Animated Solution for Chemistry - s and p-Block Elements: ${\text{Fe}}^{3+}$ is reduced to ${\text{Fe}}^{2+}$ by using -

How Hydrogen Peroxide Chooses Its Disguise

Have you ever wondered how a single molecule can play two completely opposite roles? Imagine a character in a movie who is a hero in one city but a villain in another. In the world of chemistry, hydrogen peroxide ($H_2{O}_2$) is exactly that character. Depending on the environment it finds itself in, it can either steal electrons (act as an oxidizing agent) or generously give them away (act as a reducing agent).

This classic JEE Advanced problem tests your understanding of this dual nature. We are tasked with finding the right conditions to reduce the ferric ion ($F{e}^{3+}$) to the ferrous ion ($F{e}^{2+}$). Let's dive into the fascinating mechanics of how pH controls the fate of this reaction.

Forcing a Reduction

Our goal is simple but specific:

We need to force an electron onto $F{e}^{3+}$. This means we need a reducing agent—a substance willing to sacrifice its own electrons. All our options involve hydrogen peroxide ($H_2{O}_2$) or sodium peroxide ($N{a}_2{O}_2$), which generates $H_2{O}_2$. So, the real question is: under what conditions does $H_2{O}_2$ act as a reducing agent?

The Electron Thief

Let's look at what happens when we place $H_2{O}_2$ in an acidic medium, like in the presence of sulfuric acid ($H_{2}S{O}_4$). In an acidic environment, $H_2{O}_2$ is hungry for electrons. Its standard reduction potential is a massive $+1.77\text{ V}$.

Because this potential is so high, $H_2{O}_2$ acts as a powerful oxidizing agent. If we put it with iron, it will aggressively steal electrons from $F{e}^{2+}$, oxidizing it to $F{e}^{3+}$:

This is the exact opposite of what we want! Therefore, any option with an acidic medium—like Options (C) and (D) which contain $H_{2}S{O}_4$—will fail to reduce $F{e}^{3+}$.

The Electron Donor

Now, let's flip the script. What happens if we add a base like sodium hydroxide ($NaOH$)? In an alkaline medium, the oxidation potential of $H_2{O}_2$ becomes much more favorable. It becomes willing to give up electrons and oxidize itself into oxygen gas ($O_2$).

In this basic environment, $H_2{O}_2$ transforms into a reducing agent. It will happily donate electrons to our $F{e}^{3+}$ ions:

This perfectly matches Option (A), where $H_2{O}_2$ is paired with $NaOH$.

Sodium Peroxide

What about Option (B)? It gives us sodium peroxide ($N{a}_2{O}_2$) in water. At first glance, there is no explicit base mentioned. But chemistry is full of hidden reactions!

When sodium peroxide is dropped into water, it undergoes a vigorous hydrolysis reaction:

Look closely at the products. We have generated hydrogen peroxide and sodium hydroxide in the same beaker! This is an in situ generation of a basic medium with $H_2{O}_2$. It is functionally identical to Option (A). Therefore, the mixture will successfully act as a reducing agent and convert $F{e}^{3+}$ to $F{e}^{2+}$.

By understanding the pH-dependent redox behavior of hydrogen peroxide, the problem unravels beautifully. Acidic conditions make it an oxidizer, while basic conditions make it a reducer.

Thus, the correct reagents to reduce $F{e}^{3+}$ to $F{e}^{2+}$ are (A) $H_2{O}_2$ in presence of $NaOH$ and (B) $N{a}_2{O}_2$ in water.

Always remember: in chemistry, the environment doesn't just surround the molecules; it dictates their destiny!