Consider the following redox reaction : MnO 4 - + H + + H 2 C 2 O 4 rightleftharpoons Mn 2+ + H 2 O + CO 2 The standard reduction potentials are given as below ( E red ° ) E MnO 4 - / Mn 2+ ° =+1.51 V E CO 2 / H 2 C 2 O 4 ° =-0.49 V If the equilibrium constant of the above reaction is given as K e q =10 x , then the value of x = (nearest integer)

Chemistry · Electrochemistry

JEE Main 2024 — 1 February, Shift 2 — Question 82

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

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

$\mathrm{E}_{\mathrm{MnO}_{4}^{-} / \mathrm{Mn}^{2+}}^{\circ}=+1.51 \mathrm{~V}$

$\mathrm{E}_{\mathrm{CO}_{2} / \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}}^{\circ}=-0.49 \mathrm{~V}$

If the equilibrium constant of the above reaction is given as $K_{e q}=10^{x}$ ,

then the value of $\mathrm{x}=$ $\qquad$ (nearest integer)

Answer: 338.9

Numerical answer — enter this value.

Step-by-step solution

Cell Rx ${ }^{\mathrm{n}} ; \mathrm{MnO}_{4}^{-}+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \rightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_{2}$

$E_{\text {cell }}^{\circ}=E_{\text {op }}^{\circ}$ of anode $+E_{\text {RP }}^{\circ}$ of cathode

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

At equilibrium

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

$\mathrm{E}_{\text {cell }}^{\circ}=\frac{0.059}{\mathrm{n}} \log \mathrm{K}$

(As per NCERT $\frac{\mathrm{RT}}{\mathrm{F}}=0.059$ But $\frac{\mathrm{RT}}{\mathrm{F}}=0.0591$ can also be taken.)

$2=\frac{0.059}{10} \log \mathrm{~K}$

$\log K=338.98$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 2
Subject: Chemistry
Chapter: Electrochemistry
Topic: Nernst Equation and Electrochemical Series