Question

\( \Delta G^\circ \, (\text{in kJ mol}^{-1}) \text{ for the cell reaction is} \) 
\( \text{Cu}^{2+}(aq) + \text{Fe}(s) \rightarrow \text{Fe}^{2+}(aq) + \text{Cu}(s) \) 
\( \left[\text{Given} \, E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.34 \, \text{V}, \, E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44 \, \text{V} \, \text{and} \, F = 96,500 \, \text{C mol}^{-1} \right] \)

• A. +150.5
• B. -150.5
• C. +140.2
• D. -140.2

Hint:

The sign of the Gibbs free energy change helps determine if the reaction is spontaneous. A negative \( \Delta G^\circ \) indicates that the reaction is spontaneous.

Answer 1

The Correct Option is B

The standard Gibbs free energy change (\( \Delta G^\circ \)) for a cell reaction can be calculated using the equation: \[ \Delta G^\circ = -nF E^\circ_{\text{cell}} \] Where: - \( n \) is the number of moles of electrons involved in the reaction,
- \( F \) is the Faraday constant,
- \( E^\circ_{\text{cell}} \) is the cell potential, which can be calculated as:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] For this reaction: - The cathode is the Cu\(^2+\)/Cu half-reaction: \( E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.34V \),
- The anode is the Fe\(^2+\)/Fe half-reaction: \( E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44V \).
Thus: \[ E^\circ_{\text{cell}} = 0.34V - (-0.44V) = 0.78V \] The number of electrons transferred \( n \) is 2 (as both reactions involve a 2-electron transfer). Now, substitute the values into the Gibbs free energy equation: \[ \Delta G^\circ = -2 \times 96,500 \times 0.78 = -150.5 \, \text{kJ/mol} \] Thus, the correct answer is (2) -150.5 kJ/mol.