Balancing Redox Equations

FOOL-PROOF Method Example
1. Assign oxidation numbers (o.n.) to atoms and write a net ionic equation for each half reaction. Work on each half reaction separately.

2. Balance atoms being oxidized or reduced.

3. From the change in oxidation numbers, determine the number of electrons involved and add them to the high oxidation number side of the equation.

4. Balance the charge, using either H+ (if in ACID solution) or OH- (if in BASE solution).

5. Now balance the OXYGEN by adding H2O to the appropriate side of the equation.

Check that everything balances - it will!

6. Equalize the number of electrons in each half reaction by multiplying by the appropriate factor and then add.

In acid solution, potassium permanganate can oxidize chloride ions to chlorine gas. The manganese-containing product in this reaction is
Mn2+(aq). Balance the equation for this reaction.

1.
 Mno.n. = +7 Mno.n. = +2 MnO4-(aq) -----> Mn2+(aq) Clo.n. = -1 Clo.n. = 0 Cl-(aq) -----> Cl2(aq)

2, 3.
MnO4-(aq) -----> Mn2+(aq)
The change in oxidation number is 5 and the high oxidation number side of the equation is the left side:
MnO4-(aq) + 5e- -----> Mn2+(aq)

2Cl-(aq) -----> Cl2(aq)
The change in oxidation number is 1. Because two atoms are involved, the change is 2 and the high oxidation number side of the equation is the right side:
2Cl-(aq) -----> Cl2(aq) + 2e- This half reaction is now balanced.

4.
MnO4-(aq) + 5e- + 8H+ -----> Mn2+(aq)
5.
MnO4-(aq) + 5e- + 8H+ -----> Mn2+(aq) + 4H2O(l) This half reaction is now balanced.

6.
Equalize the number of electrons in each half reaction and add:
[MnO4-(aq) + 5e- + 8H+ -----> Mn2+(aq) + 4H2O(l)] x 2
[2Cl-(aq) -----> Cl2(aq) + 2e-] x 5
The balanced equation is

 2MnO4-(aq) + 10Cl-(aq) + 16H+(aq) ----> 2Mn2+(aq) + 5Cl2(aq) + 8H2O(l)