Iron oxalate titration equation

Hey everyone, I have a prac sheet and need some help combining three half equations to get a 6:5 ratio.
MnO₄- (aq) + 8H+(aq) + 5e- ------> Mn²⁺(aq) + 4H₂O(l)
C₂O₄^2-(aq)------> 2CO₂(g) +e^-
Fe²⁺(aq)-------> Fe³⁺(aq) + e^-

"when these half equations are combined to give an overall equation for the reaction of MnO₄^-ions with Fe(C₂O₄)₂, it can be shown that 6 moles of MnO₄^-react with 5 moles of Fe₂(C₂O₄)₂ ."
I'm really stuck with this. Can someone please help?

Original post by MissDW

Hey everyone, I have a prac sheet and need some help combining three half equations to get a 6:5 ratio.
MnO₄- (aq) + 8H+(aq) + 5e- ------> Mn²⁺(aq) + 4H₂O(l)
C₂O₄^2-(aq)------> 2CO₂(g) +e^-
Fe²⁺(aq)-------> Fe³⁺(aq) + e^-

"when these half equations are combined to give an overall equation for the reaction of MnO₄^-ions with Fe(C₂O₄)₂, it can be shown that 6 moles of MnO₄^-react with 5 moles of Fe₂(C₂O₄)₂ ."
I'm really stuck with this. Can someone please help?

You know that the iron and the oxalate ions are in a fixed ratio given by the formula of iron oxalate (1:1).

Hence the iron (changing from 2+ to 3+) and the oxalate ions (losing two electrons) overall lose 3 electrons per empirical formula unit, or 6 electrons per formula unit.

Then perform the usual redox combination with the permanganate.

Reply 2

dynamicmind

5

Original post by MissDW

Hey everyone, I have a prac sheet and need some help combining three half equations to get a 6:5 ratio.
MnO₄- (aq) + 8H+(aq) + 5e- ------> Mn²⁺(aq) + 4H₂O(l)
C₂O₄^2-(aq)------> 2CO₂(g) +e^-
Fe²⁺(aq)-------> Fe³⁺(aq) + e^-

"when these half equations are combined to give an overall equation for the reaction of MnO₄^-ions with Fe(C₂O₄)₂, it can be shown that 6 moles of MnO₄^-react with 5 moles of Fe₂(C₂O₄)₂ ."
I'm really stuck with this. Can someone please help?