Balancing Redox Chemical Equations
Consider the following unbalanced reaction under basic conditions:
MnO4- + N2H4 -> Mn2+ + NO3-
The general procedure is as such: [Bold printing will emphasize the part(s) of the equation that have been "changed" with each step.]
1. Separate the unbalanced equations into half-reactions.
MnO4- -> Mn2+
N2H4 -> NO3-
2. Balance the atoms (other than O and H) in each half reaction.
MnO4- -> Mn2+
N2H4 -> 2 NO3-
3. Balance the O's with H2O. For each O present on one side of the equation add 1 H2O to the other side.
MnO4- -> Mn2+ + 4 H2O
6 H2O + N2H4 -> 2 NO3-
4. Balance the H's with H+. For each H present on one side of the equation add 1 H+ to the other side.
8 H+ + MnO4- -> Mn2+ + 4 H2O
6 H2O + N2H4 -> 2 NO3- + 16 H+
5. Add as many OH- as there are H+ to both sides of the equation.
8 OH- + 8 H+ + MnO4- -> Mn2+ + 4 H2O + 8 OH-
16 OH- + 6 H2O + N2H4 -> 2 NO3- + 16 H+ + 16 OH-