Balancing of Chemical Equation by Oxidation Number Method

The principal underlying this universally applicable of balancing equations is that electrical charge must be conserved in the course of chemical reaction like any increase in oxidation number must be compensated by a decrease.

               In brief, the method consists in selection of such coefficients for the oxidizing and reducing agents as will ensure to make the total decrease in oxidation number for the former equal to the total increase in oxidation number for the latter.

               Following rules can be adapted for it:

a) Assign oxidation numbers to the atoms that change.

b) Choose the proper ratio of oxidizing agent to the reducing agent so that the oxidation number change balanced,

c) Make appropriate change in the coefficient of the products corresponding to change in the coefficient of reactions in step (b).

d) Balance the oxygen atoms on both sides by adding H₂O to the side that is deficient in oxygen.

e) Balance the hydrogen atoms on both sides by adding H⁺ to the side that is deficient in oxygen.

f) The equation is balanced if the reaction is taking place in acidic solution. If the reaction proceeds in basic solution, and sufficient number of OH^- to get rid of H⁺ but add equal number of OH^- on both sides.

g) Cancel any duplication that might have crept in on both sides of the equation.

The following examples will cleared the method:

Example: Oxidation of ammonia by copper oxide to give copper, nitrogen and water

a) Writing skeleton equation with oxidation number of copper and nitrogen

                          Oxidation numbers      _{+2         -3           0    0}

       Skeleton                  CuO + NH₃ =   Cu + N₂ + H₂O

b)  Oxidation number of Cu change from +2 to 0 and that of N changes from -3 to 0. The Cu atom is going from +2 to 0 have to gain two units of negative charge. The nitrogen atom is going from -3 to 0 has to get rid 3 units of negative charge.

c) To balance the oxidation number change multiplies CuO by 3 and NH₃ by 2. Then provide same number of Cu and N atoms on the right. The equation will be

          3CuO + 2NH₃ = 3Cu + N₂ + H₂O.

d) Balancing oxygen by multiplying H₂O by 3, we get the required balanced equation.

   3CuO + 2NH₃ = 3Cu + N₂ + 3H₂O.