In which of the following reaction $H_{2} O_{2}$ acts as the 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 options are
B $H_{2} O_{2} - 2 e^{-} ⟶ O_{2} + 2 H^{+}$
D $H_{2} O_{2} + 2 O H^{-} - {2 e}^{-} ⟶ O_{2} + 2 H_{2} O$
A reducing agent gets oxidised while reducing the other reactant. Hence it should release electrons.
So in reactions,
$H_{2} O_{2} - 2 e^{-} ⟶ O_{2} + 2 H^{+}$ & $H_{2} O_{2} + 2 O H^{-} - 2 e^{-} ⟶ O_{2} + 2 H_{2} O$ ; $H_{2} O_{2}$ acts as reducing agent.

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Similar questions

{A g}_{2} O + H_{2} O_{2} ⟶ 2 A g + H_{2} O + O_{2}

H_{2} O_{2}

H_{2} O_{2}

H_{2} O_{2} ⟶ H_{2} O + O

In reaction

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

the reducing agent is –

H_{2} O_{2}

Z n + 2 O H^{-} + 2 H_{2} O ⟶ Z n (O H)_{4}^{2 -} + H_{2}

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