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Chemistry · Electrochemistry

JEE Main 2026 — 22 January, Evening Shift — Question 62

Consider the following electrochemical cell : Pt∣O2( g)(lbar)∣HCl(aq)∥M2+(aq,1.0M)∣M(s)\mathrm{Pt}\left|\mathrm{O}_{2}(\mathrm{~g})(\mathrm{lbar})\right| \mathrm{HCl}(\mathrm{aq}) \| \mathrm{M}^{2+}(\mathrm{aq}, 1.0 \mathrm{M}) \mid \mathrm{M}(\mathrm{s}) The pH above which, oxygen gas would start to evolve at anode is ____\_\_\_\_ (nearest integer). [ Given : EM2+/M0=0.994 VEO2/H2O0=1.23 V} standard    reduction   potential  and RTF(2.303)=0.059 V at   the   given   condition ]\left[\begin{array}{ll}\text { Given : } & \left.\begin{array}{l}\mathrm{E}_{\mathrm{M}^{2+} / \mathrm{M}}^{0}=0.994 \mathrm{~V} \mathrm{E}_{\mathrm{O}_{2} / \mathrm{H}_{2} \mathrm{O}}^{0}=1.23 \mathrm{~V}\end{array}\right\} \text { standard \; reduction\; potential } \\& \text { and } \frac{\mathrm{RT}}{\mathrm{F}}(2.303)=0.059 \mathrm{~V} \text { at \;the\; given \;condition }\end{array}\right]

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

For spontaneity Ecell >0\mathrm{E}_{\text {cell }}>0

\section*{At limiting condition :} EOxi \mathrm{E}_{\text {Oxi }} (anode) =−ERed =-\mathrm{E}_{\text {Red }} (cathode) H2O→2H++12O2+2e−\mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{H}^{+}+\frac{1}{2} \mathrm{O}_{2}+2 \mathrm{e}^{-} E=E∘−0.0592log⁡[[H+]2×PO21/21]\mathrm{E}=\mathrm{E}^{\circ}-\frac{0.059}{2} \log \left[\frac{\left[\mathrm{H}^{+}\right]^{2} \times \mathrm{P}_{\mathrm{O}_{2}}^{1 / 2}}{1}\right] −0.997=−1.23+0.059×pH-0.997=-1.23+0.059 \times \mathrm{pH} pH=3.94\mathrm{pH}=3.94 pH≃4\mathrm{pH} \simeq 4

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
22 January, Evening Shift
Subject
Chemistry
Chapter
Electrochemistry
Topic
Basics of Electrolytic Cells

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