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Question

$5$ moles of ferric oxalate are oxidised by how many number of moles of $K M n O_{4}$ in acid medium?
$6$
$4$
$2$
None of these

A

4

B

None of these

C

6

D

2

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Solution

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1 mole of ferric oxalate on dissociation gives 3 moles of oxalate ions.
$F e_{2} (C_{2} O_{4})_{3} ⇌ 2 F e^{3 +} + 3 C_{2} O_{4}^{2 -}$
The total number of electrons involved per mole oxidation of ferric oxalate is 6 moles.
The oxidation of oxalate ions to carbon dioxide involves 2 moles of electrons.
$3 C_{2} O_{4}^{2 -} \to 6 C O_{2} + 6 e^{-}$ ......(1)
In acidic medium, one mole of $K M n O_{4}$ requires 5 moles of electrons.
${M n O}_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$ ......(2)
To balance the number of electrons, multiply equation (1) with 5 and equation (2) with 6
$15 C_{2} O_{4}^{2 -} \to 30 C O_{2} + 30 e^{-}$ ......(3)
$6 {M n O}_{4}^{-} + 48 H^{+} + 30 e^{-} \to 6 M n^{2 +} + 24 H_{2} O$ ......(4)
Add equations (3) and (4)
$15 C_{2} O_{4}^{2 -} + 6 {M n O}_{4}^{-} + 48 H^{+} \to 30 C O_{2} + 6 M n^{2 +} + 24 H_{2} O$
Thus five mole of Ferric oxalate are oxidized by six moles of $K M n O_{4}$ in acid medium.

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