δ G = n F E

. In the written reaction each aluminium ion gains three electrons and

\frac{4}{3}

of them are formed, so

n = \frac{4}{3} \times 3 = 4

electrons pass. With

Δ G = 966 \times 1 0^{3}

J and

F = 96500

C mol

^{- 1}

, the least potential is

E = \frac{Δ G}{n F} = \frac{966 \times 1 0 ^{3}}{4 \times 96500} \approx 2.5

JEE Main · 2010 · 11 Apr 2010 · Offline · Single correct option