Search any question & find its solution
Question:
Answered & Verified by Expert
The standard Gibbs free energy change $\left(\Delta \mathrm{G}^{\circ}\right.$ in $\mathrm{kJ} \mathrm{mol}^{-1}$ ), in a Daniel cell $\left(\mathrm{E}_{\text {cell }}^{\circ}=1.1 \mathrm{~V}\right.$ ), when 2 moles of $\mathrm{Zn}(\mathrm{s})$ is oxidized at $298 \mathrm{~K}$, is closest to
Options:
Solution:
1220 Upvotes
Verified Answer
The correct answer is:
$-424.6$
$\mathrm{Zn}+\mathrm{Cu}^{+2} \rightarrow \mathrm{Zn}^{+2}+\mathrm{Cu}$
$2 \mathrm{Zn}+2 \mathrm{Cu}^{+2} \rightarrow 2 \mathrm{Zn}^{+2}+2 \mathrm{Cu}$
For 2 moles of $\mathrm{Zn}, \mathrm{n}=4$
$\Delta \mathrm{G}^{\circ}=-\mathrm{nFE}_{\mathrm{Cell}}^{\circ}=-4 \times 96500 \times 1.1=-424.6 \mathrm{~kJ}$
$2 \mathrm{Zn}+2 \mathrm{Cu}^{+2} \rightarrow 2 \mathrm{Zn}^{+2}+2 \mathrm{Cu}$
For 2 moles of $\mathrm{Zn}, \mathrm{n}=4$
$\Delta \mathrm{G}^{\circ}=-\mathrm{nFE}_{\mathrm{Cell}}^{\circ}=-4 \times 96500 \times 1.1=-424.6 \mathrm{~kJ}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.