[tesla]

I'm having a bit of trouble with balancing redox reactions, and I just want to confirm that I'm doing it the right way. 

MnO₄^- + H₂C₂O₄  Mn²⁺ + CO₂ + H₂O

I assigned oxidation numbers, and found that there is a transfer of 5e^-/Mn as well as 5e^-/MnO₄, and 1e^-/C and 4e^-/H₂C₂O₄  So when I balanced and added water and hydrogen ions I ended up with

22H⁺ + 4MnO₄^- + 5H₂C₂O₄  4Mn²⁺ + 10CO₂ + 16H₂O

Is that correct?

Thanks for looking

[Borek]

Too many electrons for oxalic acid, hence final answer is wrong.

[lillybeans]

Hi there, assuming this is in acidic conditions, then

MnO4- + H2C2O4 --> Mn2+ + CO2

Step 1: Identify atoms that change oxidation number; balance them.

MnO4 + H2C2O4 --> Mn2+ + 2CO2

Step 2: Recalculate the total number of electrons transferred and make them equal (multiply Mn by 2, C by 5 which I see you did)

2 MnO4- + 5 H2C2O4 --> 2 Mn2+ + 10 CO2

Step 3: Add H+ to balance the charge (Left: -2; Right: +4)

2 MnO4- + 5 H2C2O4 + 6 H+ --> 2Mn2+ + 10 CO2

Step 4: Add waters to balance the H

2 MnO4- + 5 H2C2O4 + 6 H+ ---> 2 Mn2+ + 10 CO2 + 8 H2O

Step 5: Check to make sure the charges on both sides are the same. (Left: +4, right: +4). Check for atoms to make sure they equal, which they do. voila!