On electrolyzing a solution of dilute $H_{2} {S O}_{4}$ between platinum electrodes, the gas evolved at the anode is

{S O}_{2}

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{S O}_{3}

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O_{2}

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H_{2}

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Solution

The correct option is C $O_{2}$
On electrolysing a solution of dil $H_{2} S O_{4}$ between platinum electrode, $o x y g e n$ gas evolves. The oxidation potential of $H_{2} O$ is greater than the $S O_{4}^{- 2}$ ion. So oxidation of $H_{2} O$ will take place. During the oxidation of $H_{2} O$ $o x y g e n$ $g a s$ releases at the anode.

Option C is correct.

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