Problem 117 - A Complete Resource Book in Chemistry for JEE Main

Question

In which of the following reactions, \(\mathrm{H}_{2} \mathrm{O}_{2}\) acts as an oxidant? (a) \(\mathrm{PbO}_{2}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow \mathrm{PbO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{O}_{2}(\mathrm{~g})\) (b) \(\mathrm{KNO}_{2}\) (aq) \(+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow \mathrm{KNO}_{3}\) (aq) \(+\mathrm{H}_{2} \mathrm{O}\) (l) (c) \(2 \mathrm{KI}\) (aq) \(+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow 2 \mathrm{KOH}\) (aq) \(+\mathrm{I}_{2}(\mathrm{~s})\) (d) \(\mathrm{Na}_{2} \mathrm{SO}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}_{2}\) (aq) \(\rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}\) (aq) \(+\mathrm{H}_{2} \mathrm{O}\) (l)

Step-by-Step Solution

Answer

In reactions (b), (c), and (d), \(\mathrm{H}_2 \mathrm{O}_2\) acts as an oxidant.

Step 1: Understanding Oxidizing Agent

An oxidizing agent is a substance that causes oxidation by accepting electrons from another substance. In this context, \(\mathrm{H}_2 \mathrm{O}_2\) acts as an oxidant if it is accepting electrons from another substance, causing the other substance to lose electrons (be oxidized) while itself being reduced.

Step 2: Analyze Reaction (a)

In reaction (a), \(\mathrm{PbO}_2\) is reduced to \(\mathrm{PbO}\) while \(\mathrm{H}_2 \mathrm{O}_2\) is decomposed to \(\mathrm{O}_2\). The oxidation state of Pb decreases from +4 in \(\mathrm{PbO}_2\) to +2 in \(\mathrm{PbO}\), indicating reduction, but \(\mathrm{H}_2 \mathrm{O}_2\) itself is decomposed into \(\mathrm{O}_2\), which doesn't involve electron gain by \(\mathrm{H}_2 \mathrm{O}_2\). Thus, \(\mathrm{H}_2 \mathrm{O}_2\) is not an oxidant here.

Step 3: Analyze Reaction (b)

For reaction (b), \(\mathrm{KNO}_2\) is oxidized to \(\mathrm{KNO}_3\). The nitrogen in \(\mathrm{KNO}_2\) changes from an oxidation state of +3 to +5 in \(\mathrm{KNO}_3\), indicating that electrons are taken from \(\mathrm{KNO}_2\) by \(\mathrm{H}_2 \mathrm{O}_2\). Therefore, \(\mathrm{H}_2 \mathrm{O}_2\) is acting as an oxidant in this reaction.

Step 4: Analyze Reaction (c)

In reaction (c), the iodine in \(\mathrm{KI}\) is oxidized to \(\mathrm{I}_2\), going from an oxidation state of -1 in \(\mathrm{I}^-\) to 0 in \(\mathrm{I}_2\). \(\mathrm{H}_2 \mathrm{O}_2\) is causing someone else to lose electrons, making it the oxidant in this reaction. Hence, \(\mathrm{H}_2 \mathrm{O}_2\) acts as an oxidant here.

Step 5: Analyze Reaction (d)

For reaction (d), \(\mathrm{Na}_2 \mathrm{SO}_3\) is oxidized to \(\mathrm{Na}_2 \mathrm{SO}_4\). The sulfur in \(\mathrm{SO}_3^{2-}\) changes its oxidation state from +4 to +6 in \(\mathrm{SO}_4^{2-}\), indicating oxidation. Therefore, \(\mathrm{H}_2 \mathrm{O}_2\) acts as an oxidant as it accepts electrons from \(\mathrm{Na}_2 \mathrm{SO}_3\).

Key Concepts

Oxidizing Agent

An oxidizing agent is a key player in redox reactions. It facilitates the oxidation of another substance by accepting electrons. Essentially, an oxidizing agent acts as an electron acceptor.
The substance being oxidized loses electrons, and as it does so, the oxidizing agent gains those electrons. This process is crucial in various chemical reactions, leading to changes in the oxidation state of substances involved.
For example, in a reaction where a metal oxidizes, the oxidizing agent helps facilitate this by taking electrons away.

• The oxidizing agent itself gets reduced.
• An example of a common oxidizing agent is hydrogen peroxide.
• Understanding the role of oxidizing agents is fundamental to predicting the outcomes of chemical reactions.

Recognizing oxidizing agents helps you understand which components are losing or gaining electrons, which is at the heart of redox chemistry.

Oxidation State

The oxidation state, also known as oxidation number, is a theoretical tool that helps chemists keep track of electron transfer in redox reactions. It is used to describe the degree of oxidation or reduction an atom undergoes in a chemical reaction.
Oxidation states are usually represented by integers, which can be positive, negative, or zero, depending on the electron distribution around atoms in molecules.
Here's how to determine oxidation states:

• For a simple ion, the oxidation state is the same as its charge.
• Oxygen usually has an oxidation state of -2, except in peroxides where it is -1.
• Hydrogen generally has an oxidation state of +1.

Knowing the oxidation states allows you to identify what changes occur during a reaction, indicating which atoms are being oxidized or reduced. This is crucial for balancing chemical equations and predicting reaction products.

Hydrogen Peroxide as Oxidant

Hydrogen peroxide ( H ₂ O ₂) is a versatile oxidizing agent commonly used in various chemical processes. It has the ability to accept electrons from other substances, enabling it to act as an oxidant in redox reactions.
In its role as an oxidizing agent, hydrogen peroxide can break down easily, forming water and releasing oxygen.

• It's particularly effective because it can oxidize substances in both acidic and alkaline conditions.
• It is used in disinfectants, bleaching agents, and even in environmental applications to treat contaminated water.

For example, in reactions involving potassium iodide ( KI ), hydrogen peroxide acts as an oxidant, causing iodide ions to transform into iodine molecules ( I ₂) by gaining electrons.
Recognizing when H ₂ O ₂ acts as an oxidizing agent can help in understanding its transformative effects in different chemical processes.