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) \(2 \mathrm{KI}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \longrightarrow 2 \mathrm{KOH}(\mathrm{aq})+\)

Step-by-Step Solution

Answer

H2O2 acts as a reducing agent in (a).

Step 1: Identify the Role of H2O2

A reducing agent is a substance that loses electrons and is oxidized in a chemical reaction. Therefore, to determine if \( \mathrm{H}_{2} \mathrm{O}_{2} \) acts as a reducing agent, we must investigate if it loses electrons and its oxidation state increases after the reaction.

Step 2: Analyze Reaction (a)

In reaction (a): \( \mathrm{PbO}_{2} + \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{PbO} + \mathrm{H}_{2} \mathrm{O} + \mathrm{O}_{2} \)- Initially, the oxidation state of oxygen in \( \mathrm{H}_{2} \mathrm{O}_{2} \) is \(-1\).- In \( \mathrm{O}_{2} \), the oxidation state of oxygen is \(0\).- There is an increase in the oxidation state from \(-1\) to \(0\), indicating that \( \mathrm{H}_{2} \mathrm{O}_{2} \) acts as a reducing agent.

Step 3: Analyze Reaction (b)

In reaction (b): \( \mathrm{KNO}_{2} + \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{KNO}_{3} + \mathrm{H}_{2} \mathrm{O} \)- In this case, the oxygen from \( \mathrm{H}_{2} \mathrm{O}_{2} \) does not exhibit a change indicating a transfer of electrons, primarily it acts as an oxidizing agent converting \( \mathrm{NO}_{2}^- \) to \( \mathrm{NO}_{3}^- \). Thus, \( \mathrm{H}_{2} \mathrm{O}_{2} \) is not a reducing agent.

Step 4: Analyze Reaction (c)

In reaction (c): \( 2 \mathrm{KI} + \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{KOH} + \mathrm{I}_{2} \)- We are provided with a partial equation, but we can deduce that \( \mathrm{H}_{2} \mathrm{O}_{2} \) may lose electrons as it might convert \( \mathrm{I}^- \) to \( \mathrm{I}_{2} \). Initial assessment shows \( \mathrm{H}_{2} \mathrm{O}_{2} \) could act as a reducing agent here, if complete to balance the electrons.

Step 5: Conclusion

The reaction in which \( \mathrm{H}_{2} \mathrm{O}_{2} \) acts as a reducing agent can be conclusively seen in reaction (a), where it leads to the formation of \( \mathrm{O}_{2} \) from \( \mathrm{H}_{2} \mathrm{O}_{2} \).

Key Concepts

Oxidation and Reduction

Oxidation and reduction are two fundamental concepts in chemistry that often go hand in hand. These processes are collectively known as redox reactions. In simple terms:

• Oxidation refers to the process where a substance loses electrons. This loss increases the oxidation state of the element involved.
• Reduction involves the gain of electrons, which decreases the oxidation state of an element.

This transfer of electrons from one substance to another is what constitutes a redox reaction. To identify these, look for changes in the oxidation states of elements before and after the reaction. One substance will lose electrons (oxidized), while another gains electrons (reduced). This is also why they are often called complementary or reciprocal reactions. Remember: the mnemonic "OIL RIG" can help you recall that Oxidation Is Loss and Reduction Is Gain!

Oxidation States

Understanding oxidation states is crucial when analyzing chemical reactions, especially redox reactions. The oxidation state, sometimes referred to as the oxidation number, indicates the degree of oxidation of an atom in a compound. Here's what you should keep in mind:

• Oxidation states can help predict how electrons are distributed among atoms in a compound.
• They are determined based on a set of rules. For instance, hydrogen is generally in a +1 oxidation state, while oxygen commonly has a -2 state.
• In molecules like \( \mathrm{H}_{2}\mathrm{O}_{2} \), the peroxide ion \((\mathrm{O}_{2}^{2-})\) gives oxygen an unusual oxidation state of -1.

By keeping track of these states, you can determine which atoms are oxidized or reduced. Increase in oxidation state indicates oxidation, while a decrease suggests reduction. This approach helps to easily spot reducing and oxidizing agents in reactions.

Reducing Agents

Reducing agents are essential players in redox chemistry. Despite their name, they cause the reduction of other substances but are themselves oxidized. Here's all about them:

• A reducing agent donates electrons to another species and, by doing this, itself becomes oxidized.
• This means its oxidation state increases as it loses electrons during the reaction.
• In the example of \(\mathrm{H}_{2}\mathrm{O}_{2}\), it acts as a reducing agent in some reactions by increasing its oxidation state from -1 to 0, transforming into \(\mathrm{O}_{2}\).

Understanding the role of reducing agents helps unravel the electron transfer dynamics in a reaction. Recognizing these agents in a chemical equation allows us to categorize reactions effectively and predict the products of the reaction.