JEE Main 2025 Apr 7 Shift 2, Chemistry Q20: Electrolytic Cells And Electrolysis

JEE Main2025Apr 7, Shift 2Chemistry

Given below are two statements :

1 M aqueous solution of each of Cu(NO $_{3}$ ) $_{2}$ , AgNO $_{3}$ , Hg $_{2}$ (NO $_{3}$ ) $_{2}$ ; Mg(NO $_{3}$ ) $_{2}$ are electrolysed using inert electrodes,

Given : $E_{A g^{+} / A g}^{θ} = 0.80$ V, $E_{H g_{2}^{2 +} / H g}^{θ} = 0.79$ V, $E_{C u^{2 +} / C u}^{θ} = 0.24$ V and $E_{M g^{2 +} / M g}^{θ} = - 2.37$ V

Statement (I) : With increasing voltage, the sequence of deposition of metals on the cathode will be Ag, Hg and Cu

Statement (II) : Magnesium will not be deposited at cathode instead oxygen gas will be evolved at the cathode.

In the light of the above statement, choose the most appropriate answer from the options given below

A
Both statement I and statement II are incorrect
B
Statement I is correct but statement II is incorrect
C
Both statement I and statement II are correct
D
Statement I is incorrect but statement II is correct

Solution

During electrolysis, the species with the highest (most positive) standard reduction potential is preferentially deposited first at the cathode as voltage is gradually increased.

Ranking the given reduction potentials: $E_{A g^{+} / A g}^{θ}$ (0.80 V) $>$ $E_{H g_{2}^{2 +} / H g}^{θ}$ (0.79 V) $>$ $E_{C u^{2 +} / C u}^{θ}$ (0.24 V) $>$ $E_{M g^{2 +} / M g}^{θ}$ ( $- 2.37$ V), so the order of deposition is Ag, then Hg, then Cu, matching Statement I, which is correct.

Magnesium has a very negative reduction potential, making Mg $^{2 +}$ extremely difficult to reduce compared to water. Instead of Mg $^{2 +}$ being reduced at the cathode, water is preferentially reduced there, liberating H $_{2}$ gas, not O $_{2}$ (which would form by oxidation at the anode, not the cathode). Since Statement II names the wrong gas and the wrong electrode, it is incorrect.

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JEE Main 2025 Apr 7 Shift 2, Chemistry Q20: Electrolytic Cells And Electrolysis

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