
Give appropriate reasons for each of the following:
Addition of chlorine to potassium iodide solution gives it a brown colour but excess of chlorine makes it colourness.
Answer
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Hint: Halogen displacement reaction is a reaction in which a strong halogen displaces the weaker halogen.. The most reactive halogen can displace all other halogens but a weak halogen cannot displace other halogens.
Complete step by step answer:
As we know, chlorine is a better oxidizing agent than bromine and iodine. Therefore, chlorine can displace iodine from its salt solution.
The reaction of chlorine with potassium iodine would be:
$2KI({\text{Potassium iodide)}} + C{l_2}({\text{Chlorine)}} \to 2KCl({\text{Potassium Chloride)}} + {I_2}{\text{(Brown Colour)}}$
The \[{I_2}\] gets displaced and gives brown colour to the solution.
On addition of excessive chlorine, the reaction would be:
$5C{l_2} + {I_2} + 6{H_2}O \to 10HCl + 2HI{O_3}({\text{Colourless)}}$
\[HI{O_3}\] or iodic acid is formed as a product of this reaction. \[HI{O_3}\] is colorless. Hence, addition of excess chlorine makes the solution colourless.
This type of reaction is known as displacement reaction as the more reactive chlorine displaces the less reactive iodine from the solution.
The brown colour appeared after the displacement can also appear as dark purple, which is the colour of iodine.
Additional information: The halogen displacement reactions are redox reactions, which means oxidation and reduction occurs simultaneously. In such a reaction, the halogen gains electrons and gets reduced whereas the halide loses an electron and gets oxidised.
In the reaction
$2KI({\text{Potassium iodide)}} + C{l_2}({\text{Chlorine)}} \to 2KCl({\text{Potassium Chloride)}} + {I_2}{\text{(Brown Colour)}}$
Ionic equation:
${\text{2}}{{\text{I}}^{\text{ - }}}{\text{ + C}}{{\text{l}}_{\text{2}}} \to {\text{2C}}{{\text{l}}^{\text{ - }}}{\text{ + }}{{\text{I}}_{\text{2}}}$
Where potassium is the spectator ion.
Note:
The halogen displacement reaction shown by chlorine and potassium iodine could also be shown when we react with chlorine with sodium iodine or bromide. The only requirement is that the displaying halogen should be more reactive i.e., it must have high oxidizing power.
Complete step by step answer:
As we know, chlorine is a better oxidizing agent than bromine and iodine. Therefore, chlorine can displace iodine from its salt solution.
The reaction of chlorine with potassium iodine would be:
$2KI({\text{Potassium iodide)}} + C{l_2}({\text{Chlorine)}} \to 2KCl({\text{Potassium Chloride)}} + {I_2}{\text{(Brown Colour)}}$
The \[{I_2}\] gets displaced and gives brown colour to the solution.
On addition of excessive chlorine, the reaction would be:
$5C{l_2} + {I_2} + 6{H_2}O \to 10HCl + 2HI{O_3}({\text{Colourless)}}$
\[HI{O_3}\] or iodic acid is formed as a product of this reaction. \[HI{O_3}\] is colorless. Hence, addition of excess chlorine makes the solution colourless.
This type of reaction is known as displacement reaction as the more reactive chlorine displaces the less reactive iodine from the solution.
The brown colour appeared after the displacement can also appear as dark purple, which is the colour of iodine.
Additional information: The halogen displacement reactions are redox reactions, which means oxidation and reduction occurs simultaneously. In such a reaction, the halogen gains electrons and gets reduced whereas the halide loses an electron and gets oxidised.
In the reaction
$2KI({\text{Potassium iodide)}} + C{l_2}({\text{Chlorine)}} \to 2KCl({\text{Potassium Chloride)}} + {I_2}{\text{(Brown Colour)}}$
Ionic equation:
${\text{2}}{{\text{I}}^{\text{ - }}}{\text{ + C}}{{\text{l}}_{\text{2}}} \to {\text{2C}}{{\text{l}}^{\text{ - }}}{\text{ + }}{{\text{I}}_{\text{2}}}$
Where potassium is the spectator ion.
Note:
The halogen displacement reaction shown by chlorine and potassium iodine could also be shown when we react with chlorine with sodium iodine or bromide. The only requirement is that the displaying halogen should be more reactive i.e., it must have high oxidizing power.
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