JEE Main 2024 — Chemistry Physical Chemistry

2024integereasy

Consider the following redox reaction: ${\mathrm{MnO}}_{4}^{-}+{H}^{+}+{H}_{2}{C}_{2}{O}_{4}\rightleftharpoons {\mathrm{Mn}}^{2+}+{H}_{2}O+{\mathrm{CO}}_{2}$

The standard reduction potentials are given as below $({E}_{\mathrm{red}}^{\circ})$

${{E}^{0}}_{{\mathrm{MnO}}_{4}^{-}/{\mathrm{Mn}}^{2+}}=+1.51V$ ; ${{E}^{0}}_{{\mathrm{CO}}_{2}/{H}_{2}{C}_{2}{O}_{4}}=-0.49V$

If the equilibrium constant of the above reaction is given as ${K}_{\mathrm{eq}}={10}^{x}$ , then the value of $x=$ _______ (nearest integer)

Official previous-year question

Held on 1 Feb 2024 · Verified 6 Jul 2026.

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Solution

For the reaction at equilibrium,

$2{\mathrm{MnO}}_{4}^{-}+6{H}^{+}+5{H}_{2}{C}_{2}{O}_{4}\rightleftharpoons 2{\mathrm{Mn}}^{2+}+8{H}_{2}O+10{\mathrm{CO}}_{2}.$

Given : ${E}_{{\mathrm{MnO}}_{4}^{-}/{\mathrm{Mn}}^{2+}}^{0}=1.51V\mathrm{and}{{E}^{0}}_{{H}_{2}{C}_{2}{O}_{4}^{}/{\mathrm{CO}}_{2}}=0.49V.$

So, ${E}_{\mathrm{cell}}^{0}=1.51+0.49=2V.$

Number of electrons involved in reaction can be calculated as follows:

${\mathrm{Mn}}^{+7}\mathrm{changes}{\mathrm{Mn}}^{2+}$ so, each $\mathrm{Mn}$ gain five electrons.

$n=10$

Now, ${E}_{\mathrm{cell}}^{0}=\frac{0.0591}{n}\mathrm{log}K$

$\mathrm{log}K=\frac{10\times 2}{0.0591}$

$\mathrm{log}K=338.4$

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JEE Main 2024 — Chemistry Physical Chemistry

2024integereasy

Consider the following redox reaction: ${\mathrm{MnO}}_{4}^{-}+{H}^{+}+{H}_{2}{C}_{2}{O}_{4}\rightleftharpoons {\mathrm{Mn}}^{2+}+{H}_{2}O+{\mathrm{CO}}_{2}$

The standard reduction potentials are given as below $({E}_{\mathrm{red}}^{\circ})$

${{E}^{0}}_{{\mathrm{MnO}}_{4}^{-}/{\mathrm{Mn}}^{2+}}=+1.51V$ ; ${{E}^{0}}_{{\mathrm{CO}}_{2}/{H}_{2}{C}_{2}{O}_{4}}=-0.49V$

If the equilibrium constant of the above reaction is given as ${K}_{\mathrm{eq}}={10}^{x}$ , then the value of $x=$ _______ (nearest integer)

Official previous-year question

Held on 1 Feb 2024 · Verified 6 Jul 2026.

Did you get this right?

Solution

For the reaction at equilibrium,

$2{\mathrm{MnO}}_{4}^{-}+6{H}^{+}+5{H}_{2}{C}_{2}{O}_{4}\rightleftharpoons 2{\mathrm{Mn}}^{2+}+8{H}_{2}O+10{\mathrm{CO}}_{2}.$

Given : ${E}_{{\mathrm{MnO}}_{4}^{-}/{\mathrm{Mn}}^{2+}}^{0}=1.51V\mathrm{and}{{E}^{0}}_{{H}_{2}{C}_{2}{O}_{4}^{}/{\mathrm{CO}}_{2}}=0.49V.$

So, ${E}_{\mathrm{cell}}^{0}=1.51+0.49=2V.$

Number of electrons involved in reaction can be calculated as follows:

${\mathrm{Mn}}^{+7}\mathrm{changes}{\mathrm{Mn}}^{2+}$ so, each $\mathrm{Mn}$ gain five electrons.

$n=10$

Now, ${E}_{\mathrm{cell}}^{0}=\frac{0.0591}{n}\mathrm{log}K$

$\mathrm{log}K=\frac{10\times 2}{0.0591}$

$\mathrm{log}K=338.4$

Your note

JEE Main Physical Chemistry in other years

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JEE Main 2024 — 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 2024 — Chemistry Physical Chemistry

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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