Question

Excess of $KI$ reacts with $CuS{O_4}$ solution and then $N{a_2}{S_2}{O_3}$ solution is added to it. Which of the statements is incorrect for this reaction?
a) $C{u_2}{I_2}$ Is formed.
b) $Cu{I_2}$ Is formed.
c) $N{a_2}{S_2}{O_3}$ Is oxidised.
d) Evolved ${I_2}$ is reduced.

Answer 1

The concept of oxidation and reduction reactions must be known. Knowledge about iodometric titrations is required as to answer these questions reagents of these titrations play an important role.

Complete solution:
Oxidation reactions are those where increase in oxidation state of an element is observed, whereas reduction reactions are those where decrease in oxidation state of element is observed.
Iodometric titrations are those where the amount of iodine evolved in a reaction is calculated by titrating the resulting solution with a hypo solution i.e. indirect method.
The given equation was found in the iodometric titration:
$2CuS{O_4} + 4KI \to C{u_2}{I_2} \downarrow + {I_2} + 2{K_2}S{O_4}$
In the above reaction the white precipitate formed is of the formula $C{u_2}{I_2}$ i.e. cuprous iodide.
Thus the precipitate$Cu{I_2}$, is never formed during the entire course of reaction.
After the above mentioned reaction, the liberated iodine gas is titrated with sodium thiosulphate solution to determine its strength.
$2N{a_2}{S_2}{O_3} + {I_2} \to N{a_2}{S_4}{O_6} + 2NaI$
In this reaction, the oxidation state of iodine changes from $0$to $- 1$i.e. reduction reaction occurred, whereas the oxidation state of sulphur in sodium thiosulphate changes from $- 2$ to $+ 2$ and thus oxidation reaction occurred here.
$\therefore$ $N{a_2}{S_2}{O_3}$ Is oxidised and ${I_2}$ is reduced during the course of reaction.
By calculating the amount of sodium thiosulphate used in the reaction, the amount of liberated${I_2}$ in the previous reaction can be calculated.
In the iodometric titrations, the indicator used will be starch as it can form a complex with iodine which gives characteristic blue black colour.
$\therefore$The incorrect option mentioned in the question is (b).

Note: The complex formed with starch and iodine is not very stable as when the reducing agent i.e. sodium thiosulphate is added, the colour of the solution changes as iodine from the complex is reduced to iodide.