Problem 7 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

In the following balanced reaction, \(\mathrm{X} \mathrm{MnO}_{4}^{-}+\mathrm{Y} \mathrm{C}_{2} \mathrm{O}_{4}^{2-}+\mathrm{ZH}^{+}\) $$ \rightleftharpoons \mathrm{XMn}^{2+}+2 \mathrm{Y} \mathrm{CO}_{2}+\frac{\mathrm{Z}}{2} \mathrm{H}_{2} \mathrm{O} $$ values of \(X, Y\) and \(Z\) respectively are [Main Online May 12, 2012; 2013] (a) \(2,5,16\) (b) \(8,2,5\) (c) \(5,2,16\) (d) \(5,8,4\)

Step-by-Step Solution

Answer

The correct values are \(X = 5\), \(Y = 2\), \(Z = 16\).

Step 1: Identify the Compounds and Elements

In the given chemical reaction, identify all the compounds and elements involved. The compounds are \(\mathrm{MnO}_{4}^{-}\), \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\), and \(\mathrm{H}^{+}\) on the reactant side and \(\mathrm{Mn}^{2+}\), \(\mathrm{CO}_{2}\), and \(\mathrm{H}_{2}\mathrm{O}\) on the product side.

Step 2: Write the Reaction and Balance Atoms

Write the reaction as follows: \(X\mathrm{MnO}_{4}^{-} + Y\mathrm{C}_{2}\mathrm{O}_{4}^{2-} + Z\mathrm{H}^{+} \rightarrow X\mathrm{Mn}^{2+} + 2Y\mathrm{CO}_{2} + \frac{Z}{2}\mathrm{H}_{2}\mathrm{O}\). Ensure there are equal numbers of each atom on both sides: Mn, C, O, and H.

Step 3: Balance Manganese Atoms

Ensure that the manganese atoms are balanced on both sides. We have \(X\) atoms of Mn in \(\mathrm{MnO}_{4}^{-}\) and \(X\) atoms of \(\mathrm{Mn}^{2+}\). So, no extra equation is needed for Mn.

Step 4: Balance Carbon Atoms

Balance the carbon atoms by ensuring the carbon atoms from \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) equal those in \(\mathrm{CO}_{2}\). Each \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) contributes 2 carbon atoms. Thus, \(2Y\) from \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) should match \(2Y\mathrm{CO}_{2}\). The value of \(Y\) is the same on both sides.

Step 5: Balance Oxygen Atoms

Start with the oxygen atoms from \(\mathrm{MnO}_{4}^{-}\) and \(\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\). The oxygen atoms from \(X\mathrm{MnO}_{4}^{-}\) and \(Y\mathrm{C}_{2}\mathrm{O}_{4}^{2-}\) should equate to those in \(2Y\mathrm{CO}_{2}\) and \(\frac{Z}{2}\) in \(\mathrm{H}_{2}\mathrm{O}\). Solve the equation: \(4X + 4Y = 2Y + \frac{Z}{2}\).

Step 6: Balance Hydrogen Atoms

Now balance the hydrogen atoms: \(\mathrm{ZH}^{+}\) is equal to \(\frac{Z}{2}\) in \(\mathrm{H}_{2}\mathrm{O}\). This ensures the equation is consistent for hydrogen as well.

Step 7: Solve the Equations for X, Y, Z

From the balancing process, deduce the following:- For the oxygen atoms: \(4X + 4Y = 2Y + \frac{Z}{2}\)- For the hydrogen atoms: \(Z = 2(2Y - 4X)\)Solve these equations with trial values from the given options.

Step 8: Verify & Confirm

By trying options, confirm that the option (c) \(X = 5\), \(Y = 2\), \(Z = 16\) satisfies all balance equations: 5 manganese, 2 carbon pairs, 16 hydrogens, and all oxygen atoms.

Key Concepts

Stoichiometry

Stoichiometry is the quantitative study of reactants and products in a chemical reaction. It uses a balanced chemical equation to calculate the relationships between products and reactants. This process ensures that the law of the conservation of mass is followed, meaning the total mass of reactants equals the total mass of the products.
A balanced chemical equation provides the molar ratio of the reactants to the products, which is essential for stoichiometric calculations. In the given reaction \(X\text{MnO}_{4}^{-} + Y\text{C}_{2}\text{O}_{4}^{2-} + Z\text{H}^{+} \rightarrow X\text{Mn}^{2+} + 2Y\text{CO}_{2} + \frac{Z}{2}\text{H}_{2}\text{O}\), the values of \(X, Y,\) and \(Z\) need to balance the equation.

• To achieve this, you adjust the stoichiometric coefficients \(X, Y,\) and \(Z\) so that the number of each type of atom is the same on both sides of the equation.
• Each coefficient represents the number of moles of a substance involved.

Through trial and error of the provided options, you determine the coefficients that balance the equation are \(X = 5\), \(Y = 2\), \(Z = 16\). This ensures that all atoms including manganese, carbon, oxygen, and hydrogen are balanced appropriately, showcasing the heart of stoichiometry.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, often called redox reactions, involve the transfer of electrons between species. In such reactions, one substance is oxidized, meaning it loses electrons, while another is reduced, gaining those electrons. Understanding these processes is critical for balancing redox reactions.
In the provided reaction, different species undergo oxidation or reduction:

• The \(\text{MnO}_{4}^{-}\) ion is reduced to \(\text{Mn}^{2+}\). This reduction involves a gain of electrons.
• The \(\text{C}_{2}\text{O}_{4}^{2-}\) ion undergoes oxidation, transforming into \(\text{CO}_{2}\), which involves a loss of electrons.

To balance redox reactions using the method of half-reactions:

• First, write separate half-reactions for oxidation and reduction.
• Balance each half-reaction for mass and charge.
• Finally, combine the half-reactions, ensuring the electrons lost equal the electrons gained.

The balancing of the reaction showcases the essential concepts of redox reactions, highlighting how oxidation and reduction intermingle, ensuring the conservation of charge and mass.

Mole Concept

The mole concept is a fundamental chemical concept used to measure amounts of a chemical substance. It allows chemists to count particles by weighing them. A mole is defined as the amount of substance that contains as many entities as there are atoms in 12 grams of carbon-12. This is approximately \(6.022 \times 10^{23}\) entities, known as Avogadro's number.
Understanding the concept of moles is key in stoichiometry, especially in balancing reactions, as seen in the given balanced equation. Each coefficient in a balanced chemical reaction represents moles of the substance. They allow us to directly relate the number of moles of each reactant to the moles of each product.

• For example, if \(X = 5\), it implies there are 5 moles of \(\text{MnO}_{4}^{-}\).
• If \(Y = 2\), it means there are 2 moles of \(\text{C}_{2}\text{O}_{4}^{2-}\).
• Similarly, \(Z = 16\) suggests 16 moles of \(\text{H}^{+}\).

By converting between moles and grams using molar mass, chemists can predict the quantities of reactants needed and products produced in a reaction, making the mole concept integral to quantitative chemistry.