Question

Hydrogen peroxide acts both as an oxidizing and as a reducing agent depending upon the reacting species. In which of the following cases ${H_2}{O_2}$ acts as a reducing agent in an acidic medium?
A.$MnO_4^ -$
B.$SO_3^{2 - }$
C.$KI$
D.$C{r_2}O_7^{2 - }$

Answer 1

Hydrogen peroxide acts both as an oxidizing and reducing agent. We should know the basic terms used in the reactions. If we can determine the compound as a reducing and oxidizing agent in both an acidic medium as well as a basic medium we can easily find the case in which ${H_2}{O_2}$ will be acting as a reducing agent in an acidic medium.

Complete step by step answer:
First, we will understand the terms oxidizing and reducing agent. Oxidizing agents tend to be reduced and gain electrons. On the other hand, reducing agents tend to bring reduction by being oxidized and losing electrons.
Now we will consider the following cases and then we will check in which case ${H_2}{O_2}$ acts as the reducing agent in an acidic medium. So, for option (A) $MnO_4^ -$. Consider the reaction,
$MnO_4^ - ( + 7) + {H_2}{O_2}( - 1) + {H^ + }\xrightarrow{{}}M{n^{2 + }}( + 2) + {H_2}O + {O_2}(0)$
From the above reaction, we can conclude that $Mn( + 7)$ is reduced to $Mn( + 2)$ and ${H_2}{O_2}( - 1)$ is oxidized to ${O_2}(0)$ which means oxidation. Therefore ${H_2}{O_2}$ acts as a reducing agent in an acidic medium. Acidic medium is represented by${H^ + }$. So here in the case of $MnO_4^ -$ Hydrogen peroxide acts as a reducing agent in an acidic medium.
Now for other cases, we have the reactions as follows.
$SO_3^{2 - }( + 4) + {H_2}{O_2}( - 1) + {H^ + }\xrightarrow{{}}SO_4^{2 - }( + 6) + {H_2}O( - 2)$ Here ${H_2}{O_2}$ acts as an oxidizing agent.
$KI( - 1) + {H_2}{O_2}( - 1) + {H^ + }\xrightarrow{{}}{I_2}(0) + {H_2}O( - 2) + KOH$ Here ${H_2}{O_2}$ acts as an oxidizing agent.
$C{r_2}O_7^{2 - }( + 6) + {H_2}{O_2}( - 1) + {H^ + }\xrightarrow{{}}C{r^{3 + }}( + 3) + {O_2}(0) + {H_2}O$ Here ${H_2}{O_2}$ acts as a reducing agent.
We know that ${H_2}{O_2}$ acts as a reducing agent when the oxidation state of oxygen increases. So we can conclude that in the case of $MnO_4^ -$ and $C{r_2}O_7^{2 - }$ oxidation state of oxygen is increased from $( - 1)$ to $(0)$. This means oxidation and the agent used is the reducing agent Hydrogen peroxide ${H_2}{O_2}$.
Therefore, the correct options are (A) and (D).

Note:
Oxidation state is defined as the total number of electrons that an atom either gains or losses for a chemical bond with another atom. The oxidation state can be calculated as follows. For example $MnO_4^ -$. Consider the oxidation state $Mn(x)$ as unknown. We can find it $x + 4 \times \left( { - 2} \right) = - 1$ after calculating what we get $x = + 7$. Hence, the oxidation state $Mn$ is $+ 7$.
All the reactions given above are the redox reactions.