First slide

Nernst Equation

Question

The Gibbs energy for the decomposition of Al₂O₃ at 500^oC is as follows
$\frac{2}{3} {Al}_{2} O_{3} ⟶ \frac{4}{3} Al + O_{2}; Δ_{r} G = + 960 kJ {mol}^{- 1}$
The potential difference needed for the electrolytic reduction of aluminium oxide (Al₂O₂) at 500^oC is atleast

Easy

A

4.5 V
B

3.0 V
C

2.5 V
D

5.0 V

Solution

$\begin{array}{l} 4 {Al}^{3 +} ⟶ 4 Al + 12 e^{-} \\ 12 e^{-} + 6 O^{2 -} ⟶ 3 O_{2} \end{array}$
or $\frac{4}{3} {Al}^{3 +} ⟶ \frac{4}{3} Al + 4 e^{-}$
$\begin{matrix} 4 e^{-} + 2 O^{2 -} ⟶ O_{2} \\ {ΔG}^{\circ} = - {nFE}^{\circ} \end{matrix}$
$\begin{array}{l} ΔG = + 960 kJ {mol}^{- 1} = 960 \times 1000 J {mol}^{- 1} + \\ n = 4 F = 4 \times 96500 C \end{array}$
$∴ 960 \times 1000 = - 4 \times 96500 \times E^{\circ}$
$E^{\circ} = - \frac{960000}{4 \times 96500} = - 2 .48 V$
Potential difference = 2.5 V

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