Question

The Gibbs energy change of the reaction (in kJ mol⁻¹) corresponding to the following cell
Cr | Cr³⁺(0.1M) || Fe²⁺(0.01 M) | Fe
(Given: \(E^\circ_{Cr^{3+}/Cr} = -0.75V\); \(E^\circ_{Fe^{2+}/Fe} = -0.45V\), 1F = 96,500 C mol⁻¹)

• A. -150.9
• B. +150.9
• C. -173.7
• D. +173.7

Hint:

A positive \(E_{cell}\) indicates a spontaneous reaction, which must have a negative \(\Delta G\). This helps eliminate options with the wrong sign. Be careful to use the correct number of electrons (n) in both the Nernst equation and the \(\Delta G\) formula.

Answer 1

The Correct Option is A

Step 1: Cell Reaction and n-factor: Anode (Oxidation): \( 2\text{Cr} \to 2\text{Cr}^{3+} + 6e^- \) Cathode (Reduction): \( 3\text{Fe}^{2+} + 6e^- \to 3\text{Fe} \) Overall Reaction: \( 2\text{Cr} + 3\text{Fe}^{2+} \to 2\text{Cr}^{3+} + 3\text{Fe} \) Number of electrons transferred, \( n = 6 \).
Step 2: Calculate Standard Cell Potential (\( E^\circ_{\text{cell}} \)): \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] \[ E^\circ_{\text{cell}} = E^\circ_{\text{Fe}^{2+}|\text{Fe}} - E^\circ_{\text{Cr}^{3+}|\text{Cr}} \] \[ E^\circ_{\text{cell}} = -0.45\text{V} - (-0.75\text{V}) = -0.45 + 0.75 = +0.30\text{V} \]
Step 3: Calculate Cell Potential (\( E_{\text{cell}} \)) using Nernst Equation: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log Q \] \[ Q = \frac{[\text{Cr}^{3+}]^2}{[\text{Fe}^{2+}]^3} = \frac{(0.1)^2}{(0.01)^3} = \frac{10^{-2}}{10^{-6}} = 10^4 \] \[ E_{\text{cell}} = 0.30 - \frac{0.059}{6} \log(10^4) \] \[ E_{\text{cell}} = 0.30 - \frac{0.059}{6} \times 4 \] \[ E_{\text{cell}} = 0.30 - 0.0393 = 0.2607\text{V} \]
Step 4: Calculate Gibbs Energy Change (\( \Delta G \)): \[ \Delta G = -nFE_{\text{cell}} \] \[ \Delta G = -6 \times 96500 \times 0.2607 \] \[ \Delta G = -150945.3 \, \text{J/mol} \] \[ \Delta G \approx -150.9 \, \text{kJ/mol} \]