The Student Room Group

Oxidation of ethanedioate ions by acidified potassium manganate(VII)

Hi, I am really stuck on this question about the molar ratios in this question, in fact I am not sure where to start. All I have done is work out the oxidation numbers, anyway here it is:

In acid solution, potassium manganate(VII) acts as an oxidising agent
and reacts according to the half-equation:

5e− + MnO4− + 8H+ → Mn^2+ + 4H2O

Acidified potassium manganate(VII) will oxidise ethanedioate ions (C2O42&#8722:wink:
to carbon dioxide.

What is the mole ratio (MnO4− : C2O4^2&#8722:wink: in which these ions react together?

Can anyone offer any clues? Thank you!
try writing out the oxidation of C2O42− to CO2 with electrons first, then multiply both equations so that the electrons are the same...
Reply 2
oh thank you, I finally worked it out to be 2:5 I think..
escondido
oh thank you, I finally worked it out to be 2:5 I think..


Yup :wink:

2MnO4- + 16H+ + 5C2O42- --> 2Mn2+ + 8H2O + 10CO2
Reply 4
Urgh. Nasty IUPAC-ness. What is wrong with the name oxalate?
cpchem
Urgh. Nasty IUPAC-ness. What is wrong with the name oxalate?


Yeah I know, I was quite miffed when I got to university since they don't really use the IUPAC names for a lot of things...the trivial and IUPAC versions should both be taught and tested :yep:
Reply 6
do you know the order of reaction with regards to all the reactants? I'm pretty stuck and in desperate need of help :/
Original post by jimmcc
do you know the order of reaction with regards to all the reactants? I'm pretty stuck and in desperate need of help :/


This is quite an old thread. You'd be better off making a new one asking your question there :smile: