Problem 75
Question
\(\mathrm{SO}_{2}\) acts as a temporary bleaching agent but \(\mathrm{Cl}_{2}\) acts as a permanent bleaching agent. Why? (a) \(\mathrm{Cl}_{2}\) bleaches due to oxidation but \(\mathrm{SO}_{2}\) bleaches due to reduction (b) \(\mathrm{Cl}_{2}\) bleaches due to reduction but \(\mathrm{SO}_{2}\) bleaches due to oxidation (c) both (a) and (b) (d) none of these
Step-by-Step Solution
Verified Answer
(a) Cl2 bleaches due to oxidation but SO2 bleaches due to reduction.
1Step 1: Understanding Bleaching Agents
Bleaching agents are substances used to whiten or remove the natural color of fibers, yarns, or fabrics. They typically do this through either oxidation or reduction mechanisms.
2Step 2: SO2 and its Mechanism
Sulfur dioxide (SO_2) acts as a reducing agent. It removes color by reducing the colored substances to colorless forms. However, this action can be reversed if the coloring agent is oxidized back by exposure to air, making it a temporary process.
3Step 3: Cl2 and its Mechanism
Chlorine (Cl_2) acts as an oxidizing agent. It removes color by oxidizing colored substances, turning them into colorless compounds. This reaction is not easily reversible, so the bleaching effect is permanent.
4Step 4: Matching the Explanation to the Options
The question asks which option correctly explains why SO2 is a temporary bleaching agent and Cl2 is a permanent bleaching agent. [SO[2 bleaches by reduction, while Cl2 bleaches by oxidation. This matches option (a).
Key Concepts
OxidationReductionChemical Reactions
Oxidation
Oxidation is a fundamental concept in chemistry where a substance undergoes a chemical reaction by losing electrons. When a substance is oxidized, its oxidation state increases, which typically corresponds to gaining oxygen or losing hydrogen. For example, when chlorine (\[ \mathrm{Cl}_{2} \]) is used to bleach fabrics, it acts as an oxidizing agent. By adding oxygen to the colored molecules, chlorine alters their structure, turning them into colorless compounds.
- Oxidizing agents, like chlorine, can receive electrons from other molecules, causing those molecules to become colorless.
- These kinds of reactions are usually irreversible, which is why the bleaching effect of chlorine is considered permanent.
Reduction
Reduction is the opposite process of oxidation. It involves a substance gaining electrons, which results in a decrease in oxidation state. In the context of bleaching, sulfur dioxide (\[ \mathrm{SO}_{2} \]) acts as a reducing agent. It interacts with colored compounds by donating electrons to them, reducing them to colorless forms.
- Reducing agents like \[ \mathrm{SO}_{2} \] can remove colors temporarily.
- However, the original color can return if the colorless compound is exposed to oxidizing conditions, such as air or light.
Chemical Reactions
Chemical reactions are processes where substances, known as reactants, are transformed into different substances called products. These transformations involve the making or breaking of bonds between atoms and are described by chemical equations. When it comes to bleaching, chemical reactions play a central role.
- Bleaching through oxidation involves adding oxygen or removing hydrogen from molecules.
- In contrast, bleaching through reduction involves donating electrons to a molecule to change its structure and color.
Other exercises in this chapter
Problem 72
Fluorine oxidizes \(\mathrm{HSO}_{4}^{-}\)to \(\ldots \ldots\) (a) \(\mathrm{SO}_{2}\) (b) \(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) (c) \(\mathrm{S}_{2} \mathrm{O
View solution Problem 73
\(\mathrm{Cl}_{2} \mathrm{O}\) is best prepared by passing dry (a) hydrogen chloride and oxygen over hot silver oxide (b) chlorine gas over hot \(\mathrm{HgO}\)
View solution Problem 76
Which one of the following represents the reaction between fluorine and cold dilute \(\mathrm{NaOH} ?\) (a) \(2 \mathrm{~F}_{2}+2 \mathrm{NaOH} \longrightarrow
View solution Problem 77
Fluorine is obtained by the interaction of \(\mathrm{K}_{2} \mathrm{MnF}_{6}\) with Lewis acid \(\mathrm{SbF}_{5}\) because of the (a) acidalysis of \(\mathrm{M
View solution