Question

In which of the following reactions $H_2{O}_2$ acts as a reducing agent?

$\begin{array}{c}I.H_2{O}_2+2H^{+}+2e^{-}\\ \downarrow \\ 2H_{2}O\end{array}$
II. $H_2{O}_2-2e^{-}\to O_2+2H^{+}$
III. $H_2{O}_2+2e^{-}\to 2O{H}^{-}$
$\begin{array}{c}IV.H_2{O}_2+2H^{+}+2e^{-}\\ \downarrow \\ O_2+2H_{2}O\end{array}$

• A. I and II
• B. II and IV
• C. III and IV
• D. I and III

Answer 1

The correct option is B II and IV

A reducing agent reduces another substance and itself gets oxidised.
An oxidising agent oxidises another substance and itself gets reduced. Analysing the given reactions
Reaction (I):
$H_2{O}_2+2H^{+}+2e^{-}\to 2H_{2}O$

In this reaction, after addition of $2e^{-},H_{2}O$ is obtained as product.
This implies that $H_2{O}_2$ is getting reduced by addition of 2 electrons.
So, $H_2{O}_2$ is acting as an oxidising agent .
Reaction (II):
$H_2{O}_2-2e^{-}\to O_2+2H^{+}$
Oxidation state: $H=+1O=0H=+1$
$O=-1$

In this reaction, $H_2{O}_2$ is getting oxidised since oxidation state of 𝑂 is increasing from −1 to 0.
This implies that $H_2{O}_2$ is getting oxidised. So, $H_2{O}_2$ is acts as a reducing agent.
Reaction (III):
$H_2{O}_2+2e^{-}\to 2O{H}^{-}$
In this reaction, $H_2{O}_2$ is reduced to $O{H}^{-}$ by gaining 2 electrons. Thus, it acts as an oxidising agent

Reaction (IV):
$H_2{O}_2+2O{H}^{-}+2e^{-}\to O_2+2H_{2}O$
In this reaction, after removal of $2e^{-},O_2$ and $H_{2}O$ are obtained as products.
This implies that $H_2{O}_2$ is oxidised by losing 2 electrons.
Hence, in II and IV reactions, $H_2{O}_2$ is getting oxidised thus, acts as a reducing agent.
So, the correct answer is option (C).