Why is sulphuric acid not used during the reaction of alcohols with KI?

Hint: Sulphuric acid is a very strong oxidizing agent. So, its impact on any reaction will be very different from a weak oxidizing agent. The strong oxidizing nature of sulphuric acid forms the basis of many reactions which are not shown by its dilute counterpart.

Complete step by step solution:
${H_2}{SO_4}$ being a strong oxidising agent oxidises HI and produces $I_2$ . The reaction between alcohol (OH) and HI results in production of alkyl iodide. The motive of the reaction is to form alkyl iodides. When HI gets converted to $I_2$ , the motive of the reaction remains unaccomplished. This is the reason why sulphuric acid (${H_2}{SO_4}$) is not used during the reaction of alcohol (OH) with potassium iodide (KI).
In the presence of a dilute acid, KI would produce HI.
 $ 2KI + {H_2}{SO_4}\to 2KHSO_4 + HI $
If the acid used is sulphuric acid, the HI gets used up to produce I2 gas.
 $ 2HI + {H_2}{SO_4}\to I_2 + SO_2 + H_2O $
As a result, the action of alcohol on acid to produce alkyl iodide cannot occur. Therefore, sulphuric acid is not used for this reaction. Instead, a weak oxidizing agent like ${H_3}{PO_4}$ is preferred for this reaction.

Note:
Whenever there is a reaction involving a strong oxidizing agent, always consider the difference between reaction involving dilute acid and that involving concentrated acid. Concentrated acids will result in a different product than their dilute counterparts. Also, a strong oxidizing acid may cause destruction at higher temperatures.Also consider the reaction between HI and weak oxidizing agents.