Problem 51 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A solution of \(0.2 \mathrm{~g}\) of a compound containing \(\mathrm{Cu}^{2+}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) ions on titration with \(0.02 \mathrm{M} \mathrm{KMnO}_{4}\) in presence of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) consumes \(22.6\) \(\mathrm{mL}\). of the oxidant. The resultant solution is neutralized with \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), acidified with dil. acetic acid and treated with excess KI. The liberated iodine requires \(11.3 \mathrm{~mL}\) of \(0.05 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solution for complete reduction. Find out the molar ratio of \(\mathrm{Cu}^{2+}\) to \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) in the compound. Write down the balanced redox reactions involved in the above titrations. [1991 - 5 Marks]

Step-by-Step Solution

Answer

The molar ratio of \(\text{Cu}^{2+}\) to \(\text{C}_2\text{O}_4^{2-}\) is 1:1.

Step 1: Understand the Reactions Involved

The primary reaction in sulfuric acid requires KMnO₄ to oxidize C₂O₄²⁻ to CO₂. The second step with KI forms I₂, which is titrated by Na₂S₂O₃. The last step involves the oxidation of Cu²⁺ by MnO₄⁻ to form Cu⁺.

Step 2: Calculate Moles of Oxalate Ion

For the titration with KMnO₄, use the reaction: \[ \text{5 C}_2\text{O}_4^{2-} + 2 \text{MnO}_4^- + 16 \text{H}^+ \rightarrow 2 \text{Mn}^{2+} + 10 \text{CO}_2 + 8 \text{H}_2\text{O} \]22.6 mL of 0.02 M KMnO₄ gives the moles of MnO₄⁻:\[ 0.0226 imes 0.02 = 0.000452 \text{ moles of MnO}_4^- \]Use stoichiometry: \[ \frac{5}{2} \times 0.000452 = 0.00113 \text{ moles of C}_2\text{O}_4^{2-} \]

Step 3: Calculate Moles of Copper Ion

The reaction in the presence of KI: \[ 2 \text{Cu}^{2+} + 4 \text{I}^- \rightarrow 2 \text{CuI} + \text{I}_2 \]Thiosulfate titration: \[ \text{I}_2 + 2 \text{S}_2\text{O}_3^{2-} \rightarrow 2 \text{I}^- + \text{S}_4\text{O}_6^{2-} \]11.3 mL of 0.05 M Na₂S₂O₃ gives the moles of I₂:\[ 0.0113 \times 0.05 = 0.000565 \text{ moles of I}_2 \]Thus, moles of Cu²⁺ (since 2 Cu²⁺ produce 1 I₂):\[ 2 \times 0.000565 = 0.00113 \text{ moles of Cu}^{2+} \]

Step 4: Determine Molar Ratio

The moles of C₂O₄²⁻ is 0.00113 and moles of Cu²⁺ is also 0.00113, giving a molar ratio of \(1:1\).

Step 5: Balance the Redox Reactions

The balanced reactions are:1. KMnO₄ and C₂O₄²⁻ in acidic medium: \[ 5 \text{C}_2\text{O}_4^{2-} + 2 \text{MnO}_4^- + 16 \text{H}^+ \rightarrow 2 \text{Mn}^{2+} + 10 \text{CO}_2 + 8 \text{H}_2\text{O} \]2. Cu²⁺ with I⁻: \[ 2 \text{Cu}^{2+} + 4 \text{I}^- \rightarrow 2 \text{CuI} + \text{I}_2 \]3. I₂ with Na₂S₂O₃:\[ \text{I}_2 + 2 \text{S}_2\text{O}_3^{2-} \rightarrow 2 \text{I}^- + \text{S}_4\text{O}_6^{2-} \]

Step 6: Final Step: Conclude the Molar Ratio

After understanding the stoichiometry of the reactions, count moles for each ion, and conclude that the molar ratio of \(\text{Cu}^{2+}\) to \(\text{C}_2\text{O}_4^{2-}\) is \(1:1\).

Key Concepts

Stoichiometry

Stoichiometry is the key to solving problems that involve chemical reactions, allowing us to quantitatively analyze how reactants transform into products. In our given problem, we use stoichiometry to determine the ratio of moles in the redox reactions.
In essence, stoichiometry involves the following steps:

• Balancing the Equation: Identify and balance the chemical equations involved in the reaction. This helps to ensure that the numbers of atoms for each element are the same on both sides of the reaction equation.
• Using Proportions: Proportionally convert the given volumes and concentrations of our solutions into moles using molecular weights.
• Applying Molar Ratios: Once moles are determined for each compound involved in a reaction, use the coefficients from the balanced equation to find actual ratios.

An important concept is using the mole ratio, which comes from the coefficients of the balanced equation, to link the amounts of reactants and products. In this exercise, stoichiometry allows us to determine that the ratio of copper ions (\(\text{Cu}^{2+}\)) to oxalate ions (\(\text{C}_2\text{O}_4^{2-}\)) is 1:1.

Titration

Titration is a laboratory method of quantitative chemical analysis used to determine the concentration of a known reactant. In this exercise, we perform titrations to find out the moles of substances involved.
Here’s how the titration process works in the problem:

• Acidic Titration with KMnO₄: Potassium permanganate (KMnO₄) in acidic solution is used to oxidize \(\text{C}_2\text{O}_4^{2-}\) ions. This step is quantified by measuring the volume of KMnO₄ solution needed to reach the endpoint, giving moles of oxalate ion.
• Titration with Iodine: In the second reaction, after neutralization and acidification, iodine (\(\text{I}_2\)) is liberated. This iodine is titrated using sodium thiosulfate (\(\text{Na}_2\text{S}_2\text{O}_3\)), which helps in determining the amount of iodine produced, inferring the moles of copper ions.

Each step of titration requires a careful approach to ensure accurate results. By tracking how much titrant is needed to reach the end point of each reaction, we can find the concentrations of unknown solutions.

Ionic Equations

When dealing with redox reactions, ionic equations can simplify how we represent the transformation of ions. These equations are particularly useful when looking at what happens to the ions involved in the reactions, including interactions between oxidants and reductants.
In this problem, we use ionic equations to show the detailed interactions in redox reactions:

• Ionic Equations in Acidic Medium: Potassium permanganate and oxalate ions react showcasing how MnO₄⁻ ions oxidize \(\text{C}_2\text{O}_4^{2-}\) into CO₂. This reaction is a perfect example of using ionic equations to illustrate electron transfer.
• Displacement in Redox Reactions: With iodide ions and copper ions, these equations illustrate the conversion to copper iodide (\(\text{CuI}\)) and liberation of iodine, indicating the electron acceptance by iodide.
• Subsequent Reduction: The iodine ions formed are later reduced by thiosulfate ions back to iodide, depicted beautifully by these ionic equations.

Writing ionic equations enables us to break a complex reaction into simpler parts, emphasizing the principal ions involved, thus making it easier to understand the underlying chemistry of the process.