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

(b) \(6 \times 96500\) coulomb is the required charge.

Step 1: Identify the Reduction Reaction

The reduction reaction is the transformation of dichromate ions (\(\text{Cr}_2\text{O}_7^{2-}\)) to chromium ions (\(\text{Cr}^{3+}\)). The balanced half-reaction for this process is:\[\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}\] This indicates that 6 electrons (\text{e}^-) are needed to reduce one mole of dichromate ions.

Step 2: Apply Faraday's Laws of Electrolysis

According to Faraday’s laws, the charge (Q) needed for a reaction can be calculated using:\[Q = n \cdot F\]where \(n\) is the number of moles of electrons and \(F\) is Faraday's constant (96500 coulombs per mole of electrons). Here, \(n = 6\) (from the balanced equation), so \[Q = 6 \times 96500\text{ C}\].

Step 3: Determine Charge in Faradays

To convert the charge from coulombs to Faradays:1 Faraday (F) = 96500 C. Therefore, the calculated charge of \(6 \times 96500\text{ C}\) equates to \(6 \text{ F}\).

Step 4: Choose the Correct Option

Refer back to the options. The charge required is equivalent to \(6 \text{ F}\) which corresponds to option (b) \(6 \times 96500\) coulomb.

Key Concepts

Faraday's Laws of Electrolysis

Faraday’s Laws of Electrolysis are foundational principles that help us understand how charge is used during electrochemical reactions. These laws state that:

• The amount of chemical change produced in an electrochemical reaction is proportional to the amount of electricity passed through the system.
• The amount of substance transformed at the electrode is directly proportional to the number of moles of electrons involved in the half-reaction.

In simpler terms, if you pass more electricity through a solution, more chemical reactions will happen. To put it in a formula, we use:\[ Q = n \cdot F \]where \( Q \) is the total electric charge, \( n \) is the number of moles of electrons exchanged in the reaction, and \( F \) is Faraday’s constant, which is approximately 96500 coulombs per mole of electrons. This equation helps us calculate the charge necessary for converting a specific amount of reactants, such as the dichromate ions converted to chromium ions.

Reduction Reactions

Reduction reactions are a key part of redox (reduction-oxidation) chemistry. In a reduction reaction, the oxidation state of a molecule decreases as it gains electrons. Think of electrons as tiny gifts that atoms or ions receive, making them become more stable or less reactive. For example, when dichromate ions \( \text{Cr}_2\text{O}_7^{2-} \) undergo a reduction reaction to produce chromium ions \( \text{Cr}^{3+} \), these ions are gaining electrons. The equation for this particular half-reaction is:\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]In this example, each unit of dichromate ion gains six electrons for the reaction to proceed, which highlights the reduction process. Understanding this helps us predict and balance redox reactions effectively.

Dichromate Ion Transformation

The transformation of dichromate ions is a classic example of a reduction process. Dichromate ions \( \text{Cr}_2\text{O}_7^{2-} \) contain chromium in a high oxidation state, which means they are quite reactive and can accept electrons to become more stable. When reduced, dichromate ions convert into chromium ions \( \text{Cr}^{3+} \). This change is essential in various industrial and laboratory processes due to the forming of chromium ions, which are less reactive than dichromate ions. The dichromate to chromium transformation is a balanced chemical reaction that requires the acceptance of six electrons, as we observe in the equation:\[ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \]Understanding how dichromate ions transform is crucial when solving electrochemical problems or carrying out laboratory experiments where controlling electron flow is necessary.

Half-reaction Balancing

Balancing half-reactions is an important step in dealing with redox reactions. It ensures that the number of atoms and the charge are equal on both sides of the equation. This balance is achieved by adjusting the number of electrons, so the process follows the conservation of mass and charge.When balancing a half-reaction, follow these steps:

• Identify what is being oxidized and what is being reduced.
• Write down the individual half-reactions for the oxidation and reduction processes.
• Balance all atoms except hydrogen and oxygen.
• Add water \( \text{H}_2\text{O} \) to balance oxygen atoms and \( \text{H}^+ \) ions to balance hydrogen atoms if the reaction occurs in an acidic solution.
• Finally, add electrons to one side to balance the charge. Equalize the total number of electrons in the oxidation and reduction processes.

This method is crucial when working with reactions like the transformation of dichromate ions, ensuring the process is mathematically correct and allowing chemists to quantify the reactions' impacts, like calculating how many electrons are needed or produced.