Q74 / 200 · NEET 2003 · Chemistry

On the basis of information available from the reaction $\frac{4}{3} \mathrm{Al}+\mathrm{O}_2 \rightarrow \frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3, \Delta \mathrm{G}=-827 \mathrm{~kJ} \mathrm{mol}^{-1}$ of $\mathrm{O}_2$, the minimum e.m.f required to carry out an electrolysis of $\mathrm{Al}_2 \mathrm{O}_3$ is $(\mathrm{F}=$ $\left.96500 \mathrm{C} \mathrm{mol}^{-1}\right):$

  • Correct answer
    A
    $2.14 \mathrm{~V}$
  • B
    $4.28 \mathrm{~V}$
  • C
    $6.42 \mathrm{~V}$
  • D
    $8.56 \mathrm{~V}$

Solution

For $\mathrm{O}_2, \Delta \mathrm{G}=-n \mathrm{FE}^{\circ}$ $\mathrm{E}^{\circ}=\frac{\Delta \mathrm{G}}{-n \mathrm{~F}}=\frac{-8,27,000}{-2 \times 96,500}=4.28$ Minimum EMF required to carry out electrolysis of $\mathrm{Al}_2 \mathrm{O}_3$ $=\frac{4.28}{2}=2.14 \mathrm{~V}$

Hence, the correct answer is option: 'A'.

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