1. -46.32 kJ mol ⁻¹ 2. -23.16 kJ mol ⁻¹ 3. 46.32 kJ mol ⁻¹ 4. 23.16 kJ mol ⁻¹

Answer: 1. -46.32 kJ mol ⁻¹ Formula: Δ_rG^(ominus) = -nFE^(ominus)_cell . In the reaction 2mathrmFe³⁺ + 2I^- → 2mathrmFe²⁺ + I₂ , 2 electrons are transferred, so n = 2 . Δ_rG^(ominus) = -2 × 96500 × 0.24 = -46320\ mathrmJ\ mol⁻¹ = -46.32\ mathrmkJ\ mol⁻¹ . Answer: option (1). Option (2) wrongly takes n=1 .

Topic: Cell potential and Gibbs energy· Redox Reactions and Electrochemistry

For the cell reaction2Fe^3+(aq) + 2I^-(aq) → 2Fe^2+(aq) + I_2(aq)E^_cell = 0.24 V at 298 K. The standard Gibbs energy (Δ_rG^) of the cell reaction is :[Given that Faraday constant F = 96500 C mol^-1]

1. \(-46.32\) kJ mol\(^{-1}\)

2. \(-23.16\) kJ mol\(^{-1}\)

3. 46.32 kJ mol\(^{-1}\)

4. 23.16 kJ mol\(^{-1}\)

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Answer: 1. \(-46.32\) kJ mol\(^{-1}\)

Formula: \(\Delta_rG^{\ominus} = -nFE^{\ominus}_{cell}\).

In the reaction \(2\mathrm{Fe^{3+}} + 2\mathrm{I^-} \to 2\mathrm{Fe^{2+}} + \mathrm{I_2}\), 2 electrons are transferred, so \(n = 2\).

\(\Delta_rG^{\ominus} = -2 \times 96500 \times 0.24 = -46320\ \mathrm{J\ mol^{-1}} = -46.32\ \mathrm{kJ\ mol^{-1}}\).

Answer: option (1). Option (2) wrongly takes \(n=1\).

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For the cell reaction2Fe^3+(aq) + 2I^-(aq) → 2Fe^2+(aq) + I_2(aq)E^_cell = 0.24 V at 298 K. The standard Gibbs energy (Δ_rG^) of the cell reaction is :[Given that Faraday constant F = 96500 C mol^-1]

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