# Solve this

Question:

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

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

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

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

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

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

1. $\mathrm{H}_{2} \mathrm{O}_{2}$ acts as reducing and oxidising agent respectively in equation (A) and (B)

2. $\mathrm{H}_{2} \mathrm{O}_{2}$ acts as oxidising agent in equation (A)

and (B)

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

4. $\mathrm{H}_{2} \mathrm{O}_{2}$ act as oxidizing and reducing agent respectively in equation (A) and (B)

Correct Option: , 3

Solution:

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

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