Question

Which of the following given below can liberate $Br_{2}$ from KBr?

$F_{2},Cl_{2},I_{2}$ and $Conc.\text{}\text{H}_{2}SO_{4}$

• A. $F_{2},Cl_{2},I_{2}$ and $Conc.\text{}\text{H}_{2}SO_{4}$
• B. $Cl_{2},I_{2}$ and $Conc.\text{}\text{H}_{2}SO_{4}$
• C. $F_{2},Cl_{2},I_{2}$
• D. $F_{2},Cl_{2}$ and $Conc.\text{}\text{H}_{2}SO_{4}$

Answer 1

Answer: (D)

$I_{2}$ cannot liberate $Br_{2}$ from KBr, since $I_{2}$ is weaker oxidant than $Br_{2}.$

$F_{2}$ and $Cl_{2}$ , being stronger oxidants than $Br_{2}$ , can liberate $Br_{2}$ from KBr.

$F_{2}+2KBr \rightarrow 2KF+Br_{2}$

$Cl_{2}+2KBr \rightarrow 2KCl+Br_{2}$

Reaction of conc. $H_{2}SO_{4}$ with KBr forms HBr. HBr is a strong reducing agent so it reduces $H_{2}SO_{4}$ to form a mixture of $SO_{2}$ and $Br_{2}$ .

$KBr+H_{2}SO_{4} \rightarrow HBr+KHSO_{4}$

$KHSO_{4}+KBr \rightarrow HBr+K_{2}SO_{4}$

HBr so formed reduces $H_{2}SO_{4}$ .

$H_{2}SO_{4}+2HBr \rightarrow SO_{2}+2H_{2}O+Br_{2}$