anactualmess
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Iron(II) ethanedioate dihydrate (FeC2O4.2H2O) can be analysed by titration using potassium manganate(VII) in acidic solution. In this reaction, manganate(VII) ions oxidise iron(II) ions and ethanedioate ions.
Use the half-equations given below to calculate the reacting ratio of moles of manganate(VII) ions to moles of iron(II) ethanedioate.

MnO4– +8H+ + 5e– ----> Mn2+ + 4H2O
Fe2+ ----> Fe3+ + e-
C2O42- ----> 2CO2 + 2 e-

I have only come across the first two equations in this scenario and know the ratio would be 1:5 if considering these two equations alone, however am unsure what to do with the involvement of the third ratio?
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charco
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(Original post by anactualmess)
Iron(II) ethanedioate dihydrate (FeC2O4.2H2O) can be analysed by titration using potassium manganate(VII) in acidic solution. In this reaction, manganate(VII) ions oxidise iron(II) ions and ethanedioate ions.
Use the half-equations given below to calculate the reacting ratio of moles of manganate(VII) ions to moles of iron(II) ethanedioate.

MnO4– +8H+ + 5e– ----> Mn2+ + 4H2O
Fe2+ ----> Fe3+ + e-
C2O42- ----> 2CO2 + 2 e-

I have only come across the first two equations in this scenario and know the ratio would be 1:5 if considering these two equations alone, however am unsure what to do with the involvement of the third ratio?
You just treat the ethandioate ion as an "extra" redox and add it to the other one.

You can see that the oxidation releases 2 electrons and the oxidising agent requires 5 electrons, hence 2 manganate(VII) ions react with 5 ethandioate ions.

Now you add those to the 1 manganate ion that reacts with 5 iron(II) ions.

Total: 3 manganate ions react with 5 iron(II) ethandioate formula units.
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