In which of the following reactions $H_{2} O_{2}$ is a reducing agent

2 F e C l_{2} + 2 H C l + H_{2} O_{2} \to 2 F e C l_{3} + 2 H_{2} O

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

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

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H_{2} S O_{3} + H_{2} O_{2} \to H_{2} S O_{4} + H_{2} O

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Solution

The correct option is B $C l_{2} + H_{2} O_{2} \to 2 H C l + O_{2}$
$C l_{2}^{[0]} + H_{2} O_{2} \to 2 H C l^{[- 1]} + O_{2}$ .
In this reaction chlorine gets reduced from zero to -1 oxidation state.

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H_{2} O_{2}

(a)

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

(b)

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

(c)

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

(d)

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

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