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: Understanding the Concept:
The Gibbs free energy change (\(\Delta G\)) of an electrochemical cell reaction is related to its cell potential (\(E_{cell}\)). For non-standard conditions, we first need to calculate the standard cell potential (\(E^\circ_{cell}\)) and then use the Nernst equation to find the actual cell potential (\(E_{cell}\)) under the given concentrations.
Step 2: Key Formulae:
1. Standard Cell Potential: \(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}\)
2. Nernst Equation (at 298 K): \(E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log_{10} Q\)
3. Gibbs Energy Change: \(\Delta G = -nFE_{cell}\)
Step 3: Detailed Explanation:
Identify Anode, Cathode, and Overall Reaction:
- Anode (Oxidation, left side): Cr(s) → Cr³⁺(aq) + 3e⁻
- Cathode (Reduction, right side): Fe²⁺(aq) + 2e⁻ → Fe(s)
To balance the electrons, multiply the anode reaction by 2 and the cathode reaction by 3. - Overall reaction: \(2\text{Cr}(s) + 3\text{Fe}^{2+}(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 3\text{Fe}(s)\)
- The number of moles of electrons transferred, \(n = 6\).
Calculate Standard Cell Potential (\(E^\circ_{cell}\)):
\[ E^\circ_{cell} = E^\circ_{\text{cathode (Fe)}} - E^\circ_{\text{anode (Cr)}} = (-0.45 \text{ V}) - (-0.75 \text{ V}) = +0.30 \text{ V} \] Calculate the Reaction Quotient (Q):
\[ Q = \frac{[\text{Products}]^{\text{coeff}}}{[\text{Reactants}]^{\text{coeff}}} = \frac{[\text{Cr}^{3+}]^2}{[\text{Fe}^{2+}]^3} = \frac{(0.1)^2}{(0.01)^3} = \frac{(10^{-1})^2}{(10^{-2})^3} = \frac{10^{-2}}{10^{-6}} = 10^4 \] Calculate Cell Potential (\(E_{cell}\)) using Nernst Equation:
\[ E_{cell} = 0.30 - \frac{0.0591}{6} \log_{10}(10^4) \] \[ E_{cell} = 0.30 - \frac{0.0591}{6} \times 4 = 0.30 - \frac{0.2364}{6} = 0.30 - 0.0394 = 0.2606 \text{ V} \] Calculate Gibbs Energy Change (\(\Delta G\)):
\[ \Delta G = -nFE_{cell} = -6 \times 96500 \text{ C mol}^{-1} \times 0.2606 \text{ V} \] \[ \Delta G = -579000 \times 0.2606 \text{ J mol}^{-1} = -150887.4 \text{ J mol}^{-1} \] Convert to kJ mol⁻¹:
\[ \Delta G = -150.8874 \text{ kJ mol}^{-1} \approx -150.9 \text{ kJ mol}^{-1} \] Step 4: Final Answer:
The Gibbs energy change for the reaction is -150.9 kJ mol⁻¹. Therefore, option (A) is correct.