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AIPMT PRELIMS2003Chemistry-Electrochemistry

AIPMT PRELIMS 2003 Chemistry Electrolysis MCQ Question

Type: MCQ-numerical-Medium-Class 12

On the basis of the information available from the reaction: 43Al+O2→23Al2O3,ΔG=−827kJ mol−1\frac{4}{3} \text{Al} + \text{O}_2 \rightarrow \frac{2}{3} \text{Al}_2\text{O}_3, \Delta G = -827 \text{kJ mol}^{-1} of O2\text{O}_2, the minimum e.m.f. required to carry out electrolysis of Al2O3\text{Al}_2\text{O}_3 is (F=96500 C mol−1)(F = 96500 \text{ C mol}^{-1})

A

2.14 V

B

4.28 V

C

6.42 V

D

8.56 V

Correct Answer

Option A

Detailed Explanation

The minimum e.m.f. required for electrolysis is calculated using the formula E=−ΔGnFE = \frac{-\Delta G}{nF}. Here, ΔG=−827kJ mol−1\Delta G = -827 \text{kJ mol}^{-1}, n=4n = 4, and F=96500 C mol−1F = 96500 \text{ C mol}^{-1}.

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