Question

$\text{(A)}HOCl+H_2{O}_2\to H_3{O}^{+}+C{l}^{-}+O_2$
$\text{(B)}{I}_2+H_2{O}_2+2O{H}^{-}\to 2I^{-}+2H_{2}O+O_2^{-}$

Choose the correct option.

• A. $H_2{O}_2$ acts as oxidising agent in equations (A) and (B).
• B. $H_2{O}_2$ act as oxidizing and reducing agent respectively in equation (A) and (B).
• C. $H_2{O}_2$ acts as reducing agent in equations (A) and (B).
• D. $H_2{O}_2$ acts as reducing and oxidising agent respectively in equationa (A) and (B).

Answer 1

The correct option is C $H_2{O}_2$ acts as reducing agent in equations (A) and (B).

$\text{(A)}HOCl+H_2{O}_2\to H_3{O}^{+}+C{l}^{-}+O_2$

In this equation, $H_2{O}_2$ is reducing chlorine from $+1$ to $-1$.

$\text{(B)}{I}_2+H_2{O}_2+2O{H}^{-}\to 2I^{-}+2H_{2}O+O_2^{-}$

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 and $H_2{O}_2$ 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.
Option (c) is correct