Problem 34 - A Complete Resource Book in Chemistry for JEE Main

Question

In which of the following reactions does hydrogen peroxide acts as a reducing agent? (a) \(2 \mathrm{KI}+\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{I}_{2}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{Ag}_{2} \mathrm{O}\) (moist) \(+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{Ag}+\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\) (c) \(\mathrm{Na}_{2} \mathrm{SO}_{3}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{PbS}+4 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{PbSO}_{4}+4 \mathrm{H}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

Reaction (b) involves hydrogen peroxide acting as a reducing agent.

Step 1: Identify the Oxidation States

First, determine the initial and final oxidation states of oxygen in the hydrogen peroxide (\(\text{H}_2\text{O}_2\)) for each reaction. In \(\text{H}_2\text{O}_2\), oxygen starts with an oxidation state of -1.

Step 2: Analyze Reaction (a)

In reaction (a), \(2 \text{KI} + \text{H}_2\text{O}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + \text{I}_2 + \text{H}_2\text{O}\), oxygen in \(\text{H}_2\text{O}_2\) ends up forming \(\text{H}_2\text{O}\). The oxidation state of oxygen in \(\text{H}_2\text{O}\) is -2, so it is reduced.

Step 3: Analyze Reaction (b)

In reaction (b), \(\text{Ag}_2\text{O} + \text{H}_2\text{O}_2 \rightarrow 2 \text{Ag} + \text{H}_2\text{O} + \text{O}_2\), oxygen in \(\text{H}_2\text{O}_2\) ends up as both \(\text{H}_2\text{O}\) and \(\text{O}_2\). In \(\text{O}_2\), the oxidation state is 0, indicating that \(\text{H}_2\text{O}_2\) has undergone both oxidation and reduction, known as a disproportionation reaction.

Step 4: Analyze Reaction (c)

In reaction (c), \(\text{Na}_2\text{SO}_3 + \text{H}_2\text{O}_2 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O}\), oxygen in \(\text{H}_2\text{O}_2\) is converted to \(\text{H}_2\text{O}\). The oxidation state of oxygen in \(\text{H}_2\text{O}\) is -2, so \(\text{H}_2\text{O}_2\) acts as an oxidizing agent.

Step 5: Analyze Reaction (d)

In reaction (d), \(\text{PbS} + 4 \text{H}_2\text{O}_2 \rightarrow \text{PbSO}_4 + 4 \text{H}_2\text{O}\), oxygen in \(\text{H}_2\text{O}_2\) becomes part of \(\text{PbSO}_4\) and \(\text{H}_2\text{O}\). The oxidation state has been reduced from -1 to -2, indicating oxidation.

Step 6: Identify Hydrogen Peroxide As Reducing Agent

For \(\text{H}_2\text{O}_2\) to act as a reducing agent, it must be oxidized, which means its oxidation state increases. In reaction (b), \(\text{H}_2\text{O}_2\) is oxidized to \(\text{O}_2\), so it acts as a reducing agent.

Key Concepts

Oxidation States

Understanding oxidation states is crucial for identifying the role of various substances in chemical reactions. An oxidation state indicates the degree of oxidation of an atom in a compound. It is like a bookkeeping system, where electrons get assigned to atoms. For hydrogen peroxide (02), oxygen has an oxidation state of -1.
This is an unusual state because in most compounds, oxygen typically has an oxidation state of -2. In order to identify changes in oxidation states, compare the oxidation states before and after the reaction.
If the oxidation state increases, the substance is oxidized; if it decreases, the substance is reduced. Recognizing these changes helps in determining the agent responsible for the oxidation or reduction in the reaction.

Disproportionation Reaction

Not all redox reactions are straightforward; some involve a substance acting as both an oxidizing and reducing agent in the same reaction. This type of reaction is called a disproportionation reaction. In these unique cases, the same compound is simultaneously oxidized and reduced.
Let's consider reaction (b) from the exercise involving hydrogen peroxide (02) and silver oxide (2). Here, oxygen from 02 ends up in two different products: water () and oxygen gas (2). The oxidation state of oxygen in  is -2, while in 2, it's 0.
The mixed outcomes denote that 02 underwent both an increase and decrease in oxidation states, making it a classic example of disproportionation. This concept can be complex, but remembering that both oxidation and reduction occur within the same reactant is key to identifying such reactions.

Reducing Agent Identification

Identifying the reducing agent in a chemical reaction is another important aspect of understanding redox processes. A reducing agent donates electrons, leading to the reduction of another species, and itself becomes oxidized.
In the context of the step-by-step solution, the aim was to determine in which reaction hydrogen peroxide (02) acts as a reducing agent. Oxygen in 02 typically begins with an oxidation state of -1.
By analyzing reaction (b) where 02 reacts with moist silver oxide (2), oxygen in 02 ends up as oxygen gas (2) with an oxidation state of 0. This increase in the oxidation state means 02 was oxidized. Therefore, it served as a reducing agent because it provided electrons during the formation of 2.
Remember, a reducing agent is identified by its loss of electrons and increase in oxidation state.