Balancing Redox Equations in Base
1) Divide into two half-reactions: (oxidation)
C2O42- CO32-
2) Balance
a) Atoms other than O and H
b) O atoms with water
c) H atoms with H+
d) Charge with e-
2 H2O + C2O42- 2 CO32- + 4 H+
2 H2O + C2O42- 2 CO32- + 4 H+ + 2 e-
3) Multiply each half-reaction to equalize the electrons.
oxidation = 2e-, reduction = 3e-, therefore (LCD = 6e-)
ox = 6 H2O + 3 C2O42- 6 CO32- + 12 H+ + 6 e-
red = 6 e- + 8 H+ + 2 MnO4- 2 MnO2 + 4 H2O
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