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Q.

Electrolysis of 50% $H_{2} {SO}_{4}$ produces

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a

$H_{2} {SO}_{5}$ at cathode

b

$H_{2} S_{2} O_{8}$ to anode

c

$H_{2} O_{2}$ at anode and cathode

d

$H_{2} O$ at cathode

answer is A.

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Detailed Solution

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$2 H_{2} {So}_{4} ⇌ 2 H^{+} + 2 {HSO}_{4}^{-}$
Anode:- $2 {HSo}_{4}^{-} \to H_{2} S_{2} O_{8} + 2 e^{-}$
Cathode:- ${2H}^{+} + 2 e^{-} \to H_{2} (g)$

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