Target Exam

CUET

Subject

Chemistry

Chapter

Inorganic: P Block Elements

Question:

Which of the following reactions is given by ozone \((O_3)\)?

Options:

\(PbS(s) + 4O_3(g) \longrightarrow PbSO_4 (s) + 4O_2 (g)\)

\(I_2 (s) + H_2O (l) + O_3 (g) \longrightarrow 2I^- (aq) + 2OH^- (aq) + O_2 (g)\)

\(Fe^{3+} + O_3 + H_2O \longrightarrow Fe^{2+} + 2OH^- + O_2\)

\(Sn^{4+} + O_3 + H_2O \longrightarrow Sn^{2+} + 2OH^- + O_2\)

Correct Answer:

\(PbS(s) + 4O_3(g) \longrightarrow PbSO_4 (s) + 4O_2 (g)\)

Explanation:

The correct answer is option 1. \(PbS(s) + 4O_3(g) \longrightarrow PbSO_4 (s) + 4O_2 (g)\).

The reaction:

\(\text{PbS} (s) + 4 \text{O}_3 (g) \longrightarrow \text{PbSO}_4 (s) + 4 \text{O}_2 (g) \)

is an example of an oxidation reaction where ozone \((O_3)\) acts as a powerful oxidizing agent. Here's a detailed explanation of the reaction:

Oxidation Process

Ozone (\(O_3\)): Ozone is a strong oxidizing agent due to its ability to easily release one of its oxygen atoms, converting into dioxygen \((O_2)\). The released oxygen atom is highly reactive and can oxidize other substances.

Lead Sulfide (\(PbS\)): Lead sulfide is an inorganic compound where lead is in the +2 oxidation state, and sulfur is in the -2 oxidation state.

When ozone comes into contact with lead sulfide, it oxidizes the sulfide ion \((S^{2-})\) into a sulfate ion \((SO_4^{2-})\). This process can be understood in terms of the following half-reactions:

Oxidation of \(S^{2-}\) to \(SO_4^{2-}\):

\(\text{S}^{2-} + 4 \text{O}_3 \longrightarrow \text{SO}_4^{2-} + 4 \text{O}_2\)

Reduction of ozone: During the reaction, ozone is reduced by losing one oxygen atom, transforming into molecular oxygen \((O_2)\).

The overall balanced chemical equation involves 4 molecules of ozone reacting with 1 molecule of lead sulfide to produce lead sulfate and oxygen gas.

\(\text{PbS} (s) + 4 \text{O}_3 (g) \longrightarrow \text{PbSO}_4 (s) + 4 \text{O}_2 (g)\)

Lead Sulfate (\(PbSO_4\)): The product, lead sulfate, is a white solid that is commonly formed as a result of the oxidation of lead sulfide. Lead sulfate is an insoluble salt, which precipitates out of the reaction mixture.

Oxygen Gas (\(O_2\)): The other product is oxygen gas, which is released during the reduction of ozone.

This reaction illustrates the strong oxidizing power of ozone, which can break down even relatively stable compounds like lead sulfide into their oxidized forms. It also serves as an example of how ozone can be used in processes where oxidation is required, such as in the removal of sulfur-containing compounds.

Overall, this reaction showcases the ability of ozone to convert lead sulfide into lead sulfate while producing oxygen, demonstrating ozone's strong oxidative capabilities.