Question

Which one of the following compounds does not decolourise an acidified aqueous solution of $KMn{O}_4$?

• A. Ferrous sulphate
• B. Sulphur dioxide
• C. Ferric chloride
• D. Hydrogen peroxide

Answer 1

The correct option is C Ferric chloride

Analysing the options:
Option (A), (C) & (D):
$S{O}_2,H_2{O}_2$ and $FeS{O}_4$ can act as reducing agents since they can be further oxidised to a higher oxidation state. Hence, they can reduce $Mn{O}_4^{-}$ (pink) to $M{n}^{2+}$ (almost colourless) in acidic medium.
The reactions of $S{O}_2$ , $H_2{O}_2$ and $FeS{O}_4$ with acidified $KMn{O}_4$ solution are as follows:

$2KMn{O}_4+5S{O}_2+2H_{2}O$
$\downarrow$
$K_{2}S{O}_4+2MnS{O}_4+2H_{2}S{O}_4$
$2KMn{O}_4+5H_2{O}_2+3H_{2}S{O}_4$
$\downarrow$
$K_{2}S{O}_4+2MnS{O}_4+5O_2+8H_{2}O$
$2KMn{O}_4+10FeS{O}_4+8H_{2}S{O}_4$
$\downarrow$
$5F{e}_2(S{O}_4{)}_3+K_{2}S{O}_4+2MnS{O}_4+8H_{2}O$

Option (B):
$Fe\left(III\right)$ is in its highest oxidation state in $FeC{l}_3$. Therefore, it cannot act as a reducing agent. Hence, it does not decolourise an acidified aqueous solution of $KMn{O}_4$.
So, option (B) is the correct answer.