Which of the following reactions shows reducing nature of $H_{2} O_{2}$ ?

P b S + 4 H_{2} O_{2} \to P b S O_{4} + 4 H_{2} O

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A g_{2} O + H_{2} O_{2} \to 2 A g + H_{2} O + O_{2}

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2 H C H O + H_{2} O_{2} \to 2 H C O O H + H_{2} O

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

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Solution

The correct option is B $A g_{2} O + H_{2} O_{2} \to 2 A g + H_{2} O + O_{2}$
In the following reaction $A g_{2} O$ is reduced to $A g$ hence $H_{2} O_{2}$ acts as a reducing agent
$A g_{2} O + H_{2} O_{2} \to 2 A g + H_{2} O + O_{2}$
The oxidation number of silver decreases from $+ 1$ (in $A g_{2} O$ ) to $0$ (in $A g$ ). This indicates reduction.

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Similar questions

Which of the following reaction shows the reduction reaction of $H_{2}$ $O_{2}$ ?

H_{2} O_{2}

A g_{2} O + H_{2} O_{2} \to 2 A g + H_{2} O + O_{2}

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

H_{2} O_{2} + A g_{2} O \to 2 A g + H_{2} O + O_{2}

In the reaction $A g_{2} O + H_{2} O_{2} \to 2 A g + H_{2} O + O_{2} . H_{2} O_{2}$ acts as :

(I) H_{2} O_{2} + O_{3} \to H_{2} O + 2 O_{2}

(I I) H_{2} O_{2} + A g_{2} O \to 2 A g + H_{2} O + O_{2}

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