JEE Chemistry 2025 Question with Solution

Given: $1 \, \text{M}$ aqueous solutions of $Cu(NO_3)_2$, $AgNO_3$, $Hg_2(NO_3)_2$ and $Mg(NO_3)_2$ are electrolysed using inert electrodes. The standard reduction potentials are:

$$E^0_{Ag^+/Ag} = 0.80 \, \text{V},\quad E^0_{Hg_2^{2+}/Hg} = 0.79 \, \text{V},\quad E^0_{Cu^{2+}/Cu} = 0.34 \, \text{V},\quad E^0_{Mg^{2+}/Mg} = -2.37 \, \text{V}$$

Find: Which statement is correct.

At the cathode, the species with higher reduction potential is discharged more readily.

For Statement (I), compare the given reduction potentials:

$$Ag^+ > Hg_2^{2+} > Cu^{2+} > Mg^{2+}$$

using their values

$$0.80 > 0.79 > 0.34 > -2.37$$

Therefore, with increasing voltage, deposition will occur in the order Ag, Hg and Cu. So Statement (I) is correct.

For Statement (II), magnesium is not deposited because water is reduced more easily than $Mg^{2+}$ in aqueous medium. The cathodic reaction is:

$$2H_2O + 2e^- \rightarrow H_2 + 2OH^-$$

with

$$E^0 = -0.83 \, \text{V}$$

Since $-0.83 \, \text{V} > -2.37 \, \text{V}$, water is reduced in preference to $Mg^{2+}$. Thus, hydrogen gas is evolved at the cathode, not oxygen gas.

Oxygen is evolved at the anode by oxidation of water, not at the cathode.

Therefore, Statement (I) is correct but Statement (II) is incorrect. The correct option is B.

Given: Electrolysis of aqueous nitrate solutions with inert electrodes.

Find: Correct evaluation of Statement (I) and Statement (II).

Statement (I): Metal deposition at the cathode depends on ease of reduction. A larger standard reduction potential means easier reduction.

The relevant reductions are:

$$Ag^+ + e^- \rightarrow Ag \qquad E^0 = 0.80 \, \text{V}$$ $$Hg_2^{2+} + 2e^- \rightarrow 2Hg \qquad E^0 = 0.79 \, \text{V}$$ $$Cu^{2+} + 2e^- \rightarrow Cu \qquad E^0 = 0.34 \, \text{V}$$ $$Mg^{2+} + 2e^- \rightarrow Mg \qquad E^0 = -2.37 \, \text{V}$$

Hence the order of discharge is:

$$Ag > Hg > Cu > Mg$$

So the stated sequence Ag, Hg and Cu is correct.

Statement (II): For aqueous $Mg(NO_3)_2$, magnesium does not deposit because water is reduced first:

$$2H_2O + 2e^- \rightarrow H_2 + 2OH^- \qquad E^0 = -0.83 \, \text{V}$$

This is easier than

$$Mg^{2+} + 2e^- \rightarrow Mg \qquad E^0 = -2.37 \, \text{V}$$

Therefore, the product at the cathode is hydrogen gas, not oxygen gas.

The oxygen-evolution reaction is anodic:

$$2H_2O \rightarrow O_2 + 4H^+ + 4e^-$$

So saying that oxygen is evolved at the cathode is incorrect.

Thus, Statement (I) is correct and Statement (II) is incorrect. The correct option is B.