The Half-Reaction Method for Balancing Equations for Oxidation-Reduction
Reactions Occurring in an Acidic Solution
Write the equations for the oxidation and reduction half-reactions.
For each half-reaction:
1. Balance all of the elements except hydrogen and oxygen.
2. Balance oxygen using H2O
3. Balance hydrogen using H+
4. Balance the charge using electrons
5. Cancel like-species and combine remaining species
Example:
MnO4-(aq) + Os4+(aq)  OsO4(s) + MnO2 (s)
Write each half-reaction
MnO4- 
Os4+ 
MnO2
OsO4
Step 1. Balance all the elements except H and O
MnO4- 
Os4+ 
MnO2
OsO4
Since there are the same number of Mn and Os on each side of the reaction, we
skip this step.
Step 2. Balance oxygen using H2O---we have to understand that we are adding a
new molecule “out of no-where,” but we have to remember that these reactions
are done in water, and water can be part the reaction, it must be under these
conditions.
MnO4- 
MnO2 +
2H2O
We add 2@H2O’s to the product side so we balance the O’s on each side of the
reactions.
4H2O + Os4+ 
OsO4
We do the same for the other half reaction.
Step 3. Balance hydrogen by using H+
4H+ + MnO4- 
MnO2 + 2H2O
We add 4H+’s to the reactant side to give 4@H’s on each side.
4H2O + Os4+ 
OsO4 + 8H+
Add 8H+’s to the Os half-reaction
At this point ALL atoms are balanced.
Step 4. Balance the charge using electrons

4H+ + MnO4-
4+
+
MnO2 +
1-
0
2H2O
+
= 3+
0
=0
We balance the charges of each side of the reaction using electrons. In the above
case we do the following:
3e- + 4H+ + MnO4-
3- +
4+
+

MnO2 +
1-
0
= 0
2H2O
+
0
=0
Now the charge is balanced on both sides of the reaction. We know look at the
other half-reaction.

4H2O + Os4+
0
+
8H+
OsO4 +
4+
0
+
= 4+
8+
= 8+
Balance the charges using electrons.
4H2O + Os4+
0
+
4+
= 4+

OsO4 +
0
+
8H+ + 4e8+ + 4= 4+
We are not quit done with balancing the charges. Don’t forget that this is a Redox
reaction and we need the same number of electrons for each oxidation and
reduction reaction.
4[3e- + 4H+ + MnO43[4H2O + Os4+

MnO2 +
 OsO4 +
2H2O]
8H+ + 4e-]
This goes to:
12e- + 16H+ + 4MnO412H2O + 3Os4+

4MnO2 +
 3OsO4 +
8H2O
24H+ + 12e-
We now have the same number of electrons leaving and going into this Redox
reaction.
Step 5. Cancel like-species and add the two half reactions
12e- + 16H+ + 4MnO44H2O
12H2O + 3Os4+
4H2O + 3Os4+ + 4MnO4-

4MnO2 +
 3OsO4 +

4MnO2 +
8H2O
8H+
24H+ + 12e3OsO4 + 8H+
Check to make sure all elements balance on each side of the reaction.
The Half-Reaction Method for Balancing Equations for Oxidation-Reduction
Reactions Occurring a Basic Solution
Follow the same steps above for balancing in an acidic solution, there is an extra
step at the very end.
Example:
Fe(OH)2 + CrO4-2
 Fe2O3
Fe(OH)2 
CrO4-2

+
Cr(OH)4-
Fe2O3
Cr(OH)4-
Step 1.
2Fe(OH)2  Fe2O3
CrO4-2

Cr(OH)4-
Step 2.
2Fe(OH)2  Fe2O3 + H2O
CrO4-2

Cr(OH)4-
Step 3.
2Fe(OH)2  Fe2O3 + H2O + 2H+
4H+ + CrO4-2
 Cr(OH)4-
Step 4.
2Fe(OH)2  Fe2O3 + H2O + 2H+
0
=0
0
+ 0
= +2
+ 2+
Add e2Fe(OH)2  Fe2O3 + H2O + 2H+ + 2e-
=0
=0
4H+
+ CrO4-2
 Cr(OH)4-
4+ + 2= 2+
3e- + 4H+
3-
+ 4+
= 1-
3[2Fe(OH)2 
+ CrO4-2
1= 1
+ 2-
Cr(OH)4-
1= 1-
Fe2O3 + H2O + 2H+ + 2e-]
2[3e- + 4H+ + CrO4-2

Cr(OH)4-]
This goes to:
6Fe(OH)2  3Fe2O3 + 3H2O + 6H+ + 6e6e- + 8H+ + 2CrO4-2

2Cr(OH)4-
Step 5.
6Fe(OH)2  3Fe2O3 + 3H2O + 6H+ + 6e2H+
6e- + 8H+ + 2CrO4-2
2H+ + 6Fe(OH)2 + 2CrO4-2

 2Cr(OH)43Fe2O3 + 3H2O + 2Cr(OH)4-
Step 6 (new for balancing in a basic solution)
Take the original reaction, to the side with H+, and add the same number of OH-‘s,
then add the same number of OH- to the other side of the reaction:
2H+ + 6Fe(OH)2 + 2CrO4-2
+ 2OH-
 3Fe2O3 + 3H2O + 2Cr(OH)4+ 2OH-
When we do this we now see, in this case 2H+ and 2OH- which we know forms
H2O, so rewrite the equation:
2H2O + 6Fe(OH)2 + 2CrO4-2  3Fe2O3 + 3H2O + 2Cr(OH)4- + 2OHCancel waters that appear on both sides of the reaction:
2H2O + 6Fe(OH)2 + 2CrO4
-2
H 2O
 3Fe2O3 + 3H2O + 2Cr(OH)4- + 2OH-
Then write the finished reaction:
6Fe(OH)2 + 2CrO4-2  3Fe2O3 + H2O + 2Cr(OH)4- + 2OH-
Check that you have the same number of atoms on each side.