We can also use Smart African Orangutans to balance redox reactions. I consider it a better method because it
is almost fool proof, gives you balanced half reactions before you get a net reaction and does not involve using
oxidation numbers at all.
“Smart African Orangutans Have Enormous Brain Cells”:
Smart:
Split the
African:
Balance All
Orangutans:
Balance O
reaction into except H and using H2O:
two:
O:
Have:
Enormous:
Brain:*
Cells:
Balance H
Add electrons
If basic, add
Combine
using H+:
to balance
OH- for
the
charge:
every H+
reactions
to cancel
electrons:
*Basic reactions are not tested on diploma and will not be covered in our diploma review.
Typically, I have memorized the mnemonic and write the letters SAOHEBC and cross them off as I go along:
NH3 + MnO4-
 MnO2 + NO3-
Reduction: 3e- + 4H+ + MnO4-  MnO2 + 2H2O
Oxidation: 3H2O + NH3  NO3- + 9H+ + 8e*We did not have to balance Mn or N as they were already balanced, we did nothing for step A.
*Notice that the electrons gained in the reduction equal that which we found through the “Oxidation Number
Method”. The same is true of the electrons lost in the oxidation reaction.
Combine to cancel electrons:
Reduction: 8 × (3e- + 4H+ + MnO4-  MnO2 + 2H2O)
Oxidation: 3 × (3H2O + NH3  NO3- + 9H+ + 8e-)
Net: 5H+ + 3NH3 + 8MnO4-  8MnO2 + 3NO3- + 7H2O
*We had 9 H2 O on the left side and 16 H2 O on the right side. Like in Hess’s Law, H2 O cancelled and we were
left with 7 H2 O on the right side. The same applies to cancelling H + to get 5 H + on the left side.
© Pavel Sedach 2015
Let’s do a couple of examples:
Example 1: The following is a disproportionation reaction. In a disproportionation reaction the same element
that is oxidized is also reduced.
Br2(l)  BrO3-(aq) + Br-(aq)
Oxidation Number 0
5+
1Br2(l)  BrO3-(aq) + Br-(aq)
Let’s walk through this reaction
Smart: Split the reaction into two and African and Balance All except H and O:
Oxidation: Br2(l)  2BrO3-(aq)
Reduction: Br2(l)  2Br-(aq)
Orangutans: Balance O using H2O and Have and Balance H using H+:
Oxidation: 6H2O + Br2(l)  2BrO3-(aq) + 12H+
Reduction: Br2(l)  2Br-(aq)
Enormous: Add e- to balance charge and Cells: Combine the reactions to cancel e-:
Oxidation:
Reduction:
Net:
6H2O + Br2(l)  2BrO3-(aq) + 12H+ + 10e5 × (2e- + Br2(l)  2Br-(aq))
6H2O + Br2(l) + 5Br2(l)  2BrO3-(aq) + 12H+ + 10e- + 10Br-(aq)
Example 2: Balance the following in acidic condition
Cr2O72-(aq) + Br-(aq)  Cr3+(aq) + BrO-(aq)
© Pavel Sedach 2015
15
*All examples are to be balanced in acidic solution as that is what the curriculum asks for.
*In H2 O2(aq) , Oxygen has an oxidation state of -1.
+
*Acid solution means using H(aq)
(the same way as we’ve done it all day)
−
+
2+
3+
30. MnO4(aq)
+ Fe2+
(aq) + H(aq) ⇌ Mn(aq) + Fe(aq) + H2 O(l)
+
31. HNO3(aq) + Cu(s) + H(aq)
⇌ NO2(g) + Cu2+
(aq) + H2 O(l)
15
16
−
2+
32. MnO−
4(aq) + Cl(aq) ⇌ Mn(aq) + Cl2(g)
3+
−
33. Cr2 O2−
7(aq) + Br(aq) ⇌ Cr(aq) + BrO(aq)
16
17
34. H2 O2(aq) + Cl2 O7(aq) ⇌ ClO(aq) + O2(g)
35. BrO−
(aq) ⇌ BrO3(aq) + Br2(aq)
17
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(C) Pavel Sedach Learnfaster.ca
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(C) Pavel Sedach Learnfaster.ca
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