JEE Main 2025 Jan 29 Shift 2, Chemistry Q11: Electrolytic Cells And Electrolysis

JEE Main2025Jan 29, Shift 2Chemistry

$O_{2}$ gas will be evolved as a product of electrolysis of :

(A) an aqueous solution of $A g N O_{3}$ using silver electrodes.

(B) an aqueous solution of $A g N O_{3}$ using platinum electrodes.

(D) a high concentration solution of $H_{2} S O_{4}$ using platinum electrodes.

Choose the correct answer from the options given below :

A
(B) and (C) only
B
(A) and (D) only
C
(B) and (D) only
D
(A) and (C) only

Solution

$O_{2}$ is liberated only when water is oxidised at the anode. Consider each case:

(A) Silver anode itself oxidises preferentially ( $A g \to A g^{+} + e^{-}$ ), so the electrode dissolves and no $O_{2}$ is produced.

(B) Inert Pt anode; $N O_{3}^{-}$ is not oxidised, so water is, giving $O_{2}$ .

(C) Dilute $H_{2} S O_{4}$ on Pt: $S O_{4}^{2 -}$ is hard to oxidise, water discharges and $O_{2}$ evolves.

(D) Concentrated $H_{2} S O_{4}$ : $S O_{4}^{2 -}$ is oxidised to peroxydisulphate $S_{2} O_{8}^{2 -}$ (Marshall’s acid route), so $O_{2}$ is not the product.

$∴$ Only (B) and (C) give $O_{2}$ .

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JEE Main 2025 Jan 29 Shift 2, Chemistry Q11: Electrolytic Cells And Electrolysis

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