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Chemistry Redox Titration Question

An experiment was carried out to find the value of n in hydrated iron(II) ethanedioate, FeC2O4nH2O. 1.71 g of the compound was dissolved in deionised water and made up to 250 cm3 solution in a volumetric flask. When 25.0 cm3 of this solution was acidified using an excess of dilute sulfuric acid, it was found to react with 28.50 cm3 of 0.0200 mol dm-3 potassium manganate(VII).
a) Deduce the number of moles of MnO4– that react with one mole of FeC2O4nH2O. Remember that both the Fe2+ and C2O42– ions react.

The answer is FeC2O4nH2O : MnO4- = 1 : 0.6 but I don't see how.
Thank you for your help.
I know this is an old thread, but it may be helpful for other people. I've also put brackets in to show the charge and to avoid confusion.

So you know the reaction between (C2O4)2-) and (MnO4-) is:

16(H+) + 2(MnO4-) + 5(C2O4)2-) => 2(Mn)2+) + 10(CO2) + 8H20

And you know the reaction between (Fe)2+) and (MnO4-) is

8(H+) + (MnO4-) + 5(Fe)2+) => (Mn)2+) + 5(Fe)3+) + 4H20

You need Fe and C2O4 in a 1:1 ratio because of FeC2O4nH20 (Fe and C2O4 react in equal moles) and in the equations above, there are 5 moles of both, so they are already in a 1:1 ratio.

So you just have to add both equations, to get:

24(H+) + 3(MnO4-) + 5(C2O4)2-) + 5(Fe)2+) = 3(Mn)2+) + 10CO2 + 5(Fe)3+) + 12H2O

As 5 moles of (Fe)2+) and (C2O4)2-) react with 3 moles of (MnO4-), the ratio of (Fe)2+) & (C2O2)2-) to (MnO4-) is 5:3, which simplifies to 1 : 0.6
Reply 2
Original post by ootnootzoot
I know this is an old thread, but it may be helpful for other people. I've also put brackets in to show the charge and to avoid confusion.
So you know the reaction between (C2O4)2-) and (MnO4-) is:
16(H+) + 2(MnO4-) + 5(C2O4)2-) => 2(Mn)2+) + 10(CO2) + 8H20
And you know the reaction between (Fe)2+) and (MnO4-) is
8(H+) + (MnO4-) + 5(Fe)2+) => (Mn)2+) + 5(Fe)3+) + 4H20
You need Fe and C2O4 in a 1:1 ratio because of FeC2O4nH20 (Fe and C2O4 react in equal moles) and in the equations above, there are 5 moles of both, so they are already in a 1:1 ratio.
So you just have to add both equations, to get:
24(H+) + 3(MnO4-) + 5(C2O4)2-) + 5(Fe)2+) = 3(Mn)2+) + 10CO2 + 5(Fe)3+) + 12H2O
As 5 moles of (Fe)2+) and (C2O4)2-) react with 3 moles of (MnO4-), the ratio of (Fe)2+) & (C2O2)2-) to (MnO4-) is 5:3, which simplifies to 1 : 0.6

Legend thank you

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