The Gibbs energy for the decomposition Al2O3 at 500 °C is as follows. 23Al2O3→43Al+O2; ΔrG=966 kJ mol−1 The potential difference needed to carry out the above reaction electrolytically at 500 °C is at least

A
$5.0 V$
B
$4.5 V$
C
$3.0 V$
D
$2.5 V$

Solution:

The basic reaction area: $\frac{2}{3} (2 {Al}^{3 +}) + 4 e^{-} \to \frac{4}{3} Al$ and $\frac{2}{3} (3 O^{2 -}) \to O_{2} + 4 e^{-}$

Thus, 4 electrons are involved in the redox reaction $\frac{2}{3} {Al}_{2} O_{3} \to \frac{4}{3} Al + O_{2}$ .

Thus, using the expression !:,. $Δ G = - n F E_{cell}$ we get $E_{cell} = - \frac{Δ G}{n F} = - \frac{966 \times 10^{3} J {mol}^{- 1}}{(4) (96500 C {mol}^{- 1})} = - 2.50 V$

Thus, the minimum potential difference needed for carrying out the given reaction electrolytically will be 2.50 V.

The Gibbs energy for the decomposition ${Al}_{2} O_{3}$ at 500 °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 to carry out the above reaction electrolytically at 500 °C is at least

Solution:

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