Animated Solution for Chemistry - Redox Reactions: In basic medium, $H_2{O}_2$ exhibits which of the following reactions ? A. $M{n}^{2+}\to M{n}^{4+}$ B. $I_2\to I^{-}$ C. $PbS\to PbS{O}_4$ Choose the most appropriate answer from the options given below.

Hydrogen Peroxide

Hydrogen peroxide ($H_2{O}_2$) is one of the most fascinating molecules in chemistry. It is a true chemical chameleon.

Because the oxygen atoms in $H_2{O}_2$ are in an intermediate oxidation state of $-1$, they have a choice. They can either lose electrons to become oxygen gas ($O_2$, oxidation state $0$), or they can gain electrons to become water or hydroxide ions (oxidation state $-2$).

This unique property allows $H_2{O}_2$ to act as both an oxidizing agent and a reducing agent. However, its behavior is not just about its own structure; it is heavily dictated by the environment it is in, specifically the pH of the medium.

In this problem, we are on a mission to identify which of the given reactions are exhibited by $H_2{O}_2$ specifically in a basic medium. Let's break them down one by one.

The Oxidation of Manganese

Let's look at the first candidate:

The first step in any redox analysis is to track the electrons. Here, the oxidation state of manganese increases from $+2$ to $+4$. An increase in oxidation state means a loss of electrons, which is the very definition of oxidation.

Since manganese is being oxidized, $H_2{O}_2$ must step up as the oxidizing agent. The question is: does this happen in a basic medium?

Yes, it does! In a basic environment, $H_2{O}_2$ readily accepts electrons to form hydroxide ions:

When we combine this with the oxidation of manganese, we get a perfectly balanced reaction that thrives in basic conditions:

Therefore, Reaction A is a valid process in a basic medium.

The Reduction of Iodine

Now, let's turn our attention to the second reaction:

Here, we have elemental iodine ($I_2$) with an oxidation state of $0$ converting into iodide ions ($I^{-}$) with an oxidation state of $-1$. A decrease in oxidation state indicates a gain of electrons, meaning iodine is undergoing reduction.

For iodine to be reduced, $H_2{O}_2$ must act as the reducing agent. In a basic medium, $H_2{O}_2$ reacts with hydroxide ions to release oxygen gas and electrons:

This half-reaction perfectly complements the reduction of iodine. The overall reaction is a classic example of $H_2{O}_2$ acting as a reductant in basic conditions:

Thus, Reaction B is also exhibited in a basic medium.

The Acidic Exception

Finally, let's examine the third reaction:

This is a dramatic transformation. The sulfur in lead sulfide ($PbS$) starts with an oxidation state of $-2$. In lead sulfate ($PbS{O}_4$), the sulfur has been oxidized all the way up to $+6$. This is a massive oxidation process, requiring a strong oxidizing agent.

$H_2{O}_2$ is certainly up to the task. However, there is a catch. This specific reaction—often used in art restoration to convert black lead sulfide stains back into white lead sulfate—is characteristically performed in an acidic medium.

The overall reaction looks like this:

While $H_2{O}_2$ is acting as an oxidizing agent here, the standard conditions for this specific chemical transformation are acidic, not basic. Therefore, Reaction C does not meet the criteria of our question.

We have carefully analyzed all three reactions. We found that $H_2{O}_2$ can successfully drive the oxidation of $M{n}^{2+}$ (Reaction A) and the reduction of $I_2$ (Reaction B) in a basic medium. However, the oxidation of $PbS$ (Reaction C) is a hallmark of an acidic medium.

Therefore, the reactions exhibited in a basic medium are A and B.

Looking at our options, the correct choice is (d).