Download How To Balance Redox Equations

How To Balance Redox Equations
Problem:
Redox equations are often more complex and harder to balance than simple
double replacement or acid-base equations. They require a step-by-step
procedure.
Example:
Potassium permanganate in a basic solution turns potassium cyanide into
potassium cyanate.
1)
Write reactants and products (omit spectator ions like alkali metal
ions)
MnO4- + CN-
2)
3)
4)

MnO2 + CNO-
Write oxidation numbers to determine oxidation and reduction half
reaction:
+7 –2
MnO4-

+4 -2
MnO2
+2-3
CN-

+2-1-1
CNO-
Write separate oxidation and reduction half-reactions:
Ox.:
-3
CN-

-1
CNO-
Red.:
+7
MnO4-

+4
MnO2
The change of oxidation numbers indicates the number of electrons
lost/gained. “Balance” the oxidation numbers accordingly:
Ox.:
Red.:
CN
MnO4- + 3 e- 
CNO- + 2 eMnO2
5)
Balance the charges with H+ (acidic solution) or OH- (basic solution):
CN- + 2 OH- 
MnO4- + 3 e- 
Ox.:
Red.:
6)
Balance each equation with H2O ( Law of Preservation of Mass):
Ox.:
Red.:
7)
CN- + 2 OHMnO4- + 3 e- + 2 H2O


CNO- + 2 e- + H2O
MnO2 + 4 OH-
Multiply each half-reaction so that the number of electrons are equal:
Ox.:
Red.:
8)
CNO- + 2 eMnO2 + 4 OH-
MnO4-
CN- + 2 OH+ 3 e- + 2 H2O


CNO- + 2 e- + H2O x 3
MnO2 + 4 OHx 2
Add both reactions (the number of electrons should cancel):
3 CN- + 6 OH- + 2 MnO4- + 4 H2O  3 CNO- + 3 H2O + 2 MnO2 + 8 OH-
9)
Simplify as far as possible:
3 CN- + 2 MnO4- + H2O
10)

3 CNO- + 2 MnO2 + 2 OH-
Add spectator ions, if necessary:
3 KCN + 2 KMnO4 + H2O 
3 KCNO + 2 MnO2 + 2 KOH
Sample problems:
a)
Tin (II) chloride and iron (III) chloride in acidic solution.
b)
Potassium permanganate and sodium oxalate in acidic solution.
c)
Ammonium dichromate and sodium chloride in acidic solution.
d)
Manganese (II) nitrate and sodium bismutate (NaBiO3) in acidic solution.
e)
Potassium nitrite and aluminum metal in very basic solution [forms
Al(OH)4-].
f)
Sodium chromate and sulfur dioxide in basic solution.