Which of the following equation depicts the oxidizing nature of $H_{2} O_{2}$ ?

2 I^{-} + H_{2} O_{2} + 2 H^{+} \to I_{2} + 2 H_{2} O

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K I O_{4} + H_{2} O_{2} \to K I O_{3} + H_{2} O + O_{2}

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C l_{2} + H_{2} O_{2} \to 2 H C l + O_{2}

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I_{2} + H_{2} O_{2} + 2 O H^{-} \to 2 I^{-} + 2 H_{2} O + O_{2}

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Solution

The correct option is A $2 I^{-} + H_{2} O_{2} + 2 H^{+} \to I_{2} + 2 H_{2} O$
Equation (a)
$2 I^{-} + H_{2} O_{2} + 2 H^{+} \to I_{2} + 2 H_{2} O$

In above reaction, $H_{2} O_{2}$ oxidises $I^{-}$ to $I_{2}$ and itself gets reduced to $H_{2} O$ so, this reaction depicts oxidising nature of $H_{2} O_{2}$
In equation (b) iodine gets reduced to +7 to +5 state.
In equation (c) chlorine gets reduced to 0 to -1 state
In equation (d) iodine gets reduced to +0 to -1 state.
Equations in options (b) , (c), (d) depict the reducing nature of $H_{2} O_{2}$
Option (a) is correct

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