Question

Which of the following equations depict the oxidising nature of $H_2{O}_2$?

• A. $KI{O}_4+H_2{O}_2⟶KI{O}_3+H_{2}O+O_2$
• B. $2F{e}^3++2H++H_2{O}_2\to 2F{e}^2++2H_{2}O+O_2$
• C. $2Mn{O}_4^{-}+6H^{+}+5H_2{O}_2\to 2M{n}^{2+}+8H_{2}O+5O_2$
• D. $2I^{-}+2H++H_2{O}_2\to I_2+2H_{2}O$

Answer 1

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

A. $2Mn{O}_4^{-}+6H^{+}+5H_2{O}_2\to 2M{n}^{2+}+8H_{2}O+5O_2$ In this reaction, $Mn{O}_4^{-}$ is reduced to $M{n}^{2+}$ as oxidation state decreases from $+7$ to $+2$, while $H_2{O}_2$ is getting oxidised to $O_2$. Thus, here $H_2{O}_2$ is acting as a reducing agent.

B. $2F{e}^{3+}+2H^{+}+H_2{O}_2\to 2F{e}^{2+}+2H_{2}O+O_2$ In this reaction, $F{e}^{3+}$ is reduced to $F{e}^{2+}$ as oxidation state decreases from $+3$ to $+2$, while $H_2{O}_2$ is getting oxidised to $O_2$. Thus, here $H_2{O}_2$ is acting as a reducing agent.

C.
In this reaction,$H_2{O}_2$ is oxidising $I^{-}$ to $I_2$ and itself getting reduced to $H_{2}O$. Thus, here $H_2{O}_2$ is acting as an oxidising agent.

D. $KI{O}_4+H_2{O}_2\to KI{O}_3+H_{2}O+O_2$ In this reaction, $KI{O}_4$ is reduced to $KI{O}_3$ as the oxidation state is reduced from $+7$ to $+5$, while $H_2{O}_2$ is getting oxidised to $O_2$. Thus, here $H_2{O}_2$ is acting as a reducing agent.
Hence, option (C) is correct.