Acidified ${K M n O}_{4}$ can be decolourised by ${F e}^{2 +} (o r C_{2} O_{4}^{2 -}, {S O}_{2} e t c .)$ .

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Solution

Acidified ${K M n O}_{4}$ can be decolourised by ${F e}^{2 +} (o r C_{2} O_{4}^{2 -}, {S O}_{2} e t c .)$
Potassium permanganate is a strong oxidizing agent and it oxidizes these ions/molecules and itself is reduced.. This results in decolourization.
Ferrous ion is oxidized to ferric ion and oxalate ion is oxidized to carbon dioxide.
Hence, the cation and the anion in the salt ferrous oxalate decolourizes permanganate ion in acidic medium.
$2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O$
$M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to 5 F e^{3 +} + M n^{2 +} + 4 H_{2} O$
${S O}_{2}$ can decolourises ${M n O}_{4}^{-} / H^{+}$ solution and turn ${C r}_{2} O_{7}^{2 -} / H^{+}$ solution green.
Sulphur dioxide reduces potassium permanganate to Mn(Ii). ion. Hence, potassium permanganate is decolourized.

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