Question

Which of the following equation depicts reducing nature of $H_2{O}_2$?

• A. $M{n}^{2+}+H_2{O}_2\to M{n}^{4+}+2O{H}^{-}$
• B. $PbS+4H_2{O}_2\to PbS{O}_4+4H_{2}O$
• C. $2\left[Fe\right(CN{)}_6{]}^{4-}+2H^{+}+H_2{O}_2\to 2\left[Fe\right(CN{)}_6{]}^{3-}+2H_{2}O$
• D. $I_2+H_2{O}_2+2O{H}^{-}\to 2I^{-}+2H_{2}O+O_2$

Answer 1

The correct option is D $I_2+H_2{O}_2+2O{H}^{-}\to 2I^{-}+2H_{2}O+O_2$

A.$2\left[Fe\right(CN{)}_6{]}^{4-}+2H^{+}+H_2{O}_2\to 2\left[Fe\right(CN{)}_6{]}^{3-}+2H_{2}O$ In this reaction, $\left[Fe\right(CN{)}_6{]}^{4-}$ is getting oxidised to $\left[Fe\right(CN{)}_6{]}^{3-}$ as oxidation number of Fe is getting increased from $+2$ to $+3$, while $H_2{O}_2$ is getting reduced to $H_{2}O$. Thus, here $H_2{O}_2$ is acting as an oxidising agent.

B.
In the given reaction,$I_2$ is getting reduced to $I^{-}$ , while $H_2{O}_2$ is getting oxidised to $O_2$ , thus here $H_2{O}_2$ is acting as a reducing agent.

C.$M{n}^{2+}+H_2{O}_2\to M{n}^{4+}+2O{H}^{-}$In this reaction, $M{n}^{2+}$ is getting oxidised to $M{n}^{4+}$ as oxidation number is getting increased from $+2$ to $+4$, while $H_2{O}_2$ is getting reduced to $O{H}^{-}$. Thus, here $H_2{O}_2$ is acting as an oxidising agent.

D.$PbS+4H_2{O}_2\to PbS{O}_4+4H_{2}O$ In this reaction, $PbS$ is getting oxidised to $PbS{O}_4$ , while$H_2{O}_2$ is getting reduced to $H_{2}O$. Thus, here $H_2{O}_2$ is acting as an oxidising agent.