In which of the following reaction(s) $H_{2} O_{2}$ acts as a reducing agent?

H_{2} O_{2} + 2 H^{+} + 2 e^{-} ⟶ 2 H_{2} O

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H_{2} O_{2} ⟶ - 2 e^{-} O_{2} + 2 H^{+}

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H_{2} O_{2} + 2 e^{-} ⟶ 2 O H^{-}

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H_{2} O_{2} + 2 O H^{-} ⟶ - 2 e^{-} O_{2} + 2 H_{2} O

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Solution

The correct option is D $H_{2} O_{2} + 2 O H^{-} ⟶ - 2 e^{-} O_{2} + 2 H_{2} O$
Losing or giving away electrons reduces the oxidation number/state of the species with which $H_{2} O_{2}$ reacts. Hence, in those cases, $H_{2} O_{2}$ will be acting as a reducing agent.

In reactions:
$H_{2} O_{2} ⟶ - 2 e^{-} O_{2} + 2 H^{+}$

$H_{2} O_{2} + 2 O H^{-} ⟶ - 2 e^{-} O_{2} + 2 H_{2} O$

It is clear that $H_{2} O_{2}$ is losing electrons. Hence $H_{2} O_{2}$ acts as reducing agent in the above reactions.

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H_{2} O_{2}

H_{2} O_{2}

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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