QJEEM24Ch89fc: JEE Main 2024 Chemistry - Class 12 JEEM24Ch89fc Solution: Consider the following redox reaction : The standard reduction…

Question

Verified Solution • 2025–26 Session moderate

Consider the following redox reaction :

\mathrm{MnO}_4^{-}+\mathrm{H}^{+}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightleftharpoons \mathrm{Mn}^{2+}+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2

The standard reduction potentials are given as below

\left(\mathrm{E}_{\text {red }}^0\right)

:

\begin{aligned}&amp; \mathrm{E}_{\mathrm{MnO}_4^{-} / \mathrm{Mn}^{2+}}^{\circ}=+1.51 \mathrm{~V} \\\\&amp; \mathrm{E}_{\mathrm{CO}_2 / \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4}^{\circ}=-0.49 \mathrm{~V}\end{aligned}

If the equilibrium constant of the above reaction is given as

\mathrm{K}_{\mathrm{eq}}=10^x

, then the value of

x=

__________ (nearest integer)

Accepted Answer

Cell

\mathrm{Rx}^{\mathrm{n}} ; \mathrm{MnO}_4^{-}+\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \rightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_2

\mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\text {op }}^{\circ}

of anode

+\mathrm{E}_{\mathrm{RP}}^{\circ}

of cathode

=0.49+1.51=2.00 \mathrm{~V}

At equilibrium

\mathrm{E}_{\text {cell }}=0 \text {, }

\mathrm{E}_{\text {cell }}^{\circ}=\frac{0.0591}{\mathrm{n}} \log \mathrm{K}

2=\frac{0.0591}{10} \log K

\log K=338

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