Question:

(A) \( \mathrm{HOCl + H_2O_2 \rightarrow H_3O^+ + Cl^- + O_2} \)  |  
(B) \( \mathrm{I_2 + H_2O_2 + 2OH^- \rightarrow 2I^- + 2H_2O + O_2} \). 
Choose the correct option. 

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If $H_2O_2$ produces oxygen gas ($O_2$) in a reaction, it is acting as a reducing agent. If it produces water or hydroxide, it is acting as an oxidising agent.
Updated On: Jan 9, 2026
  • H\_2O\_2 acts as oxidising agent in (A) and (B).
  • H\_2O\_2 acts as reducing agent in (A) and (B).
  • H\_2O\_2 acts as oxidizing and reducing agent respectively.
  • H\_2O\_2 acts as reducing and oxidising agent respectively.
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The Correct Option is B

Solution and Explanation

Step 1: In (A), Chlorine in HOCl (+1) is reduced to $Cl^-$ (-1). Thus $H_2O_2$ is the reducing agent.
Step 2: In (B), Iodine ($I_2$) in oxidation state 0 is reduced to $I^-$ (-1). Thus $H_2O_2$ is the reducing agent.
Step 3: In both cases, $H_2O_2$ is oxidized to $O_2$ (0).
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