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50% $H_{2} {SO}_{4}$ solution is electrolyzed using inter electrodes. The product obtained at anode is

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

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

$H_{2} S_{2} O_{5}$

$H_{2} S_{2} O_{7}$

answer is C.

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Detailed Solution

$2 H_{2} {SO}_{4} ⇌ 2 H^{+} + 2 {HSO}_{4}^{-}$

Anode

$2 {HSO}_{4}^{-} \to H_{2} S_{2} O_{8} + 2 e^{-}$

Cathode

$2 H^{+} + 2 e^{-} \to H_{2}$

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