$H_{2} O_{2}$ as a reducing agent in :-

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

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

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

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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 ${C l}_{2} + H_{2} O_{2} \to H C l + O_{2}$
$C l_{2} + H_{2} O_{2} ⟶ H C l + O_{2}$
Here, $C l$ in $C l_{2}$ is at $O$ oxidation state. In $H C l$ , $C l$ is at $- 1$ oxidation state. Its oxidation number decreases from $0$ to $- 1$ . $H_{2} O_{2}$ reduced it. So, $H_{2} O_{2}$ is a reducing agent here.

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Similar questions

What is the product of the reaction of $H_{2} O_{2}$ with $C l_{2}$ ?

Decomposition of H₂O₂yields:

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

N H_{4} C l + C a (O H)_{2} \to N H_{3} + C a C l_{2} + H_{2} O

B a C l_{2} + A l_{2} (S O_{4})_{3} \to B a S O_{4} + A l C l_{3}

N a A l O_{2} + H_{2} O + C O_{2} \to A l H_{3} + N a_{2} C O_{3}

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

Z n + H N O_{3} \to Z n (N O_{3})_{2} + N O_{2} + H_{2} O

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

P b O + C \to P b + C O_{2}

B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + H C l

C H_{4} + O_{2} \to C O_{2} + H_{2} O

In reaction

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

the reducing agent is –

C a C O_{3} + H C l \to C a C l_{2} + C O_{2} ↑ + H_{2} O

N a + H_{2} O \to N a O H + H_{2} ↑

(N H_{4})_{2} S O_{4} + C a (O H)_{2} \to C a S O_{4} + N H_{3} ↑ + H_{2} O

H g (N O_{3})_{2} \to H g + N O_{2} ↑ + O_{2} ↑

P b_{3} O_{4} + H C l \to P b C l_{2} + C l_{2} ↑ + H_{2} O

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