JEE Main 2020 — Chemistry Physical Chemistry

2020mcqmedium

For the given cell; $\mathrm{Cu}(s)|{\mathrm{Cu}}^{2+}({C}_{1}M)||{\mathrm{Cu}}^{2+}({C}_{2}M)|\mathrm{Cu}(s)$

change in Gibbs energy $(\Delta G)$ is negative, it :

Official previous-year question

Held on 6 Sept 2020 · Verified 6 Jul 2026.

Options

A
${C}_{1}={C}_{2}$
B
${C}_{2}=\frac{{C}_{1}}{\sqrt{2}}$
C
${C}_{1}=2{C}_{2}$
D
${C}_{2}=\sqrt{2}{C}_{1}$

Did you get this right?

Solution

For concentration cell ${E}_{\mathrm{cell}}^{\underset{¯}{o}}=0$

Anode : $\mathrm{Cu}(s)\rightarrow {\mathrm{Cu}}^{2+}(\mathrm{aq})A$

Cathode : ${\mathrm{Cu}}^{2+}(\mathrm{aq})c\rightarrow \mathrm{Cu}(s)$

Overall : $\bar{{\mathrm{Cu}}^{2+}(\mathrm{aq})c\rightarrow {\mathrm{Cu}}^{2+}(\mathrm{aq})}A$

As $\Delta G=-\mathrm{nF} {E}_{\mathrm{cell}}$

If $\Delta G=-\mathrm{ve}$ then ${E}_{\mathrm{cell}}$ is positive.

${E}_{\mathrm{cell}}={E}_{\mathrm{cell}}^{\underset{¯}{o}}-\frac{0.059}{2}\mathrm{log}\frac{{C}_{1}}{C2}$

${E}_{\mathrm{cell}}=\frac{-0.059}{2}\mathrm{log}\frac{{C}_{1}}{C2}$

${E}_{\mathrm{cell}}>0$

$\Rightarrow$ ${C}_{2}>{C}_{1}$

Your note

JEE Main Physical Chemistry in other years

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JEE Main 2020 — Chemistry Physical Chemistry

2020mcqmedium

For the given cell; $\mathrm{Cu}(s)|{\mathrm{Cu}}^{2+}({C}_{1}M)||{\mathrm{Cu}}^{2+}({C}_{2}M)|\mathrm{Cu}(s)$

change in Gibbs energy $(\Delta G)$ is negative, it :

Official previous-year question

Held on 6 Sept 2020 · Verified 6 Jul 2026.

Options

A
${C}_{1}={C}_{2}$
B
${C}_{2}=\frac{{C}_{1}}{\sqrt{2}}$
C
${C}_{1}=2{C}_{2}$
D
${C}_{2}=\sqrt{2}{C}_{1}$

Did you get this right?

Solution

For concentration cell ${E}_{\mathrm{cell}}^{\underset{¯}{o}}=0$

Anode : $\mathrm{Cu}(s)\rightarrow {\mathrm{Cu}}^{2+}(\mathrm{aq})A$

Cathode : ${\mathrm{Cu}}^{2+}(\mathrm{aq})c\rightarrow \mathrm{Cu}(s)$

Overall : $\bar{{\mathrm{Cu}}^{2+}(\mathrm{aq})c\rightarrow {\mathrm{Cu}}^{2+}(\mathrm{aq})}A$

As $\Delta G=-\mathrm{nF} {E}_{\mathrm{cell}}$

If $\Delta G=-\mathrm{ve}$ then ${E}_{\mathrm{cell}}$ is positive.

${E}_{\mathrm{cell}}={E}_{\mathrm{cell}}^{\underset{¯}{o}}-\frac{0.059}{2}\mathrm{log}\frac{{C}_{1}}{C2}$

${E}_{\mathrm{cell}}=\frac{-0.059}{2}\mathrm{log}\frac{{C}_{1}}{C2}$

${E}_{\mathrm{cell}}>0$

$\Rightarrow$ ${C}_{2}>{C}_{1}$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

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

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JEE Main 2020 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

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JEE Main 2020 — Chemistry Physical Chemistry

Options

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