The standard cell potential of the following cell Zn |Zn^2+(aq) | Fe^2+(aq) mid Fe is 0.32 V . Calculate the standard Gibbs energy change for the reaction : Zn(s)+Fe^2+(aq) arrow Zn^2+(aq)+Fe(s) (Given : 1 ~F=96487 C )

Correct answer: (A) -61.75 ~kJ ~mol^-1. Δ_r G^ominus=-n F E_cell^ominus For the given reaction, n=2 ∴ Δ_r G^ominus=-2 × 96487 × 0.32 =-61751.68 ~J ~mol^-1 =-61.751 ~kJ ~mol^-1

Q85 / 200 · NEET 2024 (Re-NEET) · Chemistry

The standard cell potential of the following cell $\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{aq})\right| \mathrm{Fe}^{2+}(\mathrm{aq}) \mid \mathrm{Fe}$ is 0.32 V . Calculate the standard Gibbs energy change for the reaction : $\mathrm{Zn}(\mathrm{s})+\mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Fe}(\mathrm{s})$ (Given : $1 \mathrm{~F}=96487 \mathrm{C}$ )

Correct answer
A
$-61.75 \mathrm{~kJ} \mathrm{~mol}^{-1}$
B
$+5.006 \mathrm{~kJ} \mathrm{~mol}^{-1}$
C
$-5.006 \mathrm{~kJ} \mathrm{~mol}^{-1}$
D
$+61.75 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution

$\Delta_r G^{\ominus}=-n F E_{\text {cell }}^{\ominus}$ For the given reaction, $n=2$ $\therefore \quad \Delta_{\mathrm{r}} \mathrm{G}^{\ominus}=-2 \times 96487 \times 0.32$ $=-61751.68 \mathrm{~J} \mathrm{~mol}^{-1}$ $=-61.751 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Hence, the correct answer is option: 'A'.

Solution

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