Problem 37
Question
Copper tuming when heated with conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) gives (a) \(\mathrm{O}_{3}\) (b) \(\mathrm{SO}_{2}\) (c) \(\mathrm{SO}_{3}\) (d) \(\bar{H}_{2} \mathrm{~S}\)
Step-by-Step Solution
Verified Answer
(b) \(\mathrm{SO}_{2}\) is formed when copper reacts with conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\).
1Step 1: Understand the Reactants
The problem involves copper turnings reacting with concentrated sulfuric acid,
(H_2SO_4). We need to know the products of this reaction.
2Step 2: Apply Reaction Knowledge
When copper (Cu) reacts with concentrated sulfuric acid, a redox reaction occurs. Copper acts as a reducing agent and sulfuric acid as an oxidizing agent.
The sulfur from the sulfuric acid is reduced.
3Step 3: Determine the Products
During the reaction, sulfuric acid (H_2SO_4) is reduced to form sulfur dioxide (SO_2), while copper is oxidized to copper sulfate (CuSO_4).The overall reaction is: \[Cu + 2H_2SO_4
ightarrow CuSO_4 + 2H_2O + SO_2\]
4Step 4: Verify the Correct Option
Based on the chemical reaction, the main gaseous product formed is sulfur dioxide
(SO_2).
Thus, option (b)
(SO_2)
is the correct answer.
Key Concepts
Redox ReactionSulfur Dioxide ProductionCopper Sulfate Formation
Redox Reaction
When discussing the reaction between copper and concentrated sulfuric acid, it's essential to understand the concept of redox reactions. A redox reaction is a chemical process where there is a transfer of electrons between two substances. One substance is oxidized, meaning it loses electrons.
Conversely, the other is reduced, which means it gains those electrons. In our specific reaction, copper (Cu) is oxidized. It loses electrons and turns into copper ions, creating copper sulfate in the process.
Concurrently, sulfuric acid acts as the oxidizing agent. It accepts electrons, resulting in the reduction of sulfur within the acid. This dynamic between copper and sulfuric acid highlights the fundamental principles of redox reactions, where oxidation and reduction occur simultaneously.
Conversely, the other is reduced, which means it gains those electrons. In our specific reaction, copper (Cu) is oxidized. It loses electrons and turns into copper ions, creating copper sulfate in the process.
Concurrently, sulfuric acid acts as the oxidizing agent. It accepts electrons, resulting in the reduction of sulfur within the acid. This dynamic between copper and sulfuric acid highlights the fundamental principles of redox reactions, where oxidation and reduction occur simultaneously.
Sulfur Dioxide Production
An interesting aspect of the copper and sulfuric acid reaction is the production of sulfur dioxide (SO_2). When sulfur within sulfuric acid (H_2SO_4) is reduced during the redox reaction, it leads to the release of sulfur dioxide gas as a product.
This step in the reaction is critical because the transformation of sulfuric acid into sulfur dioxide involves a reduction process, marking it as the major gaseous product formed.
The overall reaction can be summarized with the chemical equation:
This step in the reaction is critical because the transformation of sulfuric acid into sulfur dioxide involves a reduction process, marking it as the major gaseous product formed.
The overall reaction can be summarized with the chemical equation:
- Copper,
- Sulfuric Acid,
- Copper Sulfate,
- Sulfur Dioxide, and
- Water.
Copper Sulfate Formation
The formation of copper sulfate (CuSO_4) in the reaction between copper turnings and concentrated sulfuric acid is another pivotal aspect. Copper sulfate is generated when copper metal oxidizes; losing electrons this converts metallic copper into copper ions.
As sulfuric acid acts as the oxidizer, it supports the process of sulfate ion attachment to the copper ions, forming copper sulfate. This blue crystalline compound is noteworthy not only for its color but also for its role in various practical applications, such as in agriculture and industry.
The attachment of sulfate ions to the oxidized copper results in the formation of
As sulfuric acid acts as the oxidizer, it supports the process of sulfate ion attachment to the copper ions, forming copper sulfate. This blue crystalline compound is noteworthy not only for its color but also for its role in various practical applications, such as in agriculture and industry.
The attachment of sulfate ions to the oxidized copper results in the formation of
- Copper Sulfate,
- Water,
- And sulfur dioxide as part of the entire reaction scheme,
Other exercises in this chapter
Problem 35
There is \(S\)-S bond in (a) \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}\) (b) \(\mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{8}\) (c) \(\mathrm{H}_{2} \mathr
View solution Problem 36
What is the oxidation number of sulphur in \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\) ? (a) \(2 / 3\) (b) \(3 / 2\) (c) \(3 / 5\) (d) \(5 / 2\)
View solution Problem 38
P O P bond is present in (a) \(\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{6}\) (b) \(\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{5}\) (c) Both (a) and (b) (d) none
View solution Problem 34
In \(O F_{2}\) molecule, the total number of bond pairs and lone pairs of electrons present respectively are (a) 2,6 (b) 2,8 (c) 2,10 (d) 2,9
View solution