Problem 118 - A Complete Resource Book in Chemistry for JEE Main

Question

Which one of the following reactions represents the reducing property of \(\mathrm{H}_{2} \mathrm{O}_{2} ?\) (a) \(2 \mathrm{NaI}+\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{I}_{2}+2 \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{PbO}_{2}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{PbO}+\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\) (c) \(2 \mathrm{KMnO}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4}+5 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{K}_{2} \mathrm{SO}_{4}+8 \mathrm{H}_{2} \mathrm{O}\) \(+5 \mathrm{O}_{2}+2 \mathrm{MnO}_{2}\) (d) \(2 \mathrm{~K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{2}+2 \mathrm{KOH}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{~K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) \(+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\)

Step-by-Step Solution

Answer

Reactions B, C, and D show the reducing property of \(\mathrm{H}_{2} \mathrm{O}_{2}\).

Step 1: Understanding Reducing Properties

A reducing agent loses electrons and is oxidized in the process. Therefore, in a reaction where \( \mathrm{H}_{2} \mathrm{O}_{2} \) acts as a reducing agent, it will transform into a more oxidized state.

Step 2: Analyze Reaction A

The reaction is \(2 \mathrm{NaI}+\mathrm{H}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{I}_{2}+2 \mathrm{H}_{2} \mathrm{O}\). Here, \(\mathrm{H}_{2} \mathrm{O}_{2}\) is reduced to \(\mathrm{H}_{2} \mathrm{O}\), which suggests it is acting as an oxidizing agent, not a reducing agent.

Step 3: Analyze Reaction B

The reaction is \(\mathrm{PbO}_{2}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{PbO}+\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\). Here, \(\mathrm{H}_{2} \mathrm{O}_{2}\) is converted to \(\mathrm{O}_{2}\), meaning it is oxidized. Therefore, \(\mathrm{H}_{2} \mathrm{O}_{2}\) is acting as a reducing agent in this reaction.

Step 4: Analyze Reaction C

The reaction is \(2 \mathrm{KMnO}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4}+5 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{K}_{2} \mathrm{SO}_{4}+8 \mathrm{H}_{2} \mathrm{O}+5 \mathrm{O}_{2}+2 \mathrm{MnO}_{2}\). In this reaction, \(\mathrm{H}_{2} \mathrm{O}_{2}\) is converted to \(\mathrm{O}_{2}\) and appears to be oxidized. Therefore, \(\mathrm{H}_{2} \mathrm{O}_{2}\) is acting as a reducing agent.

Step 5: Analyze Reaction D

The reaction is \(2 \mathrm{K}_{3}[\mathrm{Fe}(\mathrm{CN})_{6}]^{2}+2 \mathrm{KOH}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{K}_{4}[\mathrm{Fe}(\mathrm{CN})_{6}]+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\). Here, \(\mathrm{H}_{2} \mathrm{O}_{2}\) is again converted to \(\mathrm{O}_{2}\), acting as a reducing agent.

Step 6: Conclusion

After analyzing all reactions, reactions B, C, and D show \(\mathrm{H}_{2} \mathrm{O}_{2}\) being oxidized, indicating it acts as a reducing agent. They demonstrate the reducing property of \(\mathrm{H}_{2} \mathrm{O}_{2}\).

Key Concepts

Oxidation-Reduction (Redox) Reactions

Redox reactions are chemical reactions where the oxidation states of atoms are changed. This involves the transfer of electrons between two substances. In the realm of chemistry, these reactions are crucial because they include processes that provide energy, like combustion and cellular respiration. During a redox reaction, the substance that gains electrons is reduced, while the substance that loses electrons is oxidized.
To determine which reactant is oxidized or reduced, it helps to track the movement of electrons. Oxidation involves an increase in oxidation state and loss of electrons, whereas reduction signifies a decrease in oxidation state and gain of electrons.
For example, in the given reactions, observing how hydrogen peroxide ( H_{2} O_{2}) acts reveals its behavior as either an oxidizing or reducing agent. When H_{2} O_{2} changes to O_{2} it indicates that it is losing electrons, thus undergoing oxidation in a context where it is serving as a reducing agent.

Chemical Reactivity and Mechanisms

Chemical reactivity helps us understand how different substances interact during a reaction. Mechanisms describe the step by step process by which a reaction occurs. Understanding these helps in predicting the products of chemical reactions.
Reaction mechanisms often involve intermediates and transition states that fleetingly appear and contribute to the overall transformation of reactants to products. Each step within a mechanism illustrates an electron movement, either as transfer or sharing, underpinning chemical reactions.
In the context of the reactions involving H_{2} O_{2}, its chemical behavior changes based on the substances it's reacting with. The ability of H_{2} O_{2} to function as either an oxidizing or reducing agent depends on the reaction conditions and the other reactants involved. This adaptability makes it a valuable chemical in numerous reactions and industrial applications.

Oxidizing and Reducing Agents

An oxidizing agent, or oxidant, accepts electrons from another species, thereby causing the oxidant to gain electrons and become reduced itself. A reducing agent, or reductant, donates electrons to another species, leading to its own oxidation. These agents are central to redox reactions.
For a substance to act as a reducing agent, it needs to have a high electron density or be in a state that easily loses electrons. H_{2} O_{2} is an interesting molecule because it is amphoteric, meaning it can act as both an oxidizing and a reducing agent.
In reactions B, C, and D from the exercise, H_{2} O_{2} was found to act as a reducing agent since it was oxidized to O_{2}. Its ability to oxidize indicates its electron-donating property in these reactions.