Objective C: Balancing Oxidation Reduction Reactions

Example: MnO4-1 + Cl-1 --> Mn+2 + Cl2

The basic idea is that, in each half reaction, you have to balance

(1)   the participating atoms(if necessary),

(2)   the oxidation numbers by using electrons,

(3)   the charge(if necessary)by using H+1 or OH-1,

(4)   and the number of hydrogens with water(if necessary).

(5)   Then you combine the two half reactions algebraically to get the electrons to cancel.

Steps 1 and 2

Figure out what is being oxidized and reduced. Make sure that those participating atoms are balanced and write each half reaction.

MnO4-1 + 5e--> Mn+2 (5 electrons because Mn is being reduced from +7 to 2)

2Cl-1 --> Cl2 + 2e         (2 electrons because each Cl is being oxidized from –1 to 0)

Step 3

Balance the total charge on each side of equation with H+1 if they specify acidic conditions and with OH-1 if conditions are alkaline.

Acidic situation:

MnO4-1 + 5e + 8H+1 --> Mn+2

2Cl-1 --> Cl2 + 2e

Basic(alkaline) situation:

MnO4-1 + 5e --> Mn+2 + 8OH-1

2Cl-1 --> Cl2 + 2e

Step 4

Balance the number of hydrogens on each side of the equation with water.

Acidic situation:

MnO4-1 + 5e + 8H+1 --> Mn+2 + 4H2O

2Cl-1 --> Cl2 + 2e

Basic(alkaline) situation:

MnO4-1 + 5e + 4H2O--> Mn+2 + 8OH-1

2Cl-1 --> Cl2 + 2e

Step 5

Manipulate the equations so that the sum of the two will get the electrons to cancel out.

Acidic situation:

2[MnO4-1 + 5e + 8H+1 --> Mn+2 + 4H2O] = 2MnO4-1 + 10e + 16H+1 --> 2Mn+2 + 8H2O

5[2Cl-1 --> Cl2 + 2e] = 10Cl-1 --> 5Cl2 + 10e

Final answer ( acidic solution): 2MnO4-1 + 16H+1 + 10Cl-1-->5Cl2 +2Mn+2 + 8H2O