Transcript Slide 1

Balancing Redox Equations
It is not always possible to balance a
reaction equation by trial and error.
The following method for balancing redox
equations is commonly called the ionelectron half-equation method.
Step One: (half-equation method)
Identify the oxidation and reduction reactions and the
appropriate reactant and product in each case.
For the reduction reaction the species reduced will be the
reactant with an oxidation number that decreases. The
opposite is true for the oxidation step
e.g. when a solution of potassium dichromate reacts with
iron II nitrate the species oxidised is Fe2+ and the
species reduced is Cr2O72.
Fe2+  Fe3+
Cr2O72  Cr3+
Step 2 Balance all atoms undergoing a
change in oxidation number.
Fe2+  Fe3+
Cr2O72  2Cr3+
Step 3 Balance the number of O atoms by
adding the appropriate number of water
molecules.
Fe2+  Fe3+
Cr2O72  2Cr3++ 7H2O
Step 4 Balance the H atoms by adding H+ ions.
Fe2+  Fe3+
Cr2O72 +14H+ 2Cr3++ 7H2O
Step 5 Balance the charge by adding electrons, e. This gives 2 balanced half-equations.
Fe2+  Fe3++eCr2O72 + 14H++ 6e- 2Cr3++ 7H2O
Note: In the oxidation half-equation the Fe2+ loses
electrons and in the reduction half-equation the
Cr2O72 gains electrons
Step 6
To obtain an overall balanced equation the 2 half equations must be
added together.
Before doing this the equations have to be multiplied so that the
number of electrons in each half-equation is the same.
In this way, the electrons will be eliminated in the final equation.
Fe2+  Fe3++e- (x 6)
6Fe2+  6Fe3++ 6eAdded to:
Cr2O72 + 14H++ 6e- 2Cr3++ 7H2O
gives the equation: - Cancel out items on both sides
6Fe2+ + Cr2O72 + 14H++ 6e- 2Cr3++ 7H2O + 6Fe3++ 6e-
Giving the final equation:
6Fe2+ + Cr2O72 + 14H+ 2Cr3++ 7H2O + 6Fe3+
Finally check that the equation is balanced, particularly for charge!!
Balance the following equations:
1.MnO4 + SO2 
Mn2+ + SO42
2H2O + 5SO2 + 2MnO4  4H+ + 5SO42 + 2Mn2+
2.S2O32 + I2

2S2O32 + I2
I +

S4O62
2I + S4O62
Balance the following equations:
1.MnO4 + SO2+  Mn2+ + SO42
5e- 2Mn2+ + 8
4H2O
(X2) 2 MnO4 +
168H+ +10
2+ + 10
2e2H2O  5SO42 + 204H+ + 10
(X5) 5 SO
2MnO+4- 16H++5SO2+
2
+10H2O+ 10e-
4
2+ 20H+
2Mn2++ 8H2O+ 10e+5SO
4
2MnO4-+5SO2+ +2H2O
2Mn2++ 5SO42 + 4H+