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

Answer: (C)

$H _{2} O _{2}$ acts as reducing agent in equation (A) and $( B )$

Answer 2

(A) $\quad HOCl + H _{2} O _{2} \rightarrow H _{3} O ^{+}+ Cl ^{-}+ O _{2}$

In this equation, $H _{2} O _{2}$ is reducing chlorine from +1 to -1

(B) $I _{2}+ H _{2} O _{2}+2 OH ^{-} \rightarrow 2 I ^{-}+2 H _{2} O + O _{2}$

In this equation, $H _{2} O _{2}$ is reducing iodine from 0 to $-1$

In (A) reduction of HOCl occurs so it will be a oxidising agent hence $H _{2} O _{2}$ will be a reducing agent.

In $(B)$ reduction of $I _{2}$ occurs so it will be a oxidising agent and $H _{2} O _{2}$ will be a reducing agent.