Question

A dilute ferrous sulphate solution was gradually added to the beaker containing acidified potassium permanganate solution. The light purple colour of the solution fades and finally disappears. Which of the following is the correct explanation for the observation?

• A. $KMn{O}_4$ is an oxidising agent, it oxidises $FeS{O}_4$
• B. $FeS{O}_4$ acts as an oxidising agent and oxidises $KMn{O}_4$
• C. The colour disappears due to dilution; no reaction is involved
• D. $KMn{O}_4$ is an unstable compound and decomposes in presence of $FeS{O}_4$ to a colourless compound

Answer 1

The correct option is A $KMn{O}_4$ is an oxidising agent, it oxidises $FeS{O}_4$

On addition of dilute $FeS{O}_4$ to purple coloured solution of acidified $KMn{O}_4$, the colour of the latter changes as ${Mn{O}_4}^{-}$ ion is reduced to$M{n}^{2+}$ ion, thereby acting as an oxidising agent. This can be explained as :

Potassium permanganate in the presence of an acid acts as a strong oxidising agent because of the following reaction.
$2KMn{O}_4+3H_{2}S{O}_4\to K_{2}S{O}_4+2MnS{O}_4+3H_{2}O+5O$ ..........(I)
Oxygen atom causes the oxidation of ferrous sulphate to ferric sulphate as:
[$2FeS{O}_4+H_{2}S{O}_4+O\to F{e}_2(S{O}_4{)}_3+H_{2}O$] $\times$5...........(II)
Multiplying (II) by 5 and adding to (I), the overall reaction comes out to be:
$2KMn{O}_4+8H_{2}S{O}_4+10FeS{O}_4\to K_{2}S{O}_4+2MnS{O}_4+5F{e}_2(S{O}_4{)}_3+8H_{2}O$