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Chapter 18: Problem 170

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) \(\mathrm{CuI}_{2}^{2}\) is formed (d) Evolved \(\mathrm{I}_{2}\) is reduced

Short Answer

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

Step by step solution

01

Identify the Reaction

When excess KI is added to CuSO鈧�, a redox reaction occurs. The copper in CuSO鈧� is reduced and iodide ions (I鈦�) are oxidized. This reaction leads to the formation of CuI, which exists as Cu鈧侷鈧� because CuI is unstable.
02

Analyze the Role of Na鈧係鈧侽鈧�

Na鈧係鈧侽鈧�, or sodium thiosulfate, is generally used as a reducing agent. It reacts with the iodine formed during the oxidation of I鈦� to I鈧�, reducing it back to I鈦�. It gets oxidized in the process.
03

Evaluate the Options

Now that we've reviewed the reaction: - Option (a): Cu鈧侷鈧� is indeed formed when Cu虏鈦� is reduced. - Option (b): Na鈧係鈧侽鈧� oxidizes when it reduces I鈧� to I鈦�. - Option (c): CuI鈧� does not form because copper precipitates as Cu鈧侷鈧�, hence this is incorrect. - Option (d): I鈧� is reduced 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 Sulfate and Potassium Iodide Reaction
When you mix copper sulfate (\(\mathrm{CuSO}_4\)) with potassium iodide (\(\mathrm{KI}\)), an interesting redox reaction occurs. In this reaction, \(\mathrm{Cu}^{2+}\) ions from \(\mathrm{CuSO}_4\) accept electrons from \(\mathrm{I}^-\) ions supplied by \(\mathrm{KI}\). This electron exchange results in the oxidation of iodide ions to iodine (\(\mathrm{I}_2\)).
Meanwhile, copper ions are reduced. Reduction means the copper ions gain electrons and transform into a larger solid compound known as copper(I) iodide or \(\mathrm{Cu}_2\mathrm{I}_2\).
This compound will often appear as a precipitate with a lovely white color.
  • This reaction is a typical redox process where two chemical species undergo simultaneous reduction and oxidation.
  • The copper ions act as an oxidizing agent by taking electrons.
  • The iodide ions lose electrons, making them the reducing agent in this scenario.
Sodium Thiosulfate as a Reducing Agent
Sodium thiosulfate (\(\mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3\)) plays a crucial role in reactions involving iodine because it acts as a gentle reducing agent. When it's introduced to a solution containing iodine (\(\mathrm{I}_2\)), sodium thiosulfate helps to transform it back into iodide ions (\(\mathrm{I}^-\)).
This reduction process is critical because it essentially reverse engineers the oxidation process that took place earlier.
When sodium thiosulfate reduces iodine, it itself goes through oxidation, making this an important redox reaction as well.
  • By accepting iodine and turning it back into iodide, sodium thiosulfate shows its power as a reducing agent.
  • This makes it extremely useful in analytical chemistry, especially when you want to titrate iodine.
  • Essentially, sodium thiosulfate ensures the complete and efficient removal of iodine by converting it back into a less reactive form.
Formation of Cu2I2
The formation of copper(I) iodide or \(\mathrm{Cu}_2 \mathrm{I}_2\) happens as a key step when \(\mathrm{CuSO}_4\) reacts with \(\mathrm{KI}\). Initially, one might think that cuprous iodide, \(\mathrm{CuI}\), should be the product, but in fact, it is unstable when more than a slight amount is present.
Instead, these unstable units combine to form \(\mathrm{Cu}_2 \mathrm{I}_2\), a compound that holds together more securely.
  • The resulting \(\mathrm{Cu}_2 \mathrm{I}_2\) is a key indicator that a proper redox reaction has taken place.
  • The white precipitate indicates copper ions have successfully been reduced.
  • This formation is unique because more than one \(\mathrm{CuI}\) molecule builds a stable network.
Oxidation of Iodide Ions
In the initial stages of the reaction carried out between \(\mathrm{CuSO_4}\) and \(\mathrm{KI}\), the iodide ions (\(\mathrm{I}^-\)) donate electrons to the copper ions. By giving away electrons, these iodide ions undergo an oxidation process, turning into iodine molecules (\(\mathrm{I}_2\)).
This is a classic example of oxidation and illustrates how the electrons are shifted during the reaction.
You can often observe this transformation because iodine provides a noticeable color change in solution.
  • Oxidation of iodide ions is critical for the subsequent steps of the reaction.
  • The color shift into a purplish hue signals the presence of iodine.
  • This step helps to set up further reactions like the reduction by sodium thiosulfate.

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Most popular questions from this chapter

When the sample of copper with zinc impurity is to be purified by electrolysis the appropriate electrodes are Cathode (a) pure zinc (b) pure copper (c) impure zinc (d) impure sample Anode pure copper impure sample impure sample pure copper

Match the following: List-I 1\. Bell metal 2\. Gun metal 3\. Bronze 4\. Brass List-II (i) Cu: \(60 \%-80 \%\), Zn: \(20 \%-40 \%\) (ii) Cu: \(75 \%-90 \%\), Sn: \(10 \%-25 \%\) (iii) Cu: \(88 \%\), Sn: \(10 \%\), Zn: \(2 \%\) (iv) Cu: 80\% Sn: 20\%

The fixing process of photographic film involves removal of unchanged silver bromide as (a) \(\mathrm{Na}_{2}\left[\mathrm{AgBr}\left(\mathrm{SO}_{3}\right)\right]\) (b) \(\mathrm{Na}_{3}\left[\mathrm{AgBr}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)\right]\) (c) \(\mathrm{Na}_{3}\left[\mathrm{Ag}\left(\mathrm{SO}_{3}\right)_{2}\right]\) (d) \(\mathrm{Na}_{3}\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right]\)

Galvanization is applying a coating of? (a) \(\mathrm{Cu}\) (b) \(\mathrm{Zn}\) (c) \(\mathrm{Cr}\) (d) \(\mathrm{Pb}\)

In this reaction (C) and (D) are respectively, \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O} \stackrel{100^{\circ} \mathrm{C}}{\longrightarrow}(\mathrm{A}) \stackrel{230 \mathrm{C}}{\longrightarrow}\) (B) \(\stackrel{800^{\circ} \mathrm{C}}{\longrightarrow}(\mathrm{C})+(\mathrm{D})\) (a) \(\mathrm{Cu}_{2} \mathrm{O}, \mathrm{SO}_{2}\) (b) \(\mathrm{CuO}, \mathrm{SO}_{3}\) (c) \(\mathrm{CuO}, \mathrm{SO}_{2}\) (d) \(\mathrm{Cu}_{2} \mathrm{O}, \mathrm{SO}_{3}\)
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