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Q. On the basis of the information available from the reaction,
$4 / 3 \text{Al} + \text{O}_{2} \rightarrow 2 / 3 \text{Al}_{2} \text{O}_{3} \text{;} \, \text{ΔG} = - 827 \text{kJmol}^{- 1}$ , the minimum e.m.f. required to carry out electrolysis of $Al_{2}O_{3}$ is $\left(\right.F=96500 \, Cmol^{- 1}\left.\right)$

NTA AbhyasNTA Abhyas 2022

Solution:

$\Delta G=-nFE^\circ $
$E^\circ =\frac{\Delta G}{- n F}-\frac{- 827000}{- 4 \times 96500}=2.14$