Redox Reaction Balance Method

Ion-Electron Method in Acid Solution
1)  Identify the oxidation numbers of all oxidants and reductants
2)  Identify the Oxidants and Reductants
3) Write down the half reactions
4) Balance atoms other than H and O
5) Add H2O to balance O and Add H+ to balance H
6) Balance the number of electron on the half reactions by taking common factor
7) Add up the half reactions
8) Combine and Cancel similar terms
9) Check balance on atom and charges

Example 
Balance MnO4- + H2SO3--> SO42- + Mn2+
1) Oxidation # MnO4- = +7, S in H2SO3 = +4, S in SO42- = +6, Mn2+ = +2
2) Oxidant: MnO4- and Reductant  H2SO3 +
3)  MnO4- +5e -->  Mn2+
            H2SO3 --> SO42- +2e
4) No need to balance other atoms in this case
5) 8H+ + MnO4- +5e -->  Mn2+ + 4H2O
           H2O +  H2SO3 --> SO42- +2e + 4H+
6) (8H+ + MnO4- +5e -->  Mn2+ + 4H2O) x2
           (H2O +  H2SO3 --> SO42- +2e + 4H+) x5
7) 16H+ + 2MnO4- +10e +5H2O +5H2SO3 --> 2Mn2+ + 8H2O + 5SO42- + 10e +20H+
(Cancel and Combine Highlighted similar terms)
8) 2MnO4- +5H2SO3 --> 2Mn2+ + 3H2O + 5SO42- + 4H+
9) Left hand side: 2Mn. 23O, 10H , net charge=-2
    Right hand side: 2Mn. 23O, 10H, net charge= +4 -10 +4= -2

Ion-Electron Method in Basic Solution
1)-9) the same as Acid Solution
10) Add OH- on both sides base on the H+ amount
11) Combine H+ and OH- into water/ Combine water molecule if possible
Example
Using same balanced equation
10) 4OH-+ 2MnO4- +5H2SO3 --> 2Mn2+ + 3H2O + 5SO42- + 4H+ + 4OH-
11) 4OH-+ 2MnO4- +5H2SO3 --> 2Mn2+ + 3H2O + 5SO42- + 4H2O (Combine water molecules)
       => 4OH-+ 2MnO4- +5H2SO3 --> 2Mn2+ + 7H2O + 5SO42-