The Gibbs energy for the decomposition of Al_2O_3 at 500^ is as follows: 2/3Al_2O_3 → 4/3Al + O_2, _r G = + 966 kJ mol^-1 The potential difference nee - Chemistry MCQ AIEEE 2026 | ExamTest.live

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The Gibbs energy for the decomposition of Al_{2} O_{3} at 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} at 50 0^{\circ} C is at least:

Solution & Step-by-step Explanation

The relationship between Δ G and cell potential is: Δ G = - n FE For the reaction \frac{2}{3} Al_{2} O_{3} \to \frac{4}{3} Al + O_{2} :Aluminum changes oxidation state from + 3 to 0 .Total electrons transferred for \frac{4}{3} moles of Al = \frac{4}{3} \times 3 = 4 . 966 \times 1 0^{3} = - 4 \times 96500 \times E E = - \frac{966000}{386000} = - 2.5 V The potential difference needed is 2.5 V .

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The Gibbs energy for the decomposition of Al_{2} O_{3} at 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} at 50 0^{\circ} C is at least:

4.5 V

3.0 V

2.5 V

5.0 V

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