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Electrolysis of 50% H₂SO₄ between Pt electrodes the reaction preferred at anode is

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$4 H_{2} O + 4 \bar{e} \to 2 H_{2} + 4 {OH}^{-}$

$2 H_{2} O + 2 \bar{e} \to H_{2} + 2 {OH}^{-}$

$2 H_{2} O_{(l)} \to O_{2 (g)} + 4 \overset{\oplus}{H} + 4 \bar{e}$

$2 {SO}_{4}^{2 -} \to S_{2} O_{8}^{2 -} + 2 \bar{e}$

answer is B.

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$2 {SO}_{4}^{2 -} \to S_{2} O_{8}^{2 -} + 2 \bar{e}$

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