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Chapter 19: Problem 132

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? [2004] (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

Short Answer

Expert verified
The incorrect statement is (c) CuI鈧� is formed.

Step by step solution

01

Understanding the Reaction

When excess KI is mixed with a CuSO鈧� solution, a chemical reaction takes place where copper(I) iodide (Cu鈧侷鈧�) and iodine (I鈧�) are formed, along with potassium sulfate (K鈧係O鈧�). The reaction is as follows: \[ 2 ext{CuSO}_{4} + 4 ext{KI} ightarrow 2 ext{CuI} + ext{I}_2 + 2 ext{K}_2 ext{SO}_4 \] In this step, Cu鈧侷鈧� is formed, not CuI鈧�, because Cu虏鈦� reacts with I鈦� to form Cu鈧侷鈧� and molecular iodine (I鈧�).
02

Role of Sodium Thiosulfate (Na鈧係鈧侽鈧�)

Sodium thiosulfate (Na鈧係鈧侽鈧�) is introduced into the solution containing iodine. It reacts with iodine to reduce it back to iodide ions, and the thiosulfate is oxidized in the process. The reaction is as follows:\[ ext{I}_2 + 2 ext{Na}_2 ext{S}_2 ext{O}_3 ightarrow 2 ext{NaI} + ext{Na}_2 ext{S}_4 ext{O}_6 \]This means that statement (b) is correct as thiosulfate (Na鈧係鈧侽鈧�) is indeed oxidized.
03

Checking Each Statement

Let's evaluate each statement: - (a) Cu鈧侷鈧� is formed: True, as confirmed in the reaction between CuSO鈧� and KI. - (b) Na鈧係鈧侽鈧� is oxidized: True, during its reaction with I鈧�. - (c) CuI鈧� is formed: False, CuI鈧� is not stable; Cu鈧侷鈧� is formed instead. - (d) Evolved I鈧� is reduced: True, reduced back to I鈦� by Na鈧係鈧侽鈧�.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Copper(I) Iodide Formation
When copper sulfate (\(\text{CuSO}_4\)) reacts with excess potassium iodide (\(\text{KI}\)), an interesting redox reaction occurs that leads to the formation of copper(I) iodide, represented as \(\text{Cu}_2\text{I}_2\). This compound forms because the copper ions (\(\text{Cu}^{2+}\)) in solution react with iodide ions to create the copper iodide compound.
  • Two copper ions pair up with two iodide ions to make \(\text{Cu}_2\text{I}_2\).
  • The rest of the iodide ions join together to produce iodine molecules (\(\text{I}_2\)), which are released as part of the reaction.
  • Potassium sulfate is a byproduct.
This reaction is crucial because instead of forming copper(II) iodide, which is unstable, it results in copper(I) iodide. Understanding this helps clarify why option (c) in the original exercise is incorrect.
Sodium Thiosulfate Reaction
In this chemical scenario, sodium thiosulfate (\(\text{Na}_2\text{S}_2\text{O}_3\)) plays an essential role due to its reaction with the iodine formed in the initial stages. Its primary function here is to interact with what is known as free iodine (\(\text{I}_2\)) in the solution.
  • Sodium thiosulfate adds to iodine, leading to a transformation back into iodide ions, effectively reversing the earlier reaction.
  • During this process, the thiosulfate itself undergoes oxidation, becoming tetrathionate (\(\text{S}_4\text{O}_6^{2-}\)).
  • This reaction allows the recycling of iodide ions in the mixture.
The involved oxidation of thiosulfate confirms that this statement in the textbook case is correct, emphasizing the remarkable interplay of reduction and oxidation.
Iodine Reduction
The concept of iodine reduction is integral to understanding the chemistry involved when sodium thiosulfate meets iodine. When iodine (\(\text{I}_2\)) mixes with sodium thiosulfate, it is primarily reduced:
  • This transition converts iodine molecules back to their simpler iodide ion form, noticeably changing the complexion of the chemical environment.
  • This selective reduction is significant because it allows other reactions in the solution to proceed without the interference of excess iodine.
  • Moreover, the shifting of iodine to iodide enhances the possibilities for iodine reuse, or for precisely controlling stoichiometry in chemical reactions.
Due to this reaction, statement (d) can indeed be confirmed as valid, underlining the cyclical nature of redox reactions.
Oxidation-Reduction in Chemistry
Oxidation-reduction reactions, redox for short, are fundamental principles governing chemical changes. They include two essential processes: oxidation, where a substance loses electrons, and reduction, where a substance gains electrons.
  • In our reaction context, copper ions are reduced (\(\text{Cu}^{2+} \, \rightarrow \, \text{Cu}^+\)), gaining electrons from iodide ions.
  • Iodide ions themselves are oxidized as they transform into iodine (\(\text{I}^- \, \rightarrow \, \text{I}_2\)), effectively losing electrons.
  • Moreover, the reaction undergoes further complexity when sodium thiosulfate neutralizes iodine, reducing it back to iodide.
These exchanges illustrate the beautifully reciprocal nature of redox reactions, where the balance of electron transfer drives the direction and outcome of chemical processes. Analyzing them helps in identifying which components are playing reductive or oxidative roles, crucial in problem-solving real-world chemical reactions.

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