Problem 22 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Consider the following equations: \(2 \mathrm{Fe}^{2+}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow x \mathrm{~A}+y \mathrm{~B}\) (in basic medium) \(2 \mathrm{MnO}_{4}^{-}+6 \mathrm{H}^{+}+5 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow x^{\prime} \mathrm{C}+y^{\prime} \mathrm{D}+z^{\prime} \mathrm{E}\) (in acidic medium) The sum of the stoichiometric coefficients \(x, y, x^{\prime}, y^{\prime}\) and \(z^{\prime}\) for products \(\mathrm{A}\), \(\mathrm{B}, \mathrm{C}, \mathrm{D}\) and \(\mathrm{E}\), respectively, is

Step-by-Step Solution

Answer

The sum of the stoichiometric coefficients is 14.

Step 1: Identify the Reaction Medium

The given problem involves two reaction equations under different conditions: one in basic medium and the other in acidic medium. We should balance these reactions considering their respective media.

Step 2: Balance the First Reaction (Basic Medium)

For the reaction involving \(2 \mathrm{Fe}^{2+}\) and \(\mathrm{H}_{2} \mathrm{O}_{2}\) under basic conditions, the products are typically \(\mathrm{Fe}^{3+}\) and \(\mathrm{OH}^-\). The balanced reaction is:\[2 \mathrm{Fe}^{2+} + \mathrm{H}_{2} \mathrm{O}_{2} + 2 \mathrm{OH}^- \rightarrow 2\mathrm{Fe}^{3+} + 2\mathrm{OH}^- + 2 \mathrm{e}^- = 2 \mathrm{Fe}^{3+} + 2\mathrm{OH}^-\]Thus: \(x = 2\) (for \(\mathrm{Fe}^{3+}\)), and \(y = 2\) (for \(\mathrm{OH}^-\)).

Step 3: Balance the Second Reaction (Acidic Medium)

For the second reaction, involving \(2 \mathrm{MnO}_{4}^{-}\), \(6 \mathrm{H}^{+}\), and \(5 \mathrm{H}_{2} \mathrm{O}_{2}\) in acidic medium, the products are usually \(\mathrm{Mn}^{2+}\) and water. The balanced equation is:\[2 \mathrm{MnO}_{4}^{-} + 6 \mathrm{H}^{+} + 5 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{Mn}^{2+} + 8 \mathrm{H}_{2} \mathrm{O}\]This gives: \(x' = 2\) (for \(\mathrm{Mn}^{2+}\)), \(y' = 8\) (for \(\mathrm{H}_{2} \mathrm{O}\)), and \(z' = 10\) (for electrons transferred to balance charges).

Step 4: Sum of Stoichiometric Coefficients

Now, calculate the total sum of the stoichiometric coefficients: \\(x + y + x' + y' = 2 + 2 + 2 + 8 = 14\).

Key Concepts

Stoichiometry

Understanding stoichiometry is essential for solving chemical equations, as it involves calculating the correct proportions of reactants and products in a chemical reaction. This field is vital for ensuring reactions occur as intended in both laboratory and industrial processes.
Stoichiometry relies on balanced chemical equations, which provide a clear relationship between the quantities of reactants and products. By using the coefficients from these equations, one can determine the exact amount of each substance required or produced.
Let's consider a simple case: multiplying the coefficients of a balanced equation by the molar masses of the reactants and products allows you to convert the equation into a mass-based one. For example, in the exercise given, balancing both reactions (in basic and acidic media) helps to determine the stoichiometric coefficients that describe the relationships between the reactants and products involved.

• First, write the unbalanced equation.
• Count the number of each type of atom on both sides.
• Adjust coefficients to balance the atoms.
• Finally, confirm all elements and charges are balanced.

Mastering stoichiometry will help you efficiently solve and understand chemical reactions.

Basic Medium Reactions

Basic medium reactions involve solutions where the environment is maintained at a pH greater than 7 by using bases, commonly hydroxide ions (\(OH^-\)). These reactions typically consume or produce hydroxide ions during the process of balancing.
In the given exercise, the reaction between \(2 \mathrm{Fe}^{2+}\) ions and \(\mathrm{H}_2\mathrm{O}_2\) in a basic medium must be balanced, so the products fit the medium condition. This involves ensuring that the hydroxide ions balance the charges. During this reaction, \(\mathrm{Fe}^{2+}\) is oxidized to \(\mathrm{Fe}^{3+}\), and electrons are transferred.
To balance reactions in a basic medium, follow these steps:

• Write the skeleton reaction.
• Balance the atoms other than \(O\) and \(H\).
• Balance oxygen atoms by adding \(\mathrm{H}_2\mathrm{O}\).
• Balance hydrogen by adding \(OH^-\) ions.
• Adjust charges by adding electrons.

By following these steps, you can effectively balance reactions occurring in basic media.

Acidic Medium Reactions

Reactions in acidic media occur in solutions where the pH is less than 7. This is typically achieved by the presence of an acid, such as \(H^+\) ions, which participate in the reaction. These reactions are balanced differently compared to basic medium reactions, as they involve the acid catalyst.
Consider the reaction involving \(2 \mathrm{MnO}_4^-\) with \(6 \mathrm{H}^+\) and \(5 \mathrm{H}_2\mathrm{O}_2\) from the exercise. In acidic conditions, \(\mathrm{MnO}_4^-\) is reduced to \(\mathrm{Mn}^{2+}\), and the presence of \(H^+\) ions is crucial for balancing both the elements and the charges.
Steps to balance reactions in an acidic medium include:

• Write down the unbalanced equation.
• Balance non-oxygen and non-hydrogen atoms first.
• Add \(\mathrm{H}_2\mathrm{O}\) to balance oxygen atoms.
• Add \(H^+\) to balance hydrogen atoms.
• Equalize charges by adding electrons as needed.

Balancing these reactions ensures a correct understanding of the processes in acidic environments, which is essential for accurately predicting reaction outcomes.