Question

Concentrated $H_{2}S{O}_4$ cannot be used to prepare $HBr$ or $HI$ from $KBr$ or $KI$ because it:

• A. reacts too slowly with $KBr$ or $KI$
• B. reduces $HBr$ or $HI$
• C. oxidises $HBr$ or $HI$
• D. oxidises $KBr$ to $KBr{O}_3$ or $KI$ to $KI{O}_3$

Answer 1

The correct option is C oxidises $HBr$ or $HI$

$H_{2}S{O}_4$ is a concentrated acid, and act as oxidizing agent. It oxidizes the $KBr$ and $KI$ into $B{r}_2$ and $I_2$.

$2H_{2}S{O}_4+2KBr\to K_{2}S{O}_4+B{r}_2+2H_{2}O+S{O}_2$

$2H_{2}S{O}_4+2KI\to K_{2}S{O}_4+I_2+2H_{2}O+S{O}_2$

Hence, option C is correct.