Question

Which of the following will turn colorless if treated with the appropriate reducing agent

(a) acidified potassium trioxoiodate V solution
(b) chlorine gas
(c) iodine solution
(d) acidified potassium heptaoxodichromate VI solution​

Answer 1

Answer:

ANSWER

Cl

2

is a stronger oxidising agent than I

2

, KI oxidises to I

2

which imparts brown colour to the solution.

2KI(aq)+Cl

2

(g)→2KCl(aq)+I

2

(brown color)

but excess of chlorine , the I₂ so formed gets further oxidised to HIO₃ (colourless)

5Cl

2

(g)+I

2

+6H

2

O→10HCl+2HIO

3

(colorless)

Hence, Addition of Cl

2

to KI solution gives it a brown colour but excess of chlorine turns into colourless

Answer 2

Explanation:

(c) Iodine solution

hope it helps you