(A) HOCl + H_2O_2 rightarrow H_3O^{+} + Cl^{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) In equation $(A)$: $HOCl + H_2O_2 \rightarrow H_3O^{+} + Cl^{-} + O_2$. The oxidation state of $Cl$ changes from $+1$ in $HOCl$ to $-1$ in $Cl^{-}

Vedclass · Accepted Answer

$H_2O_2$ acts as a reducing agent in equations $(A)$ and $(B)$.

Vedclass · Answer

(C) In equation $(A)$: $HOCl + H_2O_2 \rightarrow H_3O^{+} + Cl^{-} + O_2$. The oxidation state of $Cl$ changes from $+1$ in $HOCl$ to $-1$ in $Cl^{-}$. Thus,$HOCl$ is reduced and $H_2O_2$ acts as a reducing agent. In equation $(B)$: $I_2 + H_2O_2 + 2 OH^{-} \rightarrow 2 I^{-} + 2 H_2O + O_2$. The oxidation state of $I$ changes from $0$ in $I_2$ to $-1$ in $I^{-}$. Thus,$I_2$ is reduced and $H_2O_2$ acts as a reducing agent. Therefore,in both equations $(A)$ and $(B)$,$H_2O_2$ acts as a reducing agent.

$(A) \ HOCl + H_2O_2 \rightarrow H_3O^{+} + Cl^{-} + O_2$
$(B) \ I_2 + H_2O_2 + 2 OH^{-} \rightarrow 2 I^{-} + 2 H_2O + O_2$
Choose the correct option.

Similar Questions

Why does the following reaction occur?
$XeO_6^{4-}{(aq)} + 2F^{-}{(aq)} + 6H^{+}{(aq)} \rightarrow XeO_{3_{(g)}} + F_{2_{(g)}} + 3H_2O_{(l)}$
What conclusion about the compound $Na_4XeO_6$ (of which $XeO_6^{4-}$ is a part) can be drawn from the reaction?

The strongest reducing agent among the given options is:

Which of the following may act as both an oxidising and a reducing agent?

The oxide which cannot act as a reducing agent is

Which of the following is not a reducing agent?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$(A) \ HOCl + H_2O_2 \rightarrow H_3O^{+} + Cl^{-} + O_2$ $(B) \ I_2 + H_2O_2 + 2 OH^{-} \rightarrow 2 I^{-} + 2 H_2O + O_2$ Choose the correct option.