For the cell reaction:2Fe(aq)3++2I(aq)-⟶2Fe(aq)2++I2(aq)Eocell =0.24 V at 298 K. The standard Gibbs energy ΔrG°of the cell reaction is [Given that Faraday constant F=96500Cmol-1

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For the cell reaction:

$2 {Fe}_{(aq)}^{3 +} + 2 I_{(aq)}^{-} ⟶ 2 {Fe}_{(aq)}^{2 +} + I_{2 (aq)}$

${E^{o}}_{cell} = 0.24 V at 298 K$ . The standard Gibbs energy $(Δ_{r} G °)$ of the cell reaction is

$[Given that Faraday constant F = 96500 {Cmol}^{- 1}]$

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$- 46.32 kJ {mol}^{- 1}$

$46.32 kJ {mol}^{- 1}$

$23.16 kJ {mol}^{- 1}$

$- 23.16 kJ {mol}^{- 1}$

answer is B.

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Detailed Solution

$The standard Gibbs' energy (ΔG °) = - nFE °_{cell} Value of n = 2 ∴ ΔG ° = - 2 \times 96500 \times 0.24 = - 46320 J = - 46.32 kJ / mol$

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