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CUET · CHEMISTRY · PYQ PAPER 2023

On the basis of the information available from the reaction:
\(\frac{4}{3}Al+O_2\rightarrow \frac{2}{3}Al_2O_3, \Delta G = -827 kJ/mol\)
The minimum e.m.f required to carry the electrolysis of \(Al_2O_3\) is (\(F = 96500 C\ mol^{-1}\)):

  1. A \(2.14 V\)
  2. B \(4.29 V\)
  3. C \(6.42 V\)
  4. D \(8.56 V\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(2.14 V\)

Step-by-step Solution

Detailed explanation

\(\Delta G_{electrolysis} = -(-827 \text{ kJ/mol}) = 827 \text{ kJ/mol}\) \(n = 4 \text{ (from } \frac{4}{3}Al \rightarrow \frac{4}{3}Al^{3+} + 4e^- \text{ or } O_2 + 4e^- \rightarrow 2O^{2-} \text{ for formation)}\) \(E = \frac{\Delta G}{nF}\)…
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