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Consider the following reactions: (1) \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{HX} \longrightarrow \mathrm{X}_{2}+2 \mathrm{H}_{2} \mathrm{O}\) (2) \(\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{O}_{3} \longrightarrow 2 \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O}\) Here, hydrogen peroxide acts as (a) an oxidizing agent in both (1) and (2) (b) an oxidizing agent in (2) and reducing agent in (1) (c) an oxidizing agent in (1) and a reducing agent in (2) (d) a reducing agent in both (1) and (2)

Short Answer

Expert verified
(b) An oxidizing agent in (2) and reducing agent in (1).

Step by step solution

01

Understand Oxidation and Reduction

Oxidation involves the loss of electrons, and the species that undergoes oxidation is the reducing agent. Reduction involves the gain of electrons, and the species that undergoes reduction is the oxidizing agent.
02

Analyzing Reaction (1)

In reaction (1), \(\mathrm{H}_{2}\mathrm{O}_{2}+2 \mathrm{HX} \longrightarrow \mathrm{X}_{2}+2 \mathrm{H}_{2}\mathrm{O}\), we see \(\mathrm{HX}\) is typically an acid (like \(\mathrm{HCl}\), \(\mathrm{HBr}\)). Here, \(\mathrm{X}_{2}\) is formed, indicating \(\mathrm{X^-}\) ions (from \(\mathrm{HX}\)) lose electrons and \(\mathrm{H}_{2}\mathrm{O}_{2}\) gains electrons. Thus, \(\mathrm{H}_{2}\mathrm{O}_{2}\) acts as an oxidizing agent.
03

Analyzing Reaction (2)

In reaction (2), \(\mathrm{H}_{2}\mathrm{O}_{2}+\mathrm{O}_{3} \longrightarrow 2 \mathrm{O}_{2}+\mathrm{H}_{2}\mathrm{O}\), we see \(\mathrm{O}_{3}\) (ozone) is converted to \(\mathrm{O}_{2}\). This means \(\mathrm{O}_{3}\) is reduced and therefore needs an agent to donate electrons. \(\mathrm{H}_{2}\mathrm{O}_{2}\) must be the reducing agent donating electrons to \(\mathrm{O}_{3}\).
04

Identify Hydrogen Peroxide Roles in Both Reactions

Combining both observations, in reaction (1) \(\mathrm{H}_{2}\mathrm{O}_{2}\) is an oxidizing agent, and in reaction (2) \(\mathrm{H}_{2}\mathrm{O}_{2}\) acts as a reducing agent.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidizing Agent
An oxidizing agent is a chemical species that facilitates the gaining of electrons in another substance. This means it causes the other substance to become oxidized while it itself undergoes reduction. Think of it like this: an oxidizing agent "takes" electrons from other molecules.

In reaction (1) from our exercise, hydrogen peroxide (\(\mathrm{H}_{2}\mathrm{O}_{2}\)) functions as an oxidizing agent. Here's what happens:
  • The ions \(\mathrm{X}^-\) from the acid \(\mathrm{HX}\) lose electrons to form \(\mathrm{X}_{2}\), indicating oxidation of \(\mathrm{X}^-\).
  • Consequently, \(\mathrm{H}_{2}\mathrm{O}_{2}\) gains electrons, becoming reduced in the process.
Thus, the act of electron gain in \(\mathrm{H}_{2}\mathrm{O}_{2}\) classifies it as an oxidizing agent in this reaction. When you need to remember this, just think: oxidizing agents like to gain.
Reducing Agent
Reducing agents are the opposite of oxidizing agents, as they "give" or "donate" electrons to other substances. This means they themselves undergo oxidation. In simple terms, a reducing agent causes another substance to be reduced.

Considering reaction (2) from our exercise, hydrogen peroxide behaves as a reducing agent:
  • Ozone, \(\mathrm{O}_{3}\), is reduced to molecular oxygen, \(\mathrm{O}_{2}\). This indicates that \(\mathrm{O}_{3}\) is gaining electrons.
  • To achieve this, \(\mathrm{H}_{2}\mathrm{O}_{2}\) must donate the electrons needed, thus undergoing oxidation itself.
Whenever you're trying to identify a reducing agent, look for the substance that loses electrons while enabling others to gain them. Keep it simple—reducing agents love to give away electrons.
Hydrogen Peroxide Reactions
Hydrogen peroxide is an interesting molecule because it can act as both an oxidizing and a reducing agent, depending entirely on the reaction it is involved in. This dual functionality makes it versatile and common in various chemical, industrial, and even biological processes.

In reaction (1), \(\mathrm{H}_{2}\mathrm{O}_{2}\) oxidizes other substances, whereas, in reaction (2), it reduces ozone. This "double life" of \(\mathrm{H}_{2}\mathrm{O}_{2}\) is due to its unique composition:
  • Its oxygen atoms are in a higher oxidation state compared to those in water (\(\mathrm{H}_{2}\mathrm{O}\)), allowing it to accept electrons (act as an oxidizing agent).
  • Conversely, it can also lose its extra oxygen atom, donating electrons in the process (acting as a reducing agent).
This flexibility is valuable in any number of settings, from laboratory experiments to large-scale industrial applications. Understanding the exact role of hydrogen peroxide in a reaction requires careful consideration of the other reactants' behavior and the electron transfers occurring.

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Most popular questions from this chapter

Identify the two liquids \((\mathrm{X})\) and \((\mathrm{Y})\) made up of same atoms, both \((\mathrm{X})\) and \((\mathrm{Y})\) are diamagnetic. (X) turns blue litmus to red, but (Y) does not. Also \(\mathrm{X}+\mathrm{KI}+\mathrm{H}^{+} \stackrel{\text { Starch }}{\longrightarrow}\) Blue colour \(\mathrm{Y}+\mathrm{KI}+\mathrm{H}^{+} \stackrel{\text { Starch }}{\longrightarrow}\) No colour (a) \(\mathrm{X}=\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Y}=\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{X}=\mathrm{H}_{2} \mathrm{O}, \mathrm{Y}=\mathrm{H}_{2} \mathrm{O}_{2}\) (c) Both \(\mathrm{X}\) and \(\mathrm{Y}\) are \(\mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{X}=\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Y}=\mathrm{O}_{3}\)

Which of the following reactions shows the correct sequence of the Ostwald process in the manufacture of nitric acid? (a) \(4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} \stackrel{750^{\circ} \mathrm{C}-900^{\circ} \mathrm{C}, \text { catalyst }}{\longrightarrow} 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O}\) \(\mathrm{NO} \stackrel{\text { heat } \mathrm{O}_{2}}{\longrightarrow} \mathrm{NO}_{2} \stackrel{\mathrm{H}_{2} \mathrm{O}}{\mathrm{O}} \mathrm{HNO}_{3}\) (b) \(\mathrm{S}+\mathrm{O}_{2} \longrightarrow \mathrm{SO}_{2} \stackrel{\mathrm{O}_{2}}{\longrightarrow} \mathrm{SO}_{3} \frac{3}{+\mathrm{HNO}_{3}}\) \(\longrightarrow \mathrm{NaNO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{NaHSO}_{4}\) (d) both (a) and (b)

Which one of the following compounds undergo hydrolysis during distillation to yield hydrogen peroxide? (a) \(\mathrm{HNO}_{3}\) (b) \(\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7}\) (c) \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{8}\) (d) \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{6}\)

A variety of water which contains soluble salts of \(\mathrm{Ca}\) and \(\mathrm{Mg}\) is known as (a) soft water (b) heavy water (c) conductivity water (d) hard water

The dissociation constant of water \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) is \(1.0 \times 10-{ }^{14}\) and that of heavy water \(\left(\mathrm{D}_{2} \mathrm{O}\right)\) is \(3.0 \times 10^{-15}\). This is because the extent of dissociation of \(\mathrm{H}_{2} \mathrm{O}\) is about (a) \(1 / 3\) times that of \(\mathrm{D}_{2} \mathrm{O}\) (b) 3 times that of \(\mathrm{D}_{2} \mathrm{O}\) (c) 3000 times that of \(\mathrm{D}_{2} \mathrm{O}\) (d) 30 times that of \(\mathrm{D}_{2} \mathrm{O}\)

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