Download Redox Notes #3 - Balancing

Balancing Redox Equations
Consider the equation:
MnO41- + H2S ➩ S + Mn2+
Redox reactions must be balanced for not only mass, but charge as well. This is
because electron loss must equal electrons gain.
There are two ways to balance redox reactions:
1) using oxidation numbers
2) using half reactions
We will first learn how to balance in acidic aqueous solutions. Then we will learn
about basic and non-aqueous solutions.
The Oxidation Number Method
1. Assign oxidation numbers to all atoms.
2. Determine which atoms change numbers as you move from reactants to
products. (one will be oxidized, the other reduced)
3. Use coefficients to make electrons lost = electrons gained.
4. If necessary, add water (H2O) to balance for oxygen and H+ to balance
for hydrogen.
5. Balance any remaining elements by inspection.
6. Check charge to be sure both sides are equal.
MnO41- + H2S ➩ S + Mn2+
Try the following:
a) HClO2 + I1- ➩ Cl2 + HIO
b) P4 + IO31- ➩ H2PO41- + I1c) MnO41- + SO32- ➩ MnO2 + SO42-
The Half Reaction Method
1. Divide the equation into two half reactions.
2. Balance the first half reaction using the following steps:
a) balance the common element by inspection
b) add H2O to balance oxygen
c) add H+ to balance hydrogen
d) add electrons to the most positive side to balance charge
3. Repeat the process for the other half reaction.
4. Multiply each half reaction by coefficients to make electrons lost = electrons
gained.
5. Add the half reactions together.
6. Simplify if necessary.
7. Check charge to be sure both sides are equal.
MnO41- + H2S ➩ S + Mn2+
2( 5e- + 8H+ + MnO41- ➩ Mn2+ + 4H2O)
5( H2S ➩ S + 2H+ + 2e- )
10e- + 16H+ + 2MnO41- + 5H2S ➩ 2Mn2+ + 5S + 10H+ + 8H2O + 10e6H+ + 2MnO41- + 5H2S ➩ 2Mn2+ + 5S + 8H2O
Try the following:
a) HClO2 + I1- ➩ Cl2 + HIO
b) P4 + IO31- ➩ H2PO41- + I1c) MnO41- + SO32- ➩ MnO2 + SO42-