JEE Main 2021 — Chemistry Physical Chemistry

2021integerhard

The magnitude of the change in oxidising power of the ${\mathrm{MnO}}_{4}^{-}/{\mathrm{Mn}}^{2+}$ couple is $x\times {10}^{-4}V$ if the ${H}^{+}$ concentration is decreased from $1M$ to ${10}^{-4}M$ at $25^{\circ}C.$ (Assume concentration of ${\mathrm{MnO}}_{4}^{-}$ and ${\mathrm{Mn}}^{2+}$ to be same on change in ${H}^{+}$ concentration). The value of $x$ is __________.

(Rounded off to the nearest integer)

[Given: $\frac{2.303\mathrm{RT}}{F}=0.059$ ]

Official previous-year question

Held on 24 Feb 2021 · Verified 6 Jul 2026.

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Solution

Eqn is-

${\mathrm{MnO}}_{4}^{-}+{H}^{\oplus }+5{e}^{-}\rightarrow {\mathrm{Mn}}^{+2}+4{H}_{2}O$

Nernst equation:

${E}_{\mathrm{cell}}={E}_{\mathrm{Cell}}^{0}-\frac{0.059}{5}\mathrm{log}\frac{[{\mathrm{Mn}}^{+2}]}{[{\mathrm{MnO}}_{4}^{-}]}{[\frac{1}{{H}^{+}}]}^{8}$

(I) Given $[{H}^{\oplus }]=1M$

${E}_{1}={E}^{0}-\frac{0.059}{5}\mathrm{log}\frac{[{\mathrm{Mn}}^{+2}]}{[{\mathrm{MnO}}_{4}^{-}]}$

(II) Now: $[{H}^{\oplus }]={10}^{-4}M$

${E}_{2}={E}^{0}-\frac{0.059}{5}\mathrm{log}\frac{[{\mathrm{Mn}}^{+2}]}{[{\mathrm{MnO}}_{4}^{-}]}\times \frac{1}{{({10}^{-4})}^{8}}$

$={E}^{0}-\frac{0.059}{5}\mathrm{log}\frac{{\mathrm{Mn}}^{+2}}{[{\mathrm{MnO}}_{4}^{-}]}+\frac{0.059}{5}\mathrm{log}{10}^{-32}$

therefore: $|{E}_{1}-{E}_{2}|=\frac{0.059}{5}\times 32$

$=0.3776V=3776\times {10}^{-4}$

$x=3776$

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Solution

Eqn is-

${\mathrm{MnO}}_{4}^{-}+{H}^{\oplus }+5{e}^{-}\rightarrow {\mathrm{Mn}}^{+2}+4{H}_{2}O$

Nernst equation:

${E}_{\mathrm{cell}}={E}_{\mathrm{Cell}}^{0}-\frac{0.059}{5}\mathrm{log}\frac{[{\mathrm{Mn}}^{+2}]}{[{\mathrm{MnO}}_{4}^{-}]}{[\frac{1}{{H}^{+}}]}^{8}$

(I) Given $[{H}^{\oplus }]=1M$

${E}_{1}={E}^{0}-\frac{0.059}{5}\mathrm{log}\frac{[{\mathrm{Mn}}^{+2}]}{[{\mathrm{MnO}}_{4}^{-}]}$

(II) Now: $[{H}^{\oplus }]={10}^{-4}M$

${E}_{2}={E}^{0}-\frac{0.059}{5}\mathrm{log}\frac{[{\mathrm{Mn}}^{+2}]}{[{\mathrm{MnO}}_{4}^{-}]}\times \frac{1}{{({10}^{-4})}^{8}}$

$={E}^{0}-\frac{0.059}{5}\mathrm{log}\frac{{\mathrm{Mn}}^{+2}}{[{\mathrm{MnO}}_{4}^{-}]}+\frac{0.059}{5}\mathrm{log}{10}^{-32}$

therefore: $|{E}_{1}-{E}_{2}|=\frac{0.059}{5}\times 32$

$=0.3776V=3776\times {10}^{-4}$

$x=3776$

JEE Main 2021 — Chemistry Physical Chemistry

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JEE Main 2021 — 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