(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.
(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.
Show Hint
- 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.
The Correct Option is B
Solution and Explanation
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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