On the basis of the information available from the reaction
\(\frac{4}{3}Al + O_2 \longrightarrow \frac{2}{3}Al_2O_3, \Delta G = -827\text{ kJ/mol}^{-1}\)
The minimum e.m.f. required to carry the electrolysis of \(Al_2O_3\) is \((F = 96500 \text{ C mol}^{-1})\)
Answer & explanation
Correct answer: option 1
The correct answer is option 1. 2.14 V.
Given,
\(\frac{4}{3}Al + O_2 \longrightarrow \frac{2}{3}Al_2O_3, \Delta G = -827\text{ kJ/mol}^{-1}\)
\(n = 4\)
\(\Delta G = 872\text{ kJ/mol}^{-1}\) ['-' ve sign indicates exothermic raction]
or, \(\Delta G = 872000\text{ J/mol}^{-1}\)
\(F = 96500 C\)
We know that,
\(\Delta G= nFE\)
or, \(E = \frac{\Delta G}{nF}\)
or, \(E = \frac{872000}{4 × 96500}\)
or, \(E \approx 2.14 V\)