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 (\(F = 96500 C\ mol^{-1}\)):
- A \(2.14 V\)
- B \(4.29 V\)
- C \(6.42 V\)
- D \(8.56 V\)
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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