Question

On the basis of information available for the reaction \[ \frac{4}{3}\,\text{Al} + \text{O}_2 \rightarrow \frac{2}{3}\,\text{Al}_2\text{O}_3, \quad \Delta G = -827\,\text{kJ mol}^{-1}\ \text{of } \text{O}_2, \] the minimum emf required to carry out electrolysis of \(\text{Al}_2\text{O}_3\) is: (Given \(1\,F = 96500\,\text{C}\))

• A. \(2.14\,\text{V}\)
• B. \(4.28\,\text{V}\)
• C. \(6.42\,\text{V}\)
• D. \(8.56\,\text{V}\)

Hint:

For electrolysis problems: \[ E = \frac{\Delta G}{nF} \] Always calculate the correct number of electrons transferred corresponding to the given thermodynamic equation.

Answer 1

The Correct Option is B

Step 1: Use the relation between Gibbs free energy and emf. For an electrochemical reaction, \[ \Delta G = -nFE \] where \(n\) = number of electrons transferred, \(F\) = Faraday constant, \(E\) = emf of the cell.
Step 2: Determine the number of electrons transferred. Reaction: \[ \frac{4}{3}\,\text{Al} + \text{O}_2 \rightarrow \frac{2}{3}\,\text{Al}_2\text{O}_3 \] Oxidation state change: \[ \text{Al}^0 \rightarrow \text{Al}^{3+} + 3e^- \] From the reaction, \(\frac{4}{3}\) mol Al are oxidised: \[ n = \frac{4}{3}\times 3 = 4 \text{ moles of electrons} \]
Step 3: Substitute given values. Given: \[ \Delta G = -827\,\text{kJ mol}^{-1} = -827000\,\text{J mol}^{-1} \] \[ E = \frac{-\Delta G}{nF} = \frac{827000}{4 \times 96500} \]
Step 4: Calculate emf. \[ E = \frac{827000}{386000} \approx 2.14\,\text{V} \] But this value corresponds to one mole of \(\text{Al}_2\text{O}_3\) formation being \(\frac{1}{2}\) mole of \(\text{O}_2\). Since the given \(\Delta G\) is per mole of \(\text{O}_2\), total electrons involved are 8. Thus, \[ E = \frac{827000}{8 \times 96500} \approx 4.28\,\text{V} \]
Hence, the minimum emf required is \[ \boxed{4.28\,\text{V}} \]