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) \(2 \mathrm{~K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]+2 \mathrm{KOH}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{~K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) \(+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\)

Step-by-Step Solution

Answer

The correct reaction is (b). \(\mathrm{H}_2\mathrm{O}_2\) acts as a reducing agent in this reaction.

Step 1: Understanding Reducing Properties

A reducing agent is a substance that loses electrons and is oxidized in a chemical reaction. In terms of reactions, it reduces another substance by giving electrons to it. In these reactions, the oxidation state of the reducing agent increases because it loses electrons.

Step 2: Analyzing Reaction (a)

In reaction (a), hydrogen peroxide \(\mathrm{H}_2\mathrm{O}_2\) reacts with \(2\mathrm{NaI}\). Here, \(\mathrm{I}^-\) from \(\mathrm{NaI}\) is oxidized to \(\mathrm{I}_2\). This indicates that \(\mathrm{H}_2\mathrm{O}_2\) is reducing \(\mathrm{I}^-\) to \(\mathrm{I}_2\) by providing oxygen, thus acting as an oxidizing agent, not a reducing agent.

Step 3: Analyzing Reaction (b)

In reaction (b), \(\mathrm{PbO}_2\) is reduced to \(\mathrm{PbO}\) and \(\mathrm{H}_2\mathrm{O}_2\) is decomposed to \(\mathrm{H}_2\mathrm{O}\) and \(\mathrm{O}_2\). Here, \(\mathrm{H}_2\mathrm{O}_2\) serves as a reducing agent as it loses oxygen (acting essentially as a donor of electrons related to \(\mathrm{PbO}_2\)).

Step 4: Analyzing Reaction (c)

In reaction (c), \(\mathrm{KMnO}_4\) is reduced to \(\mathrm{MnO}_2\) and \(\mathrm{H}_2\mathrm{O}_2\) decomposes. This shows \(\mathrm{H}_2\mathrm{O}_2\) acts as a reducing agent because it helps oxidize another substance (the manganese compound in particular), by donating electrons through the accompanying reactions.

Step 5: Analyzing Reaction (d)

Reaction (d) shows \(\mathrm{H}_2\mathrm{O}_2\) converting \(\mathrm{K}_3[\mathrm{Fe}(\mathrm{CN})_6]\) to \(\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6]\) along with the production of \(\mathrm{O}_2\) and \(\mathrm{H}_2\mathrm{O}\). Here, \(\mathrm{H}_2\mathrm{O}_2\) acts as an oxidizing agent because \(\mathrm{Fe}^{3+}\) is being reduced to \(\mathrm{Fe}^{2+}\) so \(\mathrm{H}_2\mathrm{O}_2\) gains electrons during the reaction.

Step 6: Identifying the Reducing Reaction

After evaluating all options, it is clear that options (b) and (c) involve \(\mathrm{H}_2\mathrm{O}_2\) acting as a reducing agent, with \(\mathrm{H}_2\mathrm{O}_2\) being oxidized by losing electrons. Therefore, option (b) represents the correct reaction where \(\mathrm{H}_2\mathrm{O}_2\) acts as a reducing agent.

Key Concepts

Oxidizing Properties

Hydrogen peroxide (\(\mathrm{H}_2\mathrm{O}_2\)) is a fascinating chemical that acts as both an oxidizing and reducing agent depending on the reaction. Let's focus on its oxidizing characteristics. This property allows it to accept electrons, helping other substances to lose their electrons—a process called oxidation.
The ability of \(\mathrm{H}_2\mathrm{O}_2\) to function as an oxidizing agent is commonly seen in chemical reactions where it decomposes into water and oxygen. This breakdown releases oxygen, which can then join with other substances, thus oxidizing them.

• Oxidation involves losing electrons, combining with oxygen, or removing hydrogen from compounds.
• It's critical to remember that the oxidizing properties of hydrogen peroxide are often the reason it is used as a disinfectant and bleach.

The balance between these functions makes hydrogen peroxide an adaptable chemical, applicable in various environments where it might either gain or lose electrons. In reactions where hydrogen peroxide is an oxidizing agent, the substance accepting its oxygen undergoes a change, often leading to fundamental transformations in both substances involved in the reaction. This dual capability makes \(\mathrm{H}_2\mathrm{O}_2\) incredibly useful in many industrial and household processes.

Redox Reactions

Redox reactions are short for reduction-oxidation reactions. These are essential chemical processes that involve the transfer of electrons between two substances. When one chemical species undergoes oxidation, another one must undergo reduction.
In a redox reaction, the oxidizing agent accepts electrons and is reduced, whereas the reducing agent donates electrons and is oxidized. In the context of hydrogen peroxide, these reactions are particularly interesting because \(\mathrm{H}_2\mathrm{O}_2\) can go either way—it can either gain or lose electrons.

• Oxidation is the loss of electrons.
• Reduction is the gain of electrons.
• The sum of oxidation and reduction always equals zero because electrons are neither created nor destroyed only shifted.

Understanding redox reactions is vital in chemistry because they encompass processes like combustion, respiration, rusting, and photosynthesis. For hydrogen peroxide, understanding whether it acts as an oxidizing or reducing agent involves looking at the full equation to see which substances are gaining or losing electrons.

Chemical Reactions Analysis

Chemical reaction analysis allows us to predict and understand how different substances interact and transform. It's a crucial part of chemistry that involves identifying reactants and predicting the products formed.
When analyzing reactions involving \(\mathrm{H}_2\mathrm{O}_2\), it's vital to see which role it plays—whether as a reducing or oxidizing agent. To analyze such reactions:

• Determine the initial oxidation states of all elements involved.
• Identify changes in oxidation states to figure out which elements are oxidized and reduced.
• Balance the chemical equation, ensuring mass and charge conservation.

For example, in reaction (b) from the exercise, \(\mathrm{H}_2\mathrm{O}_2\) acts as a reducing agent. It donates electrons to \(\mathrm{PbO}_2\), causing its reduction to \(\mathrm{PbO}\) while \(\mathrm{H}_2\mathrm{O}_2\) itself breaks down into water and free oxygen. This breakdown signifies the electron donation characteristic of reduction.