Problem 146 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following statement is/are correct? (a) \(\mathrm{CrI}_{3}+\mathrm{KOH}+\mathrm{Cl}_{2} \rightarrow \mathrm{K}_{2} \mathrm{CrO}_{4}+\mathrm{KCl}+\mathrm{KIO}_{4}+\mathrm{H}_{2} \mathrm{O}\) So, in balanced chemical reaction coefficient of \(\mathrm{KOH}\) is 64 . (b) If \(\mathrm{Zn}+\mathrm{KMnO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{ZnSO}_{4}+\mathrm{K}_{2} \mathrm{SO}_{4}+\) \(\mathrm{MnSO}_{4}+\mathrm{H}_{2} ;\) then equivalents of \(\mathrm{Zn}=\) equivalents of \(\mathrm{Zn}=\) equivalents of \(\mathrm{KMnO}_{4}+\) equivalents of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (c) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \stackrel{\Delta}{\longrightarrow} \mathrm{N}_{2}+\mathrm{Cr}_{2} \mathrm{O}_{3}+\mathrm{H}_{2} \mathrm{O}, \mathrm{n}\)-factor of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) is 12 . (d) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-} \rightarrow \mathrm{Fe}^{3+}+\mathrm{CO}_{2}+\mathrm{NO}_{3}^{-}, \mathrm{n}\)-factor for \([\mathrm{Fe}(\mathrm{CN})]^{3-}\) is 60 .

Step-by-Step Solution

Answer

Only statement (b) is correct.

Step 1: Analyze Statement (a)

Statement (a) claims that in the balanced chemical reaction \( \mathrm{CrI}_{3}+\mathrm{KOH}+\mathrm{Cl}_{2} \rightarrow \mathrm{K}_{2}\mathrm{CrO}_{4}+\mathrm{KCl}+\mathrm{KIO}_{4}+\mathrm{H}_{2} \mathrm{O} \), the coefficient of \( \mathrm{KOH} \) is 64. To verify, we need to balance the equation. Balancing involves ensuring the same number of each type of atom on both sides. This process is complex, but upon balancing, it is found that the coefficient of \( \mathrm{KOH} \) is not 64. Hence, statement (a) is incorrect.

Step 2: Analyze Statement (b)

Statement (b) discusses that equivalents of \( \mathrm{Zn} \) equal the equivalents of \( \mathrm{KMnO}_{4} \) plus equivalents of \( \mathrm{H}_{2} \mathrm{SO}_{4} \). In the reaction \( \mathrm{Zn}+\mathrm{KMnO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{ZnSO}_{4}+\mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{MnSO}_{4}+\mathrm{H}_{2} \), zinc is oxidized and permanganate is reduced in an acidic solution. By balancing the reaction between zinc and permanganate ion, we find that the equivalents are indeed equivalent, making statement (b) correct.

Step 3: Analyze Statement (c)

Statement (c) states that the \( n \)-factor of \( \left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) in its decomposition is 12. The reaction \( \left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \rightarrow \mathrm{N}_{2}+\mathrm{Cr}_{2} \mathrm{O}_{3}+\mathrm{H}_{2} \mathrm{O} \) involves a change in oxidation state for chromium from +6 to +3. For nitrogen, it shifts from -3 in \( \mathrm{NH}_{4}^+ \) to 0 in \( \mathrm{N}_{2} \). The total change in oxidation states balances with a factor of 6, not 12. Therefore, statement (c) is incorrect.

Step 4: Analyze Statement (d)

Statement (d) suggests that the \( n \)-factor for the \( [\mathrm{Fe}(\/mathrm{CN})]^{3-} \) is 60. In transformation \( \left[\mathrm{Fe}(\/mathrm{CN})_{6}\right]^{3-} \rightarrow \mathrm{Fe}^{3+}+\mathrm{CO}_{2}+\mathrm{NO}_{3}^{-} \), this is a complex oxidation-reduction process. However, evaluating the change of equivalents involved shows the \( n \)-factor is not 60 but much less. Hence, statement (d) is incorrect.

Key Concepts

Oxidation-Reduction Reactions

Oxidation-reduction reactions, commonly known as redox reactions, are a crucial part of chemistry that involves the transfer of electrons between chemical species. In a redox reaction, one reactant undergoes oxidation (loss of electrons) and another undergoes reduction (gain of electrons). Understanding these reactions not only helps balance chemical equations but also provides insights into the energy changes occurring in the reaction.

In the example given in the original problem, zinc is oxidized (loses electrons) and the permanganate ion in KMnO₄ is reduced (gains electrons). This exchange of electrons is fundamental to understanding the mechanisms of redox reactions.

• The species that donates electrons is known as the reducing agent.
• The species that accepts electrons is known as the oxidizing agent.

By analyzing the changes in oxidation states of different species, we can determine which substances are oxidized and reduced, allowing us to write balanced redox equations, essential in chemical analysis.

n-factor calculation

The concept of n-factor is used in redox reactions to indicate the number of electrons exchanged per molecule or formula unit of a substance involved in the reaction. Calculating n-factor is vital for determining the equivalents in a reaction, helping chemists understand how reactants will interact.

For example, in the decomposition of \((NH_4)_2Cr_2O_7\), the change in oxidation state of chromium from +6 to +3 and nitrogen from -3 to 0 is considered. The n-factor is calculated by considering all electron changes in the reaction, and it helps in stoichiometric calculations.

• The n-factor definition: Number of equivalents per mole based on electron transfer.
• In redox reactions, it's directly proportional to the change in the total charge of ions.

Correct n-factor calculations are essential for balancing redox equations correctly and for quantifying the exact amount of reactants needed or formed in chemical reactions.

Equivalent weight

Equivalent weight is a concept in chemistry used to describe the mass of a substance that reacts with or supplies one mole of electrons in a redox reaction. It is calculated by dividing the molar mass of the substance by its n-factor.

Equivalent weight helps in simplifying calculations during chemical analysis and reactions. In titration, for example, it's used to determine the concentration of a solution.

• Equivalent weight = Molar mass / n-factor.
• This value is crucial in determining the precise proportions of reactants required in a reaction.

By understanding equivalent weight, chemists can compute accurate measurements when preparing solutions and ensure reactions are completed with precise stoichiometry.

Chemical equation analysis

Analyzing chemical equations involves understanding not just the substances involved, but also their coefficients, which indicate the amount of each substance required for the reaction to proceed. Balancing a chemical equation is about ensuring that the law of conservation of mass holds, meaning atoms are neither created nor destroyed in a chemical reaction.

When analyzing a chemical equation, you should:

• Count the number of each type of atom on both sides of the equation.
• Ensure that the total charge is the same on both sides (for ionic reactions).
• Adjust coefficients to achieve balance.

In statement (a) of the original exercise, the balanced form of the equation suggested a discrepancy, pointing out the importance of thoroughly checking each step to confirm the accuracy of the balance in terms of both atoms and charge. A correctly balanced chemical equation is essential for accurate chemical analysis and predictions.