Hydrogen iodide cannot be prepared by the act | vedclass

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Question

(C) When $KI$ reacts with concentrated $H_2SO_4$,$HI$ is initially formed as $KI + H_2SO_4 \rightarrow KHSO_4 + HI$. However,$HI$ is a strong reducing

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$H_2SO_4$ is an oxidising agent

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(C) When $KI$ reacts with concentrated $H_2SO_4$,$HI$ is initially formed as $KI + H_2SO_4 \rightarrow KHSO_4 + HI$. However,$HI$ is a strong reducing agent and concentrated $H_2SO_4$ is a strong oxidising agent. Therefore,$H_2SO_4$ oxidizes $HI$ to $I_2$ according to the reaction: $2HI + H_2SO_4 \rightarrow I_2 + SO_2 + 2H_2O$. Thus,$HI$ cannot be prepared by this method.

Hydrogen iodide cannot be prepared by the action of concentrated $H_2SO_4$ on potassium iodide because:

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