For the cell reaction: The standard Gibbs' energy of the cell reaction is [Given that Faraday constant, F = 96500 C ] [2019]

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Q.
For the cell reaction:

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

$E_{cell}^{\circ} = 0.24 V at 298 K.$ The standard Gibbs' energy $(Δ_{r} G °)$ of the cell reaction is

[Given that Faraday constant, F = 96500 C ${mol}^{- 1}$ ] [2019]

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

2 - 46.32 kJ ${mol}^{- 1}$

3 - 23.16 kJ ${mol}^{- 1}$

4 46.32 kJ ${mol}^{- 1}$

Ans.
(2)

The standard Gibbs' energy, $(Δ G °) = - n F E_{cell}^{\circ}$

Value of $n = 2$

$∴$ $Δ G ° = - 2 \times 96500 \times 0.24 = - 46320$ $J$

$= - 46.32$ $kJ/mol$

Solution

Q.
For the cell reaction:

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

$E_{cell}^{\circ} = 0.24 V at 298 K.$ The standard Gibbs' energy $(Δ_{r} G °)$ of the cell reaction is

[Given that Faraday constant, F = 96500 C ${mol}^{- 1}$ ] [2019]

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

2 - 46.32 kJ ${mol}^{- 1}$

3 - 23.16 kJ ${mol}^{- 1}$

4 46.32 kJ ${mol}^{- 1}$

Ans.
(2)

The standard Gibbs' energy, $(Δ G °) = - n F E_{cell}^{\circ}$

Value of $n = 2$

$∴$ $Δ G ° = - 2 \times 96500 \times 0.24 = - 46320$ $J$

$= - 46.32$ $kJ/mol$

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Solution

Q. For the cell reaction: 2Fe(aq)3++2I(aq)-⟶2Fe(aq)2++I2(aq) Ecell∘=0.24 V at 298 K. The standard Gibbs' energy (ΔrG°) of the cell reaction is [Given that Faraday constant, F = 96500 C mol-1] [2019]

1 23.16 kJ mol-1

2 - 46.32 kJ mol-1

3 - 23.16 kJ mol-1

4 46.32 kJ mol-1

Ans. (2) The standard Gibbs' energy, (ΔG°)=-nFEcell∘ Value of n=2 ∴ ΔG°=-2×96500×0.24=-46320 J =-46.32 kJ/mol

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Q.
For the cell reaction:

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

$E_{cell}^{\circ} = 0.24 V at 298 K.$ The standard Gibbs' energy $(Δ_{r} G °)$ of the cell reaction is

[Given that Faraday constant, F = 96500 C ${mol}^{- 1}$ ] [2019]

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

2 - 46.32 kJ ${mol}^{- 1}$

3 - 23.16 kJ ${mol}^{- 1}$

4 46.32 kJ ${mol}^{- 1}$

Ans.
(2)

The standard Gibbs' energy, $(Δ G °) = - n F E_{cell}^{\circ}$

Value of $n = 2$

$∴$ $Δ G ° = - 2 \times 96500 \times 0.24 = - 46320$ $J$

$= - 46.32$ $kJ/mol$