Question

${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ changes aqueous $\text{KI}$ solution to:
A.$\text{HI}$
B.${{\text{I}}_{\text{2}}}$
C.$\text{K}{{\text{I}}_{\text{3}}}$
D.$\text{HI}{{\text{O}}_{\text{3}}}$

Answer 1

Hint: Hydrogen peroxide is a moderately mild but efficient oxidising agent that has a range of applications in real life, while potassium iodide acts as an excellent reducing agent. The reaction results in a halogen being released.

Complete Answer:
Potassium iodide is oxidised by hydrogen peroxide under acidic conditions to lead to the formation of iodine gas and water molecules. This reaction is a classic example of oxidation-reduction reactions as here hydrogen peroxide is getting reduced to water molecules while the hydrogen iodide gets oxidized to iodine. The reaction is second order with respect to the hydrogen peroxide concentration while first order with respect to the iodide ion concentration.
Hence, potassium iodide is changed to iodine gas in the reaction.

The correct option is B.

Note: The iodine liberated as a result of the reaction is further reduced by sodium thiosulphate solution to iodide ions and hence the solution is rendered colourless. The point at which the solution turns colourless is called the neutralization point and the amount of thiosulphate consumed till this point gives the concentration of the iodine gas present in the solution. Starch is used as the indicator as it reacts with iodine turning blue-black, but it is colourless in presence of iodide anions.
The pH of the reaction and the temperature of the medium are important parameters determining the fate of this reaction. The solution must be acidic enough to maintain a constant concentration of the hydrogen ions.