Question

Excess of \[{\text{KI}}\] reacts with \[{\text{CuS}}{{\text{O}}_4}\] solution, and \[{\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_3}\] solution is added to it. Which of the following statements is incorrect for the reaction?
A ) Evolved \[{{\text{I}}_2}\] is reduced.
B ) \[{\text{Cu}}{{\text{I}}_2}\] is formed.
C ) \[{\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_3}\] is oxidised.
D ) \[{\text{C}}{{\text{u}}_2}{{\text{I}}_2}\] is formed.

Answer 1

Hint: Increase in the oxidation number is the oxidation and decrease in the oxidation number is reduction. Increase in oxidation number occurs when electrons are lost. Decrease in the oxidation number occurs when electrons are gained.

Complete step by step answer:
This is an iodometric titration. The test is called a hypo test.
The reactions involved in the titration are listed below:
> Copper sulphate reacts with potassium iodide to form potassium sulphate, cuprous iodide and iodine.
\[{\text{2 CuS}}{{\text{O}}_4}{\text{ + 4 KI }} \to {\text{ 2 }}{{\text{K}}_2}{\text{S}}{{\text{O}}_4}{\text{ + $Cu_2I_2$ + }}{{\text{I}}_2}\]
\[{\text{2 N}}{{\text{a}}_{\text{2}}}{{\text{S}}_2}{{\text{O}}_3}{\text{ + }}{{\text{I}}_2}{\text{ }} \to {\text{ N}}{{\text{a}}_2}{{\text{S}}_4}{{\text{O}}_6}{\text{ + 2 NaI}}\]
> The liberated iodine then reacts with sodium thiosulphate to form sodium iodide and sodium tetrathionate. The oxidation number of iodine in iodine metal and sodium iodide product are 0 and -1 respectively. Thus, there is a decrease in the oxidation number of iodine from 0 to 1. The decrease in the oxidation number is reduction. Hence, during the reaction,
 the liberated iodine is reduced. Hence, option A ) represents the correct statement.
> The oxidation number of sulphur in sodium thiosulphate is +2. The oxidation number of sulphur in sodium tetrathionate is +2.5. Thus, during the reaction, the oxidation number of sulphur increases from +2 to +2.5. Increase in the oxidation number is oxidation. Thus, during the reaction, sodium thiosulphate undergoes oxidation. Thus option C ) represents the correct statement.
> From the first reaction we can see $Cu_2I_2$ is formed not $CuI_2$.

Hence, option B ) represents an incorrect statement and option D ) represents the correct statement.

Note: Iodometric titration is used to determine the concentration of an oxidizing agent in solution. Indicator is a starch solution which absorbs liberated iodine. Excess of potassium iodide is added to excess of acidic solution containing oxidising agents. The liberated iodine is then titrated with sodium thiosulphate solution.