Question

Consider the following list of reagents:
$\text{Acidified}{K}_{2}C{r}_2{O}_7,\text{Alkanline}KMn{O}_4,CuS{O}_4,H_2{O}_2,C{l}_2,O_3,FeC{l}_3,HN{O}_3,andN{a}_2{S}_2{O}_3$
Which of the following reagents can oxidise aqueous iodide to iodine

• A. $\text{Acidified}{K}_{2}C{r}_2{O}_7$
• B. $N{a}_2{S}_2{O}_3$
• C. $C{l}_2$
• D. $O_3$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $\text{Acidified}{K}_{2}C{r}_2{O}_7$
C $C{l}_2$
D $O_3$

$I^{-}to{I}_2$ oxidation can be done by acidified $K_{2}C{r}_2{O}_7,CuS{O}_4,H_2{O}_2$
$K_{2}C{r}_2{O}_7+KI+H_{2}S{O}_4\to K_{2}S{O}_4+C{r}_2(S{O}_4{)}_3+I_2+H_{2}O$
$2CuS{O}_4+4KI+\to CuI\downarrow +I_2+2K_{2}S{O}_4$
$H_2{O}_2+2KI\to 2KOH+I_2$
$2KI+C{l}_2\to 2KCl+I_2$
$H_{2}O+2KI+O_3\to 2KOH+O_2+I_2$
$FeC{l}_3+2KI\to 2KCl+FeC{l}_2+I_2$
$HN{O}_3+KI\to KN{O}_3+I_2+NO\uparrow$
$N{a}_2{S}_2{O}_3+KI\to \text{no reaction}$
In $H_2{O}_2\text{and}{O}_3,I^{-}to{I}_2\text{oxidation will occur initially.}$
However as concentration of $KOH$ builds up, it will begin causing disproportination resulting $I_{2}toI{O}_3^{-}$
In alkaline $KMn{O}_4$ iodide is oxidised to iodate.