The Gibbs energy for the decomposition of Al2O3 at 5000 C is as follows 23Al2O3⟶43Al+O2;ΔrG=+960kJ The voltage needed for the electrolytic reduction of aluminium oxide Al2O3 at 5000 C is at least

The ionic reactions are 23× 2Al3++ 4e-→43Al 23×3O2-→O2+ 4e-number of electron involved are (n)= 4 △G=-nFE 960×103=-4×96500×E on solving E--2.5V

The Gibbs energy for the decomposition of Al2O3 at 5000 C is as follows 23Al2O3⟶43Al+O2;ΔrG=+960kJ The voltage needed for the electrolytic reduction of aluminium oxide Al2O3 at 5000 C is at least

The ionic reactions are 23× 2Al3++ 4e-→43Al 23×3O2-→O2+ 4e-number of electron involved are (n)= 4 △G=-nFE 960×103=-4×96500×E on solving E--2.5V

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The Gibbs energy for the decomposition of ${Al}_{2} O_{3}$ at 500⁰ C is as follows $\frac{2}{3} {Al}_{2} O_{3} ⟶ \frac{4}{3} Al + O_{2}; Δ_{r} G = + 960 kJ$ The voltage needed for the electrolytic reduction of aluminium oxide $({Al}_{2} O_{3})$ at 500⁰ C is at least

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– 5.0 V

– 4.5 V

– 3.0 V

– 2.5 V

answer is C.

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Detailed Solution

The ionic reactions are

$\frac{2}{3} \times 2 {Al}^{3 +} + 4 e^{-} \to \frac{4}{3} Al \frac{2}{3} \times 3 O^{2 -} \to O_{2} + 4 e^{-}$

number of electron involved are (n)= 4

$△ G = - n F E 960 \times 10^{3} = - 4 \times 96500 \times E o n s o l v i n g E - - 2.5 V$

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The Gibbs energy for the decomposition of Al2O3 at 5000 C is as follows 23Al2O3⟶43Al+O2;ΔrG=+960kJ The voltage needed for the electrolytic reduction of aluminium oxide Al2O3 at 5000 C is at least

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The Gibbs energy for the decomposition of ${Al}_{2} O_{3}$ at 500⁰ C is as follows $\frac{2}{3} {Al}_{2} O_{3} ⟶ \frac{4}{3} Al + O_{2}; Δ_{r} G = + 960 kJ$ The voltage needed for the electrolytic reduction of aluminium oxide $({Al}_{2} O_{3})$ at 500⁰ C is at least