- #1
- 9
- 0
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.
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.