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Hydrogen peroxide: (a) Give silver peroxide with moist silver oxide (b) Turn the mixture of aniline, \(\mathrm{KClO}_{3}\) and dil. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) violet (c) Liberate iodine from KI (d) Turn the titanium salt yellow

Short Answer

Expert verified
Hydrogen peroxide oxidizes substances: produces Ag2O2, turns aniline mix violet, liberates iodine, and forms yellow Ti complex.

Step by step solution

01

Understanding the Reaction for Silver Peroxide

Hydrogen peroxide reacts with moist silver oxide to produce silver peroxide. Silver oxide, when moist, can facilitate the oxidation, as hydrogen peroxide can act as an oxidizing agent. This can lead to the conversion of Ag2O to Ag2O2 (silver peroxide).
02

Formation of Violet Compound with Aniline

Hydrogen peroxide can oxidize aniline in the presence of potassium chlorate (KClO3) and dilute sulfuric acid (H2SO4) to form a complex that typically turns violet. The presence of these reagents creates a highly oxidative environment, encouraging the formation of colored oxidation products.
03

Liberation of Iodine from Potassium Iodide

When hydrogen peroxide is added to potassium iodide (KI), it acts as an oxidizing agent to liberate iodine, evidenced by the solution turning brown. The reaction involved is 2KI + H2O2 + H2SO4  I2 + 2KHSO4 + 2H2O, indicating the release of iodine.
04

Formation of Yellow Titanium Complex

Hydrogen peroxide reacts with titanium salts to form a yellow peroxy-titanium complex. This is commonly used as a colorimetric indicator for the presence of titanium, with the formation of a yellow color indicating a positive reaction.

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Key Concepts

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

Silver Peroxide Formation
When hydrogen peroxide meets moist silver oxide, they undergo a chemical transformation that forms silver peroxide. This process is all about oxidation. Here, hydrogen peroxide acts as an oxidizing agent. Silver oxide, when moist, is more reactive and assists in this reaction. The chemistry behind it involves the addition of an extra oxygen atom to silver oxide, creating silver peroxide (\[\text{Ag}_2\text{O} + \text{H}_2\text{O}_2 \rightarrow \text{Ag}_2\text{O}_2 + \text{H}_2\text{O}\]).
This compound, silver peroxide (Ag₂O₂), is less common but notable because of its strong oxidizing properties and potential use. Understanding this reaction is important, as it highlights how hydrogens in peroxide are key players in various reactions.
This is utilized not only in laboratory settings but can also give insights into industrial applications where silver compounds are relevant.
Aniline Oxidation
In this reaction, hydrogen peroxide is used as an oxidizing agent in a blend that involves aniline, potassium chlorate (KClO₃), and dilute sulfuric acid (H₂SO₄). When combined, these chemicals create a reaction that turns the mixture violet. The key here is hydrogen peroxide's strong oxidizing ability, transforming aniline into a complex series of oxidation products that appear as a violet color.
  • Potassium chlorate acts as a secondary source of oxygen, enhancing oxidative conditions.
  • Dilute sulfuric acid aids by providing a suitable acidic environment for the reaction to proceed efficiently.
The oxidation of aniline in this way is significant because it can lead to the formation of complex structures with potential applications in dyes and pigments. It also serves as an excellent example of organic oxidation reactions where hydrogen peroxide is pivotal.
Iodine Liberation from Potassium Iodide
Hydrogen peroxide exhibits its metal-free oxidizing power when it interacts with potassium iodide (KI). In this scenario, hydrogen peroxide oxidizes iodide ions to liberate iodine. When iodine is released, it dissolves in the solution, turning it a distinct brown color (\[2\text{KI} + \text{H}_2\text{O}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{I}_2 + 2\text{KHSO}_4 + 2\text{H}_2\text{O}\]).
This reaction is a classic demonstration in chemistry classes for illustrating redox reactions, where electrons are transferred from the iodide ion to the peroxide. The brown coloration is a visible cue for the presence of iodine.
This reaction's principles are widely applicable in various fields, including environmental analysis and iodine supplementation in biochemistry.
Titanium Yellow Complex Formation
Hydrogen peroxide can be used to test for the presence of titanium via the formation of a yellow complex. When titanium salts are mixed with hydrogen peroxide, they react to form a yellow peroxy-titanium complex. This is more than just a color change. It acts as a visual indicator in colorimetric analysis. The resulting yellow color provides an easy, recognizable sign that titanium is present. This reaction uses the peroxide's ability to form stable complexes with transition metals such as titanium.
  • This method is particularly precise and can be used to measure titanium concentrations accurately.
  • Due to this unique chemistry, it finds use in analytical chemistry, substances testing, and quality control processes.
In the broader context of scientific experiments, the yellow peroxy-titanium complex highlights how versatile hydrogen peroxide can be as an agent for both reactions and compounds identification.

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Most popular questions from this chapter

Which statement is wrong bout structure of phosphorus (V) oxide? (a) Each 'P'atom is sp \(^{3}\) hybridized (b) It has \(6 \mathrm{P}-\mathrm{O}-\mathrm{P}\) bonds. (c) IT has POP angle of \(180^{\circ}\). (d) It has two types of \(\mathrm{P}-\mathrm{O}\) bond lengths.

Aqueous solution of ammonia consists of (a) \(\mathrm{NH}_{4}^{+}\)and \(\mathrm{OH}^{-}\) (b) \(\mathrm{OH}^{-}\) (c) \(\mathrm{NH}_{4}^{+}\) (d) \(\mathrm{H}^{+}\)

A substance 'A' is obtained by boiling an aqueous solution of \(\mathrm{NH}_{3}\) with an aqueous solution of sodium hypochlorite in the presence of a little glue. 'A' forms salts with \(\mathrm{HCl}\) and \(\mathrm{H}_{2} \mathrm{SO}_{4} \cdot^{\prime} \mathrm{A}\) ' is a powerful reducing agent and reduces \(\mathrm{FeCl}_{3}\) solution and acidified \(\mathrm{KMnO}_{4}\) solution. This reaction is being accompanied by evolution of inactive gas ' \(\mathrm{B}\) '. Identify 'A' and ' \(\mathrm{B}\) '. (a) \(\mathrm{A}=\mathrm{O}_{3} ; \mathrm{B}=\mathrm{H}_{2} \mathrm{O}_{2}\) (b) \(\mathrm{A}=\mathrm{NH}_{2} \cdot \mathrm{NH}_{2} ; \mathrm{B}=\mathrm{N}_{2}\) (c) \(\mathrm{A}=\mathrm{N}_{2} ; \mathrm{B}=\mathrm{NO}_{2}\) (d) \(\mathrm{A}=\mathrm{N}_{3} \mathrm{H} ; \mathrm{B}=\mathrm{N}_{2}\)

\(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})\) is heated in a test tube. Vapours are brought in contact with red litmus paper, which changes to blue and then to red. It is because of (a) formation of \(\mathrm{NH}_{3}\) and \(\mathrm{HCl}\) (b) formation of \(\mathrm{NH}_{4} \mathrm{OH}\) and \(\mathrm{HCl}\) (c) greater diffusion of \(\mathrm{HCl}\) than \(\mathrm{NH}_{3}\) (d) greater diffusion of \(\mathrm{NH}_{3}\) than \(\mathrm{HCl}\)

When two ice cubes are pressed over each other, they unite to form one cube. Which of the following forces is responsible to hold them together? (a) van der Waals forces (b) covalent attraction (c) ionic interaction (d) hydrogen bond formation

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