Problem 35 - A Complete Resource Book in Chemistry for JEE Main

Question

For reducing \(1 \mathrm{~mol}\) of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) to \(\mathrm{Cr}^{3+}\), the charge required is (a) \(3 \times 96500\) coulomb (b) \(6 \times 96500\) coulomb (c) \(0.3 \mathrm{~F}\) (d) \(0.6 \mathrm{~F}\)

Step-by-Step Solution

Answer

The charge required is \(6 \times 96500\) coulombs.

Step 1: Identify the reaction equation

The reduction half-reaction of dichromate ion to chromium (III) ion can be written as \[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]. This reaction tells us that 6 electrons are needed to reduce 1 mole of \( \text{Cr}_2\text{O}_7^{2-} \) to \( \text{Cr}^{3+} \).

Step 2: Determine charge per electron

The charge of one mole of electrons is equivalent to one Faraday, or 96500 coulombs. Thus, 1 electron corresponds to 96500 coulombs of charge.

Step 3: Calculate total charge required

Since 6 moles of electrons are needed, the total charge required is calculated by multiplying the Faraday constant by 6: \[ 6 \times 96500 \text{ coulombs} = 579000 \text{ coulombs} \].

Key Concepts

Redox Reactions

Redox reactions are fundamental chemical reactions involving the transfer of electrons between two substances. The term "redox" is an abbreviation for "reduction-oxidation." In these types of reactions, one substance undergoes oxidation and another undergoes reduction.

- **Oxidation** refers to the loss of electrons. When a substance is oxidized, it gives up electrons to another substance. - **Reduction** involves the gain of electrons. The substance that gains electrons becomes reduced.

Redox reactions are pivotal in many systems, including biological processes like respiration and photosynthesis, as well as in industrial applications. To balance these reactions, you need to ensure that the number of electrons lost in oxidation equals the number of electrons gained in reduction. This balance is crucial for determining the stoichiometry of the reaction and the charge transfer involved. In the case of the reduction of dichromate ions, each mole requires 6 moles of electrons, as outlined in the step-by-step solution. Understanding redox reactions is essential for analyzing any electrochemical process, including electrolysis and battery functioning.

Faraday's Laws of Electrolysis

Faraday's laws of electrolysis are fundamental in understanding the relationship between the amount of electric charge passed through an electrolyte and the amount of substance that is either dissolved or deposited at an electrode. These laws are named after Michael Faraday, who established the quantitative basis of electrochemistry.

- **First Law**: This law states that the amount of substance liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. The more electricity you pass, the greater the amount of material produced or consumed. - **Second Law**: According to this law, the mass of substances produced or consumed at an electrode when a certain amount of electricity is passed through is directly proportional to their equivalent weights.

Electrolysis plays a crucial role in industries where elements are extracted from compounds and in the manufacturing of chemical products. These laws help in calculating the amount of charge necessary to bring about a desired chemical change.

In the context of dichromate ion reduction, knowing that 6 faradays are required helps predict how much electric charge is needed to fully reduce the ion to chromium ions.

Dichromate Ion Reduction

The reduction of dichromate ions (\( ext{Cr}_2 ext{O}_7^{2-}\) involves a powerful redox reaction where the ions are reduced to chromium ions (\( ext{Cr}^{3+}\). This reaction occurs in acidic media, and it serves as an important example in teaching redox chemistry.

The half-reaction provided shows that each dichromate ion requires 6 moles of electrons to convert to two moles of chromium ions. The reaction is expressed as:

\[\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\]
Understanding this concept is crucial for chemistry students because it illustrates the interplay between charge transfer and chemical transformation. Besides its academic interest, dichromate reduction has practical applications in industries like electroplating and in treating waste-water. Recognizing the role of the equivalent number of electrons assures the right amount of reactants are used, which is essential for efficiency and cost-effectiveness in various chemical processes.