JEE Chemistry 2025 Question with Solution

Given: We must identify in which cases oxygen gas is evolved during electrolysis.

Find: The correct option among the given combinations.

For each case, examine the anode reaction.

(A) An aqueous solution of AgNO$_3$ using silver electrodes: Silver electrode is active and gets oxidized at the anode, so silver dissolves instead of water giving oxygen. Therefore, O$_2$ is not evolved.

(B) An aqueous solution of AgNO$_3$ using platinum electrodes: Platinum is an inert electrode, so water is oxidized at the anode.

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

Therefore, O$_2$ is evolved.

(C) A dilute solution of H$_2$SO$_4$ using platinum electrodes: In dilute sulfuric acid, water is oxidized at the anode.

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

Therefore, O$_2$ is evolved.

(D) A high concentration solution of H$_2$SO$_4$ using platinum electrodes: In concentrated sulfuric acid, sulfate ions are oxidized to peroxydisulfate instead of oxygen.

$$2SO_4^{2-} \rightarrow S_2O_8^{2-} + 2e^-$$

Therefore, O$_2$ is not evolved.

So oxygen gas is evolved in (B) and (C) only. Therefore, the correct option is A.

Given: Four electrolysis setups are listed.

Find: In which setups oxygen gas is obtained as a product.

Principle: Oxygen is evolved at the anode when water is oxidized and the anode itself does not preferentially participate. Active electrodes can dissolve, while inert electrodes force oxidation of water or anions depending on the medium.

1. Case (A): aqueous AgNO$_3$ with silver electrodes At the cathode:

$$Ag^+ + e^- \rightarrow Ag$$

At the anode:

$$Ag \rightarrow Ag^+ + e^-$$

Since the silver anode dissolves, oxygen is not liberated.

2. Case (B): aqueous AgNO$_3$ with platinum electrodes At the cathode:

$$Ag^+ + e^- \rightarrow Ag$$

At the anode, platinum is inert, so water is oxidized:

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

Hence, oxygen is evolved.

3. Case (C): dilute H$_2$SO$_4$ with platinum electrodes At the cathode:

$$2H^+ + 2e^- \rightarrow H_2$$

At the anode:

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

Hence, oxygen is evolved.

4. Case (D): concentrated H$_2$SO$_4$ with platinum electrodes In concentrated acid, sulfate can undergo oxidation:

$$2SO_4^{2-} \rightarrow S_2O_8^{2-} + 2e^-$$

So oxygen is not evolved in this case.

Conclusion: Only (B) and (C) give oxygen gas. The correct option is A.

The solution states the correct answer as (1) (B) and (C) only, which corresponds to option A.