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Need help using Ion Electron Method in acidic conditions.

 
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Oct26-11, 10:20 PM   #1
 

Need help using Ion Electron Method in acidic conditions.


Using the Ion Electron Method (acidic conditions) I need to solve:

MnO4- + C2O42- -> Mn2+ + CO2

So I tried doing:

MnO4- -> Mn2+

And:

C2O42- -> CO2 + 2H2O

Then I balanced the oxygen atoms with H2O

MnO4- -> Mn2+ + 4H2O

And:

C2O42- -> CO2 + 2H2O

Then I added the hydrogen atoms:

MnO4- + 8H+ -> Mn2+ + 4H2O

And:

C2O42- + 4H+ -> CO2 + 2H2O

Then I tried balancing the electrons and I got -3e for the first equation and -2e for the second. Then I tried combining the two:

2MnO4- + 16H+ +
3C2O42- + 12H+ -> 2Mn2+ + 8H2O + 3CO2 + 6H2O

Cancel out the H+ and H2O:

2MnO4- + 28H+ +
3C2O42- -> 2Mn2+ + 14H2O + 3CO2

I don't know if there is something else I should do (which I suspect there is but I can't tell), and I think I am doing something wrong because my answer didn't turn out right. Any help would be appreciated.
 
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Oct27-11, 02:45 AM   #2
 
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Quote by thyrgle View Post
C2O42- -> CO2 + 2H2O
You should balance carbon first. H2O - if needed - should appear in the next step.
 
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