Problem 35 - A Complete Resource Book in Chemistry for JEE Main

Question

In which of the following reactions does \(\mathrm{H}_{2} \mathrm{O}_{2}\) acts as a reducing agent? (a) \(\mathrm{PbO}_{2}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \longrightarrow \mathrm{PbO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) \(+\mathrm{O}_{2}(\mathrm{~g})\) (b) \(\mathrm{KNO}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \longrightarrow \mathrm{KNO}_{3}(\mathrm{aq})+\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (c) \(2 \mathrm{KI}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \longrightarrow 2 \mathrm{KOH}(\mathrm{aq})+\) (d) \(\mathrm{Na}_{2} \mathrm{SO}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \stackrel{\mathrm{I}_{2}(\mathrm{~s})}{\longrightarrow} \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) \(+\overline{\mathrm{H}}_{2} \mathrm{O}(\mathrm{l})\)

Step-by-Step Solution

Answer

\(\mathrm{H}_2 \mathrm{O}_2\) acts as a reducing agent in reaction (a).

Step 1: Definition of Reducing Agent

A reducing agent is a substance that donates electrons to another substance in a chemical reaction and is oxidized in the process. To determine where \( \mathrm{H}_2 \mathrm{O}_2 \) acts as a reducing agent, we must find where it is oxidized.

Step 2: Determine Oxidation States of Reactants and Products

For each reaction, identify the oxidation states of \( \mathrm{H}_2 \mathrm{O}_2 \) in the reactants and products, focusing on changes in oxidation states.

Step 3: Analyze Reaction (a)

In reaction (a), \( \mathrm{H}_2 \mathrm{O}_2 \) is a reactant. The oxidation state of oxygen in \( \mathrm{H}_2 \mathrm{O}_2 \) is -1. In \( \mathrm{H}_2 \mathrm{O} \), it's -2 and in \( \mathrm{O}_2 \), it's 0. Thus, \( \mathrm{H}_2 \mathrm{O}_2 \) is both oxidized and reduced. Therefore, \( \mathrm{H}_2 \mathrm{O}_2 \) acts as a reducing agent.

Step 4: Analyze Reaction (b)

In reaction (b), the oxidation state of oxygen in \( \mathrm{H}_2 \mathrm{O}_2 \) changes from -1 to -2 in \( \mathrm{H}_2 \mathrm{O} \). \( \mathrm{H}_2 \mathrm{O}_2 \) is only reduced, not oxidized.

Step 5: Analyze Reaction (c)

In reaction (c), the oxidation state of oxygen in \( \mathrm{H}_2 \mathrm{O}_2 \) remains the same as it forms water, indicating that it is only reduced instead of being oxidized.

Step 6: Analyze Reaction (d)

In reaction (d), the oxidation state of oxygen in \( \mathrm{H}_2 \mathrm{O}_2 \) changes from -1 to -2 in \( \mathrm{H}_2 \mathrm{O} \) and \( \mathrm{O}_3 \), indicating it is only reduced. No oxidation of \( \mathrm{H}_2 \mathrm{O}_2 \) occurs.

Step 7: Conclusion

Compile the findings: \( \mathrm{H}_2 \mathrm{O}_2 \) acts as a reducing agent in reaction (a) because it is oxidized to form \( \mathrm{O}_2 \) where the oxidation state of oxygen increases.

Key Concepts

Oxidizing and reducing agents

Oxidizing and reducing agents are key players in oxidation-reduction (redox) reactions, which involve the transfer of electrons from one substance to another. In a redox reaction, the reducing agent is the substance that donates electrons, becoming oxidized in the process. Conversely, the oxidizing agent gains electrons and is reduced. These agents effectively "swap" oxidation states to sustain the reaction.

Understanding the role of these agents is crucial when analyzing chemical reactions because it helps in identifying how substances interact on an atomic level. In the context of our exercise, hydrogen peroxide, \( \mathrm{H}_2\mathrm{O}_2 \), can act as both an oxidizing agent and a reducing agent, depending on the specific reaction it participates in.

In reaction (a) from our example, \( \mathrm{H}_2\mathrm{O}_2 \) acts as a reducing agent as it donates electrons when transforming to \( \mathrm{O}_2 \), demonstrating its dual role and reinforcing how versatile it can be in chemical reactions.

Oxidation states

Oxidation states, also known as oxidation numbers, are an essential part of understanding chemical reactions, as they provide insight into the loss or gain of electrons by elements in a compound. The oxidation state is an integer value that represents the total number of electrons that an atom either gains or loses to form a chemical bond.

Here are some key pointers:

• For elemental oxygen, like in \( \mathrm{O}_2 \), the oxidation state is 0.
• In \( \mathrm{H}_2\mathrm{O}_2 \), each oxygen atom has an oxidation state of -1, due to its intermediate nature where it can either donate or accept electrons.
• In water \( \mathrm{H}_2\mathrm{O} \), oxygen has an oxidation state of -2, reflecting its complete reduction.

By studying these oxidation changes, as seen where \( \mathrm{H}_2\mathrm{O}_2 \) transforms into \( \mathrm{O}_2 \), we can track if the molecule undergoes oxidation or reduction, revealing the roles they play within the reaction.

Chemical reaction analysis

Chemical reaction analysis involves examining the transformation of reactants into products, exploring changes in their chemical structures and energy levels. This fundamental process helps to identify whether the transformations involve redox processes, combustion reactions, or synthesis, among others.

Let's delve into how this applies to our exercise amongst the reactions:

• In reaction (a), \( \mathrm{H}_2\mathrm{O}_2 \) decomposes, and both its reduction (forming water) and oxidation (forming oxygen gas) are noted, marking it as a classic redox reaction where \( \mathrm{H}_2\mathrm{O}_2 \) acts as a reducing agent.
• For reactions (b), (c), and (d), \( \mathrm{H}_2\mathrm{O}_2 \) does not undergo oxidation; it is only reduced, indicating that \( \mathrm{H}_2\mathrm{O}_2 \) is only receiving electrons rather than donating them across these processes.

By analyzing the oxidation states and the nature of the reaction, we can determine the specific roles the reactants play within the reactions.

Hydrogen peroxide reactions

Hydrogen peroxide is a fascinating compound with the chemical formula \( \mathrm{H}_2\mathrm{O}_2 \). It serves multiple purposes and has different roles in chemical reactions due to its instability and tendency to decompose into water and oxygen.

In redox reactions, \( \mathrm{H}_2\mathrm{O}_2 \) can act both as an oxidizing agent and a reducing agent, a feature that reflects its intermediate oxidation state of -1 in each oxygen atom. Its behavior varies based on the chemicals it interacts with.

For instance, in our exercise, reaction (a) showcases \( \mathrm{H}_2\mathrm{O}_2 \) acting as a reducing agent resulting in the release of oxygen gas. This dual functionality is harnessed in various applications, from bleaching and antiseptics to propulsion in rocketry.

• This versatility makes \( \mathrm{H}_2\mathrm{O}_2 \) unique, standing out among other compounds in its ability to donate and accept electrons, influencing reactions by changing the oxidation state of other substances.
• Understanding this behavior is critical for predicting the outcomes of reactions where \( \mathrm{H}_2\mathrm{O}_2 \) is involved.