Balancing the reaction?

Question

Balance the follwing redox reaction in Basic solution:

MnO4^-1 + C2O4^-2 -----> MnO2 + CO2

Please use Half cell method and show step by step method plz plz plz

Answer 1

MnO4^-1 + 3 e- taken -----> MnO2   (Mn+7 reduces to Mn+4)

C2O4^-2 ----->  2CO2 & 2 e- lost    (carbon oxidizes from C+3 to C+4)

balance the electrons by doubling the Mn, & tripling the C2O4:

2MnO4^-1 + 6 e- taken -----> 2MnO2  

3 C2O4^-2 ----->  6CO2 &   6 e- lost  

combine them:

2MnO4^-1 &  3 C2O4^-2    -----> 2MnO2  & 6CO2

we will balance the charges by adding OH- 's,... right now the left adds up to -8 , while the right adds up to zero:

2MnO4^-1 &  3 C2O4^-2    -----> 2MnO2  & 6CO2  & 8 OH-

now we will add neutral water to balance the H's:

2MnO4^-1 &  3 C2O4^-2  & 4 H2O   -----> 2MnO2  & 6CO2  & 8 OH-

last, the oxygens should already be balanced , as a check on your work:

24 oxygens --->  24 oxygens

it checks out,

yourt answer is:

2MnO4^-1 &  3 C2O4^-2  & 4 H2O---> 2MnO2  & 6CO2  & 8 OH-