Question:
The standard Gibbs energy for the given cell reaction in $\mathrm{kJ}$ $\mathrm{mol}^{-1}$ at $298 \mathrm{~K}$ is:
$\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$
$\mathrm{E}^{\circ}=2 \mathrm{~V}$ at $298 \mathrm{~K}$
(Faraday's constant, $\mathrm{F}=96000 \mathrm{C} \mathrm{mol}^{-1}$ )
Correct Option: 1
Solution:
$\Delta \mathrm{G}^{\circ}=-\mathrm{nFE}_{\text {cell }}^{\circ}$
$=-2(96000) 2 \mathrm{~V}=-384000 \mathrm{~J} / \mathrm{mol}$
$=-384 \mathrm{~kJ} / \mathrm{mol}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.