[unleash10]

Here is the overal reaction:
H2C2O4 + AuCl4- --------> H2CO2 + Au + Cl2
So the reduction part is:
H2C204----->2H2CO2; Where carbon is being reduced
Im guessing the oxidation part is
AuCl4-----> Au + 2Cl2
The question i have is Au is being reduced by 3e- but Cl is being oxidized by 4e-. Should i put only 4e- on the right hand, or should i put both 3e- on the left and 4e- on the right and have it cancel out to become 1e- on the right hand?

[Albert]

Try writing the three half-reactions.

[AWK]

I doubt if this reaction can proceed this way
3H2C2O4 + 2AuCl4- + 2H+--------> 3CO2 + 2Au + 8HCl
is more reliable