Problem 132 - A Complete Resource Book in Chemistry for JEE Main

Question

Excess of \(\mathrm{KI}\) react with \(\mathrm{CuSO}_{4}\) solution and then \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solution is added to it. Which of the following statements is incorrect for this reaction? (a) \(\mathrm{Cu}_{2} \mathrm{I}_{2}\) is formed (b) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) is oxidized (c) \(\mathrm{CuI}_{2}\) is formed (d) evolved \(\mathrm{I}_{2}\) is reduced

Step-by-Step Solution

Answer

The incorrect statement is (c): CuI2 is formed.

Step 1: Identify the Reaction Between KI and CuSO4

When excess KI is mixed with CuSO4, a double displacement reaction occurs. Copper(II) iodide, CuI2, is initially formed, but it is unstable and decomposes to form CuI and I2.The reaction can be expressed as:\[2 ext{Cu}^{2+} + 4 ext{I}^- \rightarrow 2 ext{CuI} + ext{I}_2\]Thus, the precipitate that forms is Cu2I2 (due to the decomposition of CuI2), along with the liberation of I2.

Step 2: Reaction of Iodine with Na2S2O3

The Na2S2O3 solution reacts with the liberated iodine. This is a classic iodine-thiosulfate titration reaction where iodine is reduced to iodide by the thiosulfate, and thiosulfate is oxidized.The reaction is:\[ ext{I}_2 + 2 ext{S}_2 ext{O}_3^{2-} \rightarrow 2 ext{I}^- + ext{S}_4 ext{O}_6^{2-}\]Thus, Na2S2O3 acts as a reducing agent and iodine (I2) is reduced.

Step 3: Analyze Each Statement

(a) \(\text{Cu}_2\text{I}_2\) is indeed formed as a result of the decomposition of \(\text{CuI}_2\). This statement is correct.(b) Na2S2O3 is oxidized to form \(\text{S}_4\text{O}_6^{2-}\). This statement is correct.(c) \(\text{CuI}_2\) is not stable and is not observed in the final product; only \(\text{Cu}_2\text{I}_2\) forms as a stable compound. Thus, this statement is incorrect.(d) The evolved \(\text{I}_2\) gets reduced back to \(\text{I}^-\) through reaction with \(\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3\). This statement is correct.

Key Concepts

Copper(I) Iodide Formation

Understanding the formation of Copper(I) Iodide, or CuI, begins with a double displacement reaction. Here, excess potassium iodide (KI) reacts with copper(II) sulfate (CuSO₄). Initially, copper(II) iodide (CuI₂) is formed in this reaction. However, CuI₂ is unstable and decomposes. This instability leads to its breakdown into Copper(I) Iodide (CuI) and iodine (I₂).

This process can be understood by considering that copper, a metal in the reaction, seeks a stable electron configuration. By reducing its oxidation state from +2 to +1, copper achieves stability in the form of CuI. The transformation of CuI₂ into CuI and I₂ can be depicted by the equation: \[2 \text{Cu}^{2+} + 4 \text{I}^- \rightarrow 2 \text{CuI} + \text{I}_2\].

It's important to remember that CuI₂ does not remain as it is; rather, it decomposes, and we observe Cu₂I₂ along with free iodine.

Iodine-Thiosulfate Reaction

The Iodine-Thiosulfate reaction is a key step in understanding how iodine can be reduced back to iodide ions. When iodine is released as a product of the decomposition of CuI₂, it does not remain in its elemental state in this particular reaction scenario. Instead, it engages in a titration reaction with thiosulfate ions from sodium thiosulfate (Na₂S₂O₃).

This reaction is known for its role in iodometric titration, where iodine acts as an oxidizing agent, accepting electrons from thiosulfate, which is a reducing agent. The balanced chemical equation for this reaction is: \[ \text{I}_2 + 2 \text{S}_2\text{O}_3^{2-} \rightarrow 2 \text{I}^- + \text{S}_4\text{O}_6^{2-} \].

Here, iodine (I₂) is reduced to iodide ions (I⁻), while thiosulfate is oxidized to tetrathionate (S₄O₆²⁻). This elegant exchange showcases classic redox chemistry principles, where electron transfer plays a critical role.

Displacement Reaction in Chemistry

The concept of displacement reactions is central to understanding various chemical transformations. In general, a displacement reaction involves the replacement of one element from a compound by another element. Here, the reaction between KI and CuSO₄ operates via a double displacement mechanism.

Initially, the components of these compounds swap partners, resulting in the formation of temporary products. However, due to the inherent instability of CuI₂ in the product mix, it further decomposes to copper(I) iodide and free iodine. This nature of swapping and subsequent decomposition exemplifies a unique aspect of displacement reactions, where the initial product formed does not necessarily stabilize the reaction system.

Displacement reactions often feature metals replacing other metals or, as in this case, dissociating into more stable components. This is tightly linked to the chemical reactivity and stability of overlap products that form during such reactions.