Q21 / 75 · JEE Main 2026 (22 January Shift 2) · Chemistry
Consider the following electrochemical cell : $\mathrm{Pt}\left|\mathrm{O}_{2}(\mathrm{~g})(1 \mathrm{bar})\right| \mathrm{HCl}(\mathrm{aq}) \| \mathrm{M}^{2+}(\mathrm{aq}, 1.0 \mathrm{M}) \mid \mathrm{M}(\mathrm{~s})$ The pH above which, oxygen gas would start to evolve at anode is $\_\_\_\_$ (nearest integer).
$\left[\begin{array}{ll}\text { Given: } & \mathrm{E}^{\mathrm{o}}{ }_{\mathrm{M}^{2+} / \mathrm{M}}=0.994 \mathrm{~V} \\ & \mathrm{E}^{\mathrm{o}}{ }_{\mathrm{O}_{2} / \mathrm{H}_{2} \mathrm{O}}=1.23 \mathrm{~V}\end{array}\right\}$ standard reduction potential $]$
$\left[\begin{array}{ll}\text { Given: } & \mathrm{E}^{\mathrm{o}}{ }_{\mathrm{M}^{2+} / \mathrm{M}}=0.994 \mathrm{~V} \\ & \mathrm{E}^{\mathrm{o}}{ }_{\mathrm{O}_{2} / \mathrm{H}_{2} \mathrm{O}}=1.23 \mathrm{~V}\end{array}\right\}$ standard reduction potential $]$
Correct answer: 4
Solution
In this electrochemical cell, at the anode oxygen evolution occurs from water oxidation. Using Nernst equation for the O₂/H₂O couple: $E_{anode} = E°_{O_2/H_2O} - \frac{0.059}{4} \log \frac{1}{[H^+]^4} = 1.23 - 0.059 \times pH$
At the cathode: $E_{cathode} = 0.994$ V (standard potential for M²⁺/M reduction)
For oxygen to start evolving, the cell potential becomes zero: $E_{cell} = E_{cathode} - E_{anode} = 0$
$0.994 = 1.23 - 0.059 \times pH$
$0.059 \times pH = 0.236$
$pH = 4.0$
At the cathode: $E_{cathode} = 0.994$ V (standard potential for M²⁺/M reduction)
For oxygen to start evolving, the cell potential becomes zero: $E_{cell} = E_{cathode} - E_{anode} = 0$
$0.994 = 1.23 - 0.059 \times pH$
$0.059 \times pH = 0.236$
$pH = 4.0$
Hence, the correct answer is 4.