\(\mathrm{O}_{2}\) gas will be evolved as a product of electrolysis of : (A) an aqueous solution of \(\mathrm{AgNO}_{3}\) using silver electrodes. (B) an aqueous solution of \(\mathrm{AgNO}_{3}\) using platinum electrodes. (C) a dilute solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) using platinum electrodes. (D) a high concentration solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) using platinum electrodes. Choose the correct answer from the options given below :

Redox Reactions and Electrochemistry — JEE Main · 2025 · 29 Jan 2025 · Shift 2 Question with Solution

O_{2}

gas will be evolved as a product of electrolysis of :
(A) an aqueous solution of

AgNO_{3}

using silver electrodes.
(B) an aqueous solution of

AgNO_{3}

using platinum electrodes.
(C) a dilute solution of

H_{2} SO_{4}

using platinum electrodes.
(D) a high concentration solution of

H_{2} SO_{4}

using platinum electrodes.

Choose the correct answer from the options given below :

Answer: (A)

O_{2}

appears at the anode only when water itself is oxidized there,

2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-}

, which happens only when no other species is preferentially oxidized. With silver electrodes in

A g N O_{3}

, the anode metal itself dissolves (

A g \to A g^{+} + e^{-}

) since it is more easily oxidized than water, so no

O_{2}

forms — (A) fails. With inert platinum electrodes in

A g N O_{3}

N O_{3}^{-}

cannot be oxidized, so water is oxidized at the anode, releasing

O_{2}

— (B) holds. In dilute

H_{2} S O_{4}

with platinum electrodes,

S O_{4}^{2 -}

also resists oxidation, so water is again oxidized and

O_{2}

is liberated — (C) holds. In concentrated

H_{2} S O_{4}

with platinum electrodes, the high concentration of

H S O_{4}^{-}

favors anodic oxidation to peroxydisulfate ion rather than water oxidation, so

O_{2}

is not the anode product — (D) fails.

O_{2}

is evolved only in cases (B) and (C).

JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Single correct option