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Which of the following is not correct regarding the electrolytic preparation of \(\mathrm{H}_{2} \mathrm{O}_{2}\) ? (a) sulphuric acid undergoes oxidation (b) hydrogen is liberated at anode (c) \(50 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) is used (d) lead is used as cathode

Short Answer

Expert verified
Option (b) is not correct.

Step by step solution

01

Understand the Concept of Electrolytic Preparation

The electrolysis of sulfuric acid ( H_2SO_4) helps prepare hydrogen peroxide ( H_2O_2). This involves oxidizing sulfuric acid at the anode and the reduction reactions happening at the cathode.
02

Analyze Option (a) - Oxidation of Sulfuric Acid

In the electrolytic preparation of H_2O_2, sulfuric acid ( H_2SO_4) is indeed oxidized at the anode to form peroxydisulfuric acid, which eventually leads to the formation of hydrogen peroxide. Thus, option (a) is a correct statement.
03

Analyze Option (b) - Liberation of Hydrogen at Anode

During the electrolysis process, hydrogen is liberated at the cathode as a result of reduction, not at the anode, where oxidation occurs. Therefore, option (b) is not correct regarding the preparation.
04

Analyze Option (c) - Use of 50% Sulfuric Acid

The common practice involves using 50% sulfuric acid for the electrolysis process to create suitable conditions for H_2O_2 formation. Thus, option (c) is a correct statement.
05

Analyze Option (d) - Use of Lead as Cathode

Lead is often used as a material for the cathode in the electrolytic preparation of H_2O_2 due to its desirable properties and resistance to sulfuric acid. Therefore, option (d) is also correct.

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Key Concepts

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

Sulfuric Acid Oxidation
In the process of electrolysis for producing hydrogen peroxide, sulfuric acid plays a crucial role. When it's subjected to electrolysis, oxidation occurs at the anode.
The chemical reaction involved results in the transformation of sulfuric acid into peroxydisulfuric acid. This compound, over time, will further transform into hydrogen peroxide, the desired product.
During this process, the oxidation reaction at the anode can be represented as: \[2 ext{H}_2 ext{SO}_4 ightarrow ext{H}_2 ext{S}_2 ext{O}_8 + 2 ext{H}^+ + 2e^-\] Understanding this step highlights the importance of sulfuric acid and its reaction pathway, crucial for the successful production of hydrogen peroxide.
Hydrogen Liberation
As part of the electrolysis, hydrogen is released but not at the anode. Instead, hydrogen liberation occurs at the cathode during this process.
This distinction is important because while oxidation happens at the anode, reduction occurs at the cathode where hydrogen gas is evolved.
The cathode reduction process can be represented with the equation:\[2 ext{H}^+ + 2e^- ightarrow ext{H}_2\]This hydrogen liberation is a significant aspect of the electrolysis process, ensuring that the byproducts are clearly understood and accounted for during the conversion procedures of sulfuric acid.
Electrolysis Conditions
Creating optimum conditions during the electrolysis process is crucial for efficiency and effectiveness. The concentration of sulfuric acid plays a key role in this setup.
Typically, 50% sulfuric acid is used as the electrolyte solution. This concentration of sulfuric acid provides just the right conditions; it balances conductivity and cost, while ensuring efficient conversion into hydrogen peroxide.
These conditions facilitate the generation of the peroxide, maximizing yield and ensuring the stability of the process. Adjusting parameters such as current density and temperature also aids in refining the electrolysis for better results.
Cathode Material Selection
The selection of cathode material is another critical factor that influences the electrolysis process. Lead is often chosen as the cathode material due to its compatibility and resistance to sulfuric acid.
Lead's inertness and ability to withstand corrosive environments makes it ideal for this setup. Its use helps to ensure that the reactions proceed without unwanted side reactions or degradation of materials involved.
This careful material selection minimizes maintenance and extends the lifespan of the electrolytic cell, thereby enhancing the reliability and efficiency of hydrogen peroxide production.

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Most popular questions from this chapter

Ammonia can be dried by (a) conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (b) \(\mathrm{P}_{4} \mathrm{O}_{10}\) (c) \(\mathrm{CaO}\) (d) anhydrous \(\mathrm{CaCl}_{2}\)

\(2 \mathrm{~g}\) of aluminium is treated separately with excess of dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and excess of \(\mathrm{NaOH}\). The ratio of the volumes of hydrogen evolved is (a) \(1: 1\) (b) \(1: 2\) (c) \(2: 1\) (d) \(2: 3\)

Nitrogen is a relatively inactive element because (a) dissociation energy of its molecule is fairly high (b) its electronegativity is fairly high (c) it has low atomic radius (d) its atom has a stable electronic configuration

What are the products obtained when ammonia is reacted with excess chlorine? (a) \(\mathrm{N}_{2}\) and \(\mathrm{NH}_{4} \mathrm{Cl}\) (b) \(\mathrm{N}_{2}\) and \(\mathrm{NCl}_{3}\) (c) \(\mathrm{NCl}_{3}\) and \(\mathrm{HCl}\) (d) \(\mathrm{N}_{2}\) and \(\mathrm{HCl}\)

Find \([\mathrm{A}],[\mathrm{B}]\) and \([\mathrm{C}]\) respectively in the reaction given below: \([\mathrm{A}]+\mathrm{Na}\left[\mathrm{Cr}(\mathrm{OH})_{4}\right] \stackrel{\mathrm{H}_{2} \mathrm{O}_{2}, \text { boil }}{\longrightarrow}[\mathrm{B}] \frac{\mathrm{H}_{2} \mathrm{SO}_{4}}{\text { Yellow }}-[\mathrm{C}]\) compound (a) \(\mathrm{NaOH}, \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}, \mathrm{Na}_{2} \mathrm{CrO}_{4}\) (b) \(\mathrm{NaOH}, \mathrm{Na}_{2} \mathrm{CrO}_{4}, \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (c) \(\mathrm{H}_{2} \mathrm{O}, \mathrm{Na}_{2} \mathrm{CrO}_{4}, \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (d) \(\mathrm{NaOH}, \mathrm{K}_{2} \mathrm{CrO}_{4}, \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\)

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