Question

$CHC{l}_3$ and $H_{2}O$ from two separate immiscible layers. $B{r}_2$ (orange) and $I_2$ (violet) are soluble in organic layer to a greater extent. In test tube $A$ ($KI+CHC{l}_3+H_{2}O$) and $B$ $(KBr+CHC{l}_3+H_{2}O)$, $C{l}_2$ gas is passed then:
Given:
$\frac{1}{2}C{l}_2+e^{-}⟶C{l}^{-}$ $E^o$= 1.40 V

$\frac{1}{2}B{r}_2+e^{-}⟶B{r}^{-}$ $E^o$= 1.09 V

$\frac{1}{2}{I}_2+e^{-}⟶I^{-}$ $E^o$ = 0.62 V

• A. organic layer in $A$ appear violet
• B. organic layer in $A$ appear orange
• C. organic layer in $B$ appear orange
• D. organic layer in $B$ first appear violet then orange

Answer 1

Answer: (A), (C)

The correct options are
A organic layer in $A$ appear violet
C organic layer in $B$ appear orange

A. and B. Chlorine will displace iodine from potassium iodide. The violet coloured iodine will dissolve in the organic layer. hence, the organic layer in A will appear violet. Hence, the option A is correct and the option B is incorrect.
C. and D. Chlorine will displace bromine from potassium bromide. The bromine will dissolve in the organic layer. Hence, the organic layer in B will appear orange. Hence, the option C is correct and the option D is incorrect.