Balancing Redox Equations in Base
4) Add half-reactions and cancel substances appearing on both sides.
oxidation = 6 H2O + 3 C2O42- 6 CO32- + 12 H+ + 6 e-
reduction = 6 e- + 8 H+ + 2 MnO4- 2 MnO2 + 4 H2O
2 MnO4- + 2 H2O + 3 C2O42- 2 MnO2 + 6 CO32- + 4 H+
4a) Base: Add OH - to both sides of the equation to neutralize H+ .
4 OH - + 2 MnO4- + 2 H2O + 3 C2O42- _ 2 MnO2 + 6CO32- + 4H+ + 4 OH-
4 H2O
Subtract out the water that is on both sides:
4 OH - + 2 MnO4- + 3 C2O42- 2 MnO2 + 6 CO32- + 2 H2O
2 MnO4-(aq) + 3 C2O42-(aq) + 4 OH -(aq) _ 2 MnO2(s) + 6CO32-(aq) + 2 H2O(l)
5) Check: _
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