Number of moles of KMnO₄ required to oxidize one mole of Fe(C₂O₄) in acidic medium is

Solution:

Total number of electrons involved per mole oxidation of Fe(C₂​O₄)​ to Fe³⁺ and CO₂​ is 3 moles.
The oxidation of ferrous ions to ferric ions involves 1 mole of electrons.
Fe²⁺→Fe³⁺+e⁻
The oxidation of oxalate ions to carbon dioxide involves 2 moles of electrons.
${\mathrm{C}}_2{\mathrm{O}}_4^{2-}$​→2CO₂​+2e⁻
Thus,
Fe²⁺+${\mathrm{C}}_2{\mathrm{O}}_4^{2-}$​→Fe³⁺+2CO₂​+3e⁻......(1)
In acidic medium, one mole of KMnO₄​ requires 5 moles of electrons.
${\mathrm{MnO}}_4^{-}$​+8H⁺+5e⁻→Mn²⁺+4H₂​O......(2)
To balance the number of electrons, multiply equation (1) with 5 and equation (2) with 3.
5Fe²⁺+5${\mathrm{C}}_2{\mathrm{O}}_4^{2-}$​→5Fe³⁺+10CO₂​+15e⁻......(3)
$3{\mathrm{MnO}}_4^{-}$​+24H⁺+15e⁻→3Mn²⁺+12H₂​O......(4)
Add equations (3) and (4)
5Fe²⁺+$5{\mathrm{C}}_2{\mathrm{O}}_4^{2-}$​+$3{\mathrm{MnO}}_4^{-}$​+24H⁺→5Fe³⁺+10CO₂​+3Mn²⁺+12H₂​O
Thus, five mole of Ferrous oxalate are oxidized by three moles of KMnO4​ in acidic medium.
The number of moles of KMnO₄​ required to oxidize 1 mole of Fe(C₂​O₄​) is $\frac{3}{5}=0.6.$