Question

(A) $HOCl + H _{2} O _{2} \rightarrow H _{3} O ^{+}+ Cl ^{-}+ O _{2}$
(B) $I _{2}+ H _{2} O _{2}+2 OH ^{-} \rightarrow 2 I ^{-}+2 H _{2} O + O _{2}$
Choose the correct option.

• A. $H _{2} O _{2}$ acts as reducing and oxidising agent respectively in equation (A) and (B)
• B. $H _{2} O _{2}$ acts as oxidising agent in equation (A) and $( B )$
• C. $H _{2} O _{2}$ acts as reducing agent in equation (A) and $( B )$
• D. $H _{2} O _{2}$ act as oxidizing and reducing agent respectively in equation (A) and (B)

Answer 1

The Correct Option is C

To determine the role of $H_2O_2$ in each equation, we need to analyze whether it undergoes oxidation or reduction.

1. In equation (A): $HOCl + H_2O_2 \rightarrow H_3O^+ + Cl^- + O_2$
   • The oxidation state of oxygen in $H_2O_2$ is -1.
   • In $O_2$, the oxidation state is 0. This increase in the oxidation state indicates that $H_2O_2$ is oxidized (losing electrons), hence acting as a reducing agent.
2. In equation (B): $I_2 + H_2O_2 + 2 OH^- \rightarrow 2 I^- + 2 H_2O + O_2$
   • Similarly, in this reaction, the oxidation state of oxygen in $H_2O_2$ goes from -1 to 0 in $O_2$.
   • This again indicates that $H_2O_2$ is oxidized, acting as a reducing agent.

Hence, in both equations (A) and (B), $H_2O_2$ acts as a reducing agent. Therefore, the correct answer is: $H_2O_2$ acts as a reducing agent in equation (A) and (B).