Question 6 of 29intermediate🔧 ApplyNumerical3 marks

The cell in which the following reaction occurs: 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(s) has E°cell = 0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Correct Answer

Standard Gibbs energy (ΔrG°) = -45.54 kJ mol⁻¹ Equilibrium constant (Kc) = 9.62×109.62 \times 10⁷

Exercise: Intext Questions | Q: 2.6 | (Page Page 11)
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Explanation

This numerical problem tests the relationship between standard cell potential, Gibbs energy, and equilibrium constant. The key formulas from the chapter are: ΔrG° = -nFE°(cell) and ΔrG° = -RT ln K. The number of electrons (n) is determined from the balanced redox reaction, where 2Fe³⁺ gains 2 electrons and 2I⁻ loses 2 electrons, making n = 2.

Solution Steps

  1. Step 1: Determine the number of electrons (n) transferred From the reaction: 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(s) Fe³⁺ + e⁻ → Fe²⁺ (reduction, 1 electron per Fe³⁺) 2I⁻ → I₂ + 2e⁻ (oxidation, 2 electrons) Therefore, n = 2

  2. Step 2: Calculate Standard Gibbs Energy Using formula: ΔrG° = -nFE°(cell) Given: E°(cell) = 0.236 V, F = 96487 C mol⁻¹ ΔrG° = -2×964872 \times 96487 C mol⁻¹ × 0.236 V ΔrG° = -45541.86 J mol⁻¹ ΔrG° = -45.54 kJ mol⁻¹

  3. Step 3: Calculate Equilibrium Constant (Kc) Using formula: ΔrG° = -RT ln K -45541.86 = -8.314×2988.314 \times 298 × ln K ln K = 45541.86 / (8.314×2988.314 \times 298) ln K = 18.38 Kc = e^(18.38) Kc = 9.62×109.62 \times 10⁷

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