1. half reaction method for balancing in an acid or base
electrochem part 2
half reaction method for balancing in an acid or base

2. MnO4-(aq) + C2O42-(aq)  Mn2+(aq) + CO2(g) (acid)
Half-Reaction Method :
MnO4-(aq) + C2O42-(aq)  Mn2+(aq) + CO2(g) (acid)

3. Half-Reaction Method (in acid):
MnO4-(aq) + C2O42-(aq)  Mn2+(aq) + CO2(g)
Step 1: Divide into two half-reactions,
MnO4-  Mn2+
C2O42-  CO2
Step 2: Balance main element
MnO4-  Mn2+
C2O42-  2CO2

4. MnO4-  Mn2+ + 4H2O C2O42-  2CO2 MnO4-  Mn2+ + 4H2O 8H+ +
Step 3: Balance the O atoms by adding H2O : :
MnO4-  Mn2+
+ 4H2O
C2O42-  2CO2
Step 4: Balance the H atoms by adding H+
8H+ +
MnO4-  Mn2+ + 4H2O
C2O42-  2CO2
Step 5: Balance charge with electrons +7 +2
5e- +
8H++ MnO4-  Mn2+ + 4H2O -2
C2O42-  2CO2
+ 2e-

5. 5e- + 8H++ MnO4-  Mn2+ + 4H2O C2O42-  2CO2 + 2e-
Step 6: Make the electrons balance :
5e- + 8H++ MnO4-  Mn2+ + 4H2O
C2O42-  2CO2 + 2e-
Step 7: Cancel and add
10e H++ 2MnO4-  2Mn2+ + 8H2O +
5C2O42-  10CO2 +10e-
16H+ + 2MnO4- + 5C2O42-  2Mn CO2 + 8H2O

6. CN- + MnO4-  CNO- + MnO2 -1 +1 H2O + CN-  CNO- + 2H+ + 2e- 3e-+
Balancing in base
CN- + MnO4-  CNO- + MnO2 -1 +1 3 3 6 3 6
H2O +
CN-  CNO-
+ 2H+
+ 2e- 6 2 8 2 4
3e-+
4H+ +
MnO4-  MnO2
+ 2H2O +3
2H+ + 3CN- + 2MnO4-  3CNO- + 2MnO2 + H2O
+ 2OH-
2OH-
2H2O 1

7. Cr(OH)3 + ClO  CrO4-2 + Cl2 Have at it Balancing in base
4Cr(OH)3 + 6ClO + 8OH- 4CrO Cl2 + 10H2O