The standard cell potential of the following cell is 0.32 V. Calculate the standard Gibbs energy change for the reaction (Given : 1 F = 96487 C) [2024]

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Q.
The standard cell potential of the following cell $Zn | {Zn}_{(a q)}^{2 +} {||Fe}_{(a q)}^{2 +} | Fe$ is 0.32 V. Calculate the standard Gibbs energy change for the reaction ${Zn}_{(s)} + {Fe}_{(a q)}^{2 +} ⟶ {Zn}_{(a q)}^{2 +} + {Fe}_{(s)}$ (Given : 1 F = 96487 C) [2024]

1 $- 61.75$ $kJ$ ${mol}^{- 1}$

2 $+ 5.006$ $kJ$ ${mol}^{- 1}$

3 $- 5.006$ $kJ$ ${mol}^{- 1}$

4 $+ 61.75$ $kJ$ ${mol}^{- 1}$

Ans.
(1)

$Δ G ° = - n F E_{cell}^{\circ}$

Given, $E_{cell}^{\circ} = 0.32$ $V$

and 1F = 96487 C

For the given reaction, $n = 2$

$Δ G ° = - 2 \times 96487 \times 0.32 = - 61751.68$ $J$ ${mol}^{- 1}$ $= - 61.75$ $kJ$ ${mol}^{- 1}$

Solution

Q.
The standard cell potential of the following cell $Zn | {Zn}_{(a q)}^{2 +} {||Fe}_{(a q)}^{2 +} | Fe$ is 0.32 V. Calculate the standard Gibbs energy change for the reaction ${Zn}_{(s)} + {Fe}_{(a q)}^{2 +} ⟶ {Zn}_{(a q)}^{2 +} + {Fe}_{(s)}$ (Given : 1 F = 96487 C) [2024]

1 $- 61.75$ $kJ$ ${mol}^{- 1}$

2 $+ 5.006$ $kJ$ ${mol}^{- 1}$

3 $- 5.006$ $kJ$ ${mol}^{- 1}$

4 $+ 61.75$ $kJ$ ${mol}^{- 1}$

Ans.
(1)

$Δ G ° = - n F E_{cell}^{\circ}$

Given, $E_{cell}^{\circ} = 0.32$ $V$

and 1F = 96487 C

For the given reaction, $n = 2$

$Δ G ° = - 2 \times 96487 \times 0.32 = - 61751.68$ $J$ ${mol}^{- 1}$ $= - 61.75$ $kJ$ ${mol}^{- 1}$

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Solution

Q. The standard cell potential of the following cell Zn|Zn(aq)2+||Fe(aq)2+|Fe is 0.32 V. Calculate the standard Gibbs energy change for the reaction Zn(s)+Fe(aq)2+⟶Zn(aq)2++Fe(s) (Given : 1 F = 96487 C) [2024]

1 -61.75 kJ mol-1

2 +5.006 kJ mol-1

3 -5.006 kJ mol-1

4 +61.75 kJ mol-1

Ans. (1) ΔG°=-nFEcell∘ Given, Ecell∘=0.32 V and 1F = 96487 C For the given reaction, n=2 ΔG°=-2×96487×0.32=-61751.68 J mol-1=-61.75 kJ mol-1

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Q.
The standard cell potential of the following cell $Zn | {Zn}_{(a q)}^{2 +} {||Fe}_{(a q)}^{2 +} | Fe$ is 0.32 V. Calculate the standard Gibbs energy change for the reaction ${Zn}_{(s)} + {Fe}_{(a q)}^{2 +} ⟶ {Zn}_{(a q)}^{2 +} + {Fe}_{(s)}$ (Given : 1 F = 96487 C) [2024]

1 $- 61.75$ $kJ$ ${mol}^{- 1}$

2 $+ 5.006$ $kJ$ ${mol}^{- 1}$

3 $- 5.006$ $kJ$ ${mol}^{- 1}$

4 $+ 61.75$ $kJ$ ${mol}^{- 1}$

Ans.
(1)

$Δ G ° = - n F E_{cell}^{\circ}$

Given, $E_{cell}^{\circ} = 0.32$ $V$

and 1F = 96487 C

For the given reaction, $n = 2$

$Δ G ° = - 2 \times 96487 \times 0.32 = - 61751.68$ $J$ ${mol}^{- 1}$ $= - 61.75$ $kJ$ ${mol}^{- 1}$