Problem 24
Question
In which reaction does \(\mathrm{SO}_{2}\) act as oxidizing agent (a) Acidified \(\mathrm{KMnO}_{4}\) (b) Acidified \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{1}\) (c) Acidified \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (d) \(\mathrm{HS}\)
Step-by-Step Solution
Verified Answer
\(\mathrm{SO}_{2}\) acts as an oxidizing agent in reaction with acidified \(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}\) (option c) and \(\mathrm{HS}\) (option d).
1Step 1: Understand the Role of an Oxidizing Agent
An oxidizing agent is a substance that gains electrons in a chemical reaction, and in doing so, it causes another substance to lose electrons. Essentially, the oxidizing agent gets reduced.
2Step 2: Consider Acidified \(\mathrm{KMnO}_{4}\) with \(\mathrm{SO}_{2}\)
In the reaction of \(\mathrm{SO}_{2}\) with acidified \(\mathrm{KMnO}_{4}\), sulfur dioxide acts as a reducing agent. It gets oxidized to \(\mathrm{SO}_{4}^{2-}\) while \(\mathrm{Mn}^{7+}\) in \(\mathrm{KMnO}_{4}\) gets reduced to \(\mathrm{Mn}^{2+}\). Thus, \(\mathrm{SO}_{2}\) does not act as an oxidizing agent here.
3Step 3: Consider Acidified \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\) with \(\mathrm{SO}_{2}\)
Similarly, in the reaction of \(\mathrm{SO}_{2}\) with acidified \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\), sulfur dioxide acts as a reducing agent. It gets oxidized to sulfate (\(\mathrm{SO}_{4}^{2-}\)), and chromium (from \(\mathrm{Cr}^{6+}\) in \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\)) gets reduced to \(\mathrm{Cr}^{3+}\). Hence, \(\mathrm{SO}_{2}\) does not act as an oxidizing agent here either.
4Step 4: Consider Acidified \(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}\) with \(\mathrm{SO}_{2}\)
In the presence of acidified \(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}\), \(\mathrm{SO}_{2}\) can oxidize ethanol to aldehydes, such as acetaldehyde. Here, \(\mathrm{SO}_{2}\) acts as an oxidizing agent as it aids in the oxidation of ethanol by gaining electrons.
5Step 5: Consider Reaction with \(\mathrm{HS}\)
When \(\mathrm{SO}_{2}\) reacts with hydrogen sulfide (\(\mathrm{HS}\)), \(\mathrm{SO}_{2}\) oxidizes \(\mathrm{HS}\) to sulfur or sulfate ions. In this case, \(\mathrm{SO}_{2}\) acts as an oxidizing agent as it helps in oxidizing \(\mathrm{HS}\) to higher oxidation states.
Key Concepts
Chemical ReactionsSulfur DioxideOxidation and ReductionElectrochemistry
Chemical Reactions
Chemical reactions are processes in which substances, known as reactants, transform into different substances, called products. This transformation involves making or breaking chemical bonds, and it leads to a change in properties. Here's a simpler breakdown:
In oxidation and reduction reactions, electron transfer plays a critical role. This type of reaction is crucial in processes like combustion, respiration, and electrolysis.
- The molecules of reactants break apart.
- New bonds form to create different molecules, called products.
- These products often have different properties compared to the reactants.
In oxidation and reduction reactions, electron transfer plays a critical role. This type of reaction is crucial in processes like combustion, respiration, and electrolysis.
Sulfur Dioxide
Sulfur dioxide (SO₂) is a colorless gas with a pungent smell synonymous with burnt matches. It's produced naturally from volcanic eruptions and industrially, mainly from burning fossil fuels.
Sulfur dioxide serves multiple roles in chemical reactions, often acting as both a reducing and oxidizing agent depending on the other substances involved. This dual behavior is due to its ability to gain or lose electrons. For instance:
Sulfur dioxide serves multiple roles in chemical reactions, often acting as both a reducing and oxidizing agent depending on the other substances involved. This dual behavior is due to its ability to gain or lose electrons. For instance:
- When reacting with stronger oxidizing agents like acidified KMnO₄, SO₂ acts as a reducing agent.
- When paired with certain reducing agents or compounds like hydrogen sulfide, SO₂ acts as an oxidizing agent.
Oxidation and Reduction
Oxidation and reduction, often abbreviated as redox reactions, are crucial in chemistry. These reactions are all about the transfer of electrons.
Here's how it works:
Here's how it works:
- Oxidation: This is the process where a substance loses electrons. The substance that gets oxidized increases its oxidation state.
- Reduction: This is the process of gaining electrons. When a substance is reduced, it decreases its oxidation state.
- The oxidizing agent is the substance that gets reduced, while the reducing agent is the one that gets oxidized.
Electrochemistry
Electrochemistry is the branch of chemistry that examines the relationship between electricity and chemical reactions. It plays a fundamental role in processes that involve electron transfers.
Key concepts in electrochemistry include:
Key concepts in electrochemistry include:
- Electrolytic Cells: These use electrical energy to cause chemical changes.
- Galvanic Cells: These convert chemical energy into electrical energy.
- Redox Reactions: The core of electrochemical processes, involving the flow of electrons.
- Batteries: converting chemical energy to electrical energy for use in electronic devices.
- Electrolysis: often used in metal plating and water splitting. Sulfur dioxide's behavior as an oxidizing agent in certain scenarios illustrates electrochemical principles at work, showcasing how redox reactions facilitate the movement of electrons.
Other exercises in this chapter
Problem 22
The number of \(\mathrm{P}-\mathrm{O}\) - \(\mathrm{P}\) bonds in cyclic metaphosphoric acid is (a) Zero (b) Two (c) Three (d) Four
View solution Problem 23
Which acid on heating produces phosphine? (a) phosphoric acid (b) phosphorous acid (c) peroxymonophosphorie acit' (d) metaphosphoric acid
View solution Problem 28
Which one of the following is not a true peroxide? (d) \(\overline{\mathrm{H}_{2} \mathrm{SO}_{4}}\) (a) \(\overline{\mathrm{PbO}_{2}}\) (b) \(\mathrm{BaO}_{2}\
View solution Problem 30
The number of hydroxyl groups in pyrophosphoric acid is (a) \(\underline{3}\) (b) \(\overline{4}\) (c) 5
View solution