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Question: Consider the reaction: $I_2 + H_2O_2 + 2OH^- \rightarrow 2I^- + 2H_2O + O_2$. Which of the following...

Consider the reaction: I2+H2O2+2OH2I+2H2O+O2I_2 + H_2O_2 + 2OH^- \rightarrow 2I^- + 2H_2O + O_2. Which of the following statements are correct?

A

I2I_2 is oxidized.

B

H2O2H_2O_2 is reduced.

C

I2I_2 acts as an oxidizing agent.

D

H2O2H_2O_2 acts as a reducing agent.

Answer

(C) I2I_2 acts as an oxidizing agent. (D) H2O2H_2O_2 acts as a reducing agent.

Explanation

Solution

The reaction involves changes in oxidation states: Iodine in I2I_2 (oxidation state 0) is reduced to II^- (oxidation state -1). A species that is reduced acts as an oxidizing agent, therefore, I2I_2 is the oxidizing agent. Oxygen in H2O2H_2O_2 (oxidation state -1) is oxidized to O2O_2 (oxidation state 0). A species that is oxidized acts as a reducing agent, therefore, H2O2H_2O_2 is the reducing agent. The presence of OHOH^- indicates that the reaction is taking place in a basic medium.