JEE Main 2023 — Chemistry Physical Chemistry

2023integermedium

At $298K$ , a $1$ litre solution containing $10\mathrm{mmol}$ of ${\mathrm{Cr}}_{2}{O}_{7}^{2-}$ and $100\mathrm{mmol}$ of ${\mathrm{Cr}}^{3+}$ shows a $\mathrm{pH}$ of $3.0$ .

Given : ${\mathrm{Cr}}_{2}{O}_{7}\rightarrow 2-{\mathrm{Cr}}^{3+};{E}^{0}=1.330V$ and $\frac{2.303\mathrm{RT}}{F}=0.059V$

The potential for the half cell reaction is $x\times {10}^{-3}V$ . The value of $x$ is _____ .

Official previous-year question

Held on 24 Jan 2023 · Verified 6 Jul 2026.

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Solution

The balanced half-reaction can be written as follows,

${\mathrm{Cr}}_{2}{O}_{7}^{2-}+14{H}^{+}+6{e}^{-}\rightarrow 2{\mathrm{Cr}}^{3+}+7{H}_{2}O$

Using Nernst equation

${E}_{\mathrm{cell}}={E}_{\mathrm{cell}}^{o}-\frac{0.059}{n}\mathrm{log}\frac{[\mathrm{products}]}{[\mathrm{reactants}]}$

For the above reaction, number of electrons involved n = 6.

$E=1.33-\frac{0.059}{6}\mathrm{log}\frac{{(0.1)}^{2}}{({10}^{-2}){({10}^{-3})}^{14}}$

$E=1.33-\frac{0.059}{6}\times 42=0.917$

$E=917\times {10}^{-3}$

$x=917$ .

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JEE Main 2023 — Chemistry Physical Chemistry

2023integermedium

At $298K$ , a $1$ litre solution containing $10\mathrm{mmol}$ of ${\mathrm{Cr}}_{2}{O}_{7}^{2-}$ and $100\mathrm{mmol}$ of ${\mathrm{Cr}}^{3+}$ shows a $\mathrm{pH}$ of $3.0$ .

Given : ${\mathrm{Cr}}_{2}{O}_{7}\rightarrow 2-{\mathrm{Cr}}^{3+};{E}^{0}=1.330V$ and $\frac{2.303\mathrm{RT}}{F}=0.059V$

The potential for the half cell reaction is $x\times {10}^{-3}V$ . The value of $x$ is _____ .

Official previous-year question

Held on 24 Jan 2023 · Verified 6 Jul 2026.

Did you get this right?

Solution

The balanced half-reaction can be written as follows,

${\mathrm{Cr}}_{2}{O}_{7}^{2-}+14{H}^{+}+6{e}^{-}\rightarrow 2{\mathrm{Cr}}^{3+}+7{H}_{2}O$

Using Nernst equation

${E}_{\mathrm{cell}}={E}_{\mathrm{cell}}^{o}-\frac{0.059}{n}\mathrm{log}\frac{[\mathrm{products}]}{[\mathrm{reactants}]}$

For the above reaction, number of electrons involved n = 6.

$E=1.33-\frac{0.059}{6}\mathrm{log}\frac{{(0.1)}^{2}}{({10}^{-2}){({10}^{-3})}^{14}}$

$E=1.33-\frac{0.059}{6}\times 42=0.917$

$E=917\times {10}^{-3}$

$x=917$ .

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 2023 — Chemistry Physical Chemistry

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2023 — 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