1. Balancing Redox Reactions using the ½ Reaction Method

2. Half Reaction Half reaction – one of the two parts of a redox reaction
One half will be oxidation
One half will be reduction

3. Step to Balancing Half Reaction Method
Separate the reactions into 2 half reactions (one for oxidation & one for reduction)
Balance each ½ reaction in the following order:
Balance elements other than H & O
Balance O by adding H2O
Balance H by adding H+
Balance the charge to make each side equal

4. Step to Balancing Half Reaction Method
Multiply each ½ reaction by a number to make the number of electrons gained = the number of electrons lost
Add the ½ reactions
Simplify
Check your work
Tip – Break things into ions or binary compounds

5. Example Fe + CuSO4  Cu + Fe2(SO4)3 ( ) 3
Fe  2Fe +3
2Fe  2Fe +3
0  +6
0  (+6e-)
2Fe  2Fe +3 +6e-
Cu +2  Cu
+2  0
e-  0
Cu e-  Cu
3Cu e-  3Cu
( ) 3
2Fe + 3Cu e-  2Fe +3 +6e- + 3Cu x x
2Fe + 3Cu +2  2Fe Cu

6. Another Example AsO4 -3 + Zn  H3As + Zn +2 ( )4 Zn  Zn +2 0 +2
AsO4 -3  H3As
AsO4 -3  H3As + 4H2O
11 H+ +AsO4 -3  H3As + 4H2O
(+8 e-)
11 H+ +AsO e-  H3As + 4H2O
Zn  Zn +2
(+2 e-)
Zn  Zn e-
4Zn  4Zn e-
( )4
11H+ +AsO e- + 4Zn H3As + 4H2O + 4Zn e-
11H+ +AsO Zn H3As + 4H2O + 4Zn +2

7. Try This one… MnO4 -1 + C2O4 -2  Mn +2 + CO2
2MnO H++ 5C2O4-2  2Mn+2 +8H2O + 10CO2

8. Last one… KMnO4 + HCl  KCl + MnCl2 + H2O + Cl2
16 H+ + 2MnO Cl-  2Mn +2 +8H2O + 5Cl2