JEE Main 2022 — Chemistry Physical Chemistry

2022integerhard

$\mathrm{Cu}(s)+{\mathrm{Sn}}^{2+}(0.001M)\rightarrow {\mathrm{Cu}}^{2+}(0.01M)+\mathrm{Sn}(s)$

The Gibbs free energy change for the above reaction at $298K$ is $x\times {10}^{-1}\mathrm{kJ}{\mathrm{mol}}^{-1}$ . The value of $x$ is____[nearest integer]

[Given : ${E}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^{\ominus }=0.34V;{E}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn}}^{\ominus }=-0.14V;F=96500{\mathrm{Cmol}}^{-1}$ ]

Official previous-year question

Held on 26 Jun 2022 · Verified 6 Jul 2026.

Did you get this right?

Solution

${E}_{\mathrm{cell}}^{^{\circ}}={E}_{{\mathrm{Sn}}^{2+}\mathrm{Sn}}^{^{\circ}}-{E}_{{\mathrm{Cu}}^{2+}\mathrm{Cu}}^{^{\circ}}=-0.14-0.34=-0.48$

${E}_{\mathrm{cell}}={E}_{\mathrm{cell}}^{^{\circ}}-\frac{0.059}{2}\mathrm{log}\frac{[{\mathrm{Cu}}^{2+}]}{[{\mathrm{Sn}}^{2+}]}$

$=-0.48-0.0295[\mathrm{log}\frac{{10}^{-2}}{{10}^{-3}}]$

$=-0.48-0.0295$

$=-0.5095$

$\Delta G=-{\mathrm{nFE}}_{\mathrm{cell}}$

$=-2\times 96500(-0.5095)$

$=98333.5=98.33\mathrm{kJ}=983.3\times {10}^{-1}\mathrm{kJ}$

$x=983.3 \mathrm{Nearest}\mathrm{integer}=983$

Your note

JEE Main Physical Chemistry in other years

Physical Chemistry 2026 Physical Chemistry 2025 Physical Chemistry 2024 Physical Chemistry 2023 Physical Chemistry 2021 Physical Chemistry 2020 Physical Chemistry 2019 Physical Chemistry 2018 Physical Chemistry 2017 Physical Chemistry 2016 Physical Chemistry 2015 Physical Chemistry 2014

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2022 — Chemistry Physical Chemistry

2022integerhard

$\mathrm{Cu}(s)+{\mathrm{Sn}}^{2+}(0.001M)\rightarrow {\mathrm{Cu}}^{2+}(0.01M)+\mathrm{Sn}(s)$

The Gibbs free energy change for the above reaction at $298K$ is $x\times {10}^{-1}\mathrm{kJ}{\mathrm{mol}}^{-1}$ . The value of $x$ is____[nearest integer]

[Given : ${E}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^{\ominus }=0.34V;{E}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn}}^{\ominus }=-0.14V;F=96500{\mathrm{Cmol}}^{-1}$ ]

Official previous-year question

Held on 26 Jun 2022 · Verified 6 Jul 2026.

Did you get this right?

Solution

${E}_{\mathrm{cell}}^{^{\circ}}={E}_{{\mathrm{Sn}}^{2+}\mathrm{Sn}}^{^{\circ}}-{E}_{{\mathrm{Cu}}^{2+}\mathrm{Cu}}^{^{\circ}}=-0.14-0.34=-0.48$

${E}_{\mathrm{cell}}={E}_{\mathrm{cell}}^{^{\circ}}-\frac{0.059}{2}\mathrm{log}\frac{[{\mathrm{Cu}}^{2+}]}{[{\mathrm{Sn}}^{2+}]}$

$=-0.48-0.0295[\mathrm{log}\frac{{10}^{-2}}{{10}^{-3}}]$

$=-0.48-0.0295$

$=-0.5095$

$\Delta G=-{\mathrm{nFE}}_{\mathrm{cell}}$

$=-2\times 96500(-0.5095)$

$=98333.5=98.33\mathrm{kJ}=983.3\times {10}^{-1}\mathrm{kJ}$

$x=983.3 \mathrm{Nearest}\mathrm{integer}=983$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2022 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry

JEE Main 2022 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry