SCH 4U1
Balancing Redox Equations:
The Oxidation Number Method
When given a molecular equation one method of balancing is using
oxidation numbers.
K2Cr2O7 + SnCl2 + HCl  CrCl3 + SnCl4 + KCl + H2O
This equation can be simplified if you can locate the elements that change
oxidation number. If this is difficult – assign oxidation numbers to all of
them.
6+
2+
3+
4+
K2Cr2O7 + SnCl2 + HCl  CrCl3 + SnCl4 + KCl + H2O
1) Assign oxidation numbers (it is clear that Cr6+ is reduced and Sn2+ is
oxidized).
2) Find the change in oxidation numbers and determine the number of
electrons transferred in each case. (In the case of Cr you must double
the number of electrons because there are two Cr atoms in the
reactants).
____reduction of 6e-_____
|
|
6+
2+
3+
4+
K2Cr2O7 + SnCl2 + HCl  2 CrCl3 + SnCl4 + KCl + H2O
|____________________|
oxidation of 2e3) Determine the coefficients used to bring the number of electrons in
each reaction to the same number.
1 K2Cr2O7 + 3 SnCl2 + HCl  2 CrCl3 + 3 SnCl4 + KCl + H2O
4) Balance the rest by inspection.
1 K2Cr2O7 + 3 SnCl2 + _ HCl  2 CrCl3 + 3 SnCl4 + _ KCl + H2O