Respuesta :

Dreil40
Assuming the manganese IV oxide is being reduced to a manganese II ion, the answer is 10 moles of electrons.

Oxidation: 5[ C2O4(2-) -> 2 CO2 + 2e- ]
Reduction: 2[ MnO4 + 5e- + 8H+ -> Mn(2+)
+ 4H2O ]