The Gibbs energy for the decomposition of Al2O3 at 500oC is as follows 23Al2O3→43Al+O2 ΔrG=+966 kJ mol−1The potential difference needed for electrolytic reduction of Al2O3 at 500oC is at least

23Al2O3→43Al+O2Thus, 23×3(O2−)i.e., 2O2−→O2+4e−[∵ n=4]ΔG=+966 kJ mol−1=966×103 J mol−1ΔG=−nFEcell966×103=−4×96500×EcellEcell=2.5 V

The Gibbs energy for the decomposition of Al2O3 at 500oC is as follows 23Al2O3→43Al+O2 ΔrG=+966 kJ mol−1The potential difference needed for electrolytic reduction of Al2O3 at 500oC is at least

23Al2O3→43Al+O2Thus, 23×3(O2−)i.e., 2O2−→O2+4e−[∵ n=4]ΔG=+966 kJ mol−1=966×103 J mol−1ΔG=−nFEcell966×103=−4×96500×EcellEcell=2.5 V

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The Gibbs energy for the decomposition of ${Al}_{2} O_{3}$ at $500^{o} C$ 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 ${Al}_{2} O_{3}$ at $500^{o} C$ is at least

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2.5 V

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$\frac{2}{3} {Al}_{2} O_{3} \to \frac{4}{3} Al + O_{2}$

Thus, $\frac{2}{3} \times 3 (O^{2 -})$

i.e., $2 O^{2 -} \to O_{2} + 4 e^{-} [∵ n = 4]$

$\begin{array}{l} ΔG = + 966 {kJ mol}^{- 1} = 966 \times 10^{3} {J mol}^{- 1} \\ ΔG = - {nFE}_{cell} \\ 966 \times 10^{3} = - 4 \times 96500 \times E_{cell} \\ E_{cell} = 2 .5 V \end{array}$

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The Gibbs energy for the decomposition of Al2O3 at 500oC is as follows 23Al2O3→43Al+O2 ΔrG=+966 kJ mol−1The potential difference needed for electrolytic reduction of Al2O3 at 500oC is at least

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The Gibbs energy for the decomposition of ${Al}_{2} O_{3}$ at $500^{o} C$ 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 ${Al}_{2} O_{3}$ at $500^{o} C$ is at least