Question

Which of the following given below can liberate $B{r}_2$ from $KBr$ ?
$F_2,C{l}_2,I_2$ and $Conc.H_{2}S{O}_4$

• A. $F_2,C{l}_2,I_2$ and $Conc.H_{2}S{O}_4$
• B. $C{l}_2,I_2$ and $Conc.H_{2}S{O}_4$
• C. $F_2,C{l}_2,I_2$
• D. $F_2,C{l}_2$ and $Conc.H_{2}S{O}_4$

Answer 1

Answer: (D)

$I_2$ cannot liberate $B{r}_2$ from KBr, since $I_2$ is weaker oxidant than $B{r}_2.$
$F_2$ and $C{l}_2$ , being stronger oxidants than $B{r}_2$ , can liberate $B{r}_2$ from KBr.
$F_2+2KBr\to 2KF+B{r}_2$
$C{l}_2+2KBr\to 2KCl+B{r}_2$
Reaction of conc. $H_{2}S{O}_4$ with KBr forms HBr. HBr is a strong reducing agent so it reduces $H_{2}S{O}_4$ to form a mixture of $S{O}_2$ and $B{r}_2$ .
$KBr+H_{2}S{O}_4\to HBr+KHS{O}_4$
$KHS{O}_4+KBr\to HBr+K_{2}S{O}_4$
HBr so formed reduces $H_{2}S{O}_4$ .
$H_{2}S{O}_4+2HBr\to S{O}_2+2H_{2}O+B{r}_2$