Change in oxidation number in a redox reaction and balancing

The change in oxidation number and balancing a redox reaction

Calculating change in oxidation number:

Following are the steps to follow when you calculate change in oxidation number in a redox reaction. Once you practice two to three times than that becomes super easy to calculate the change in O.N.

Take an example of the following reaction (First and easy method).

H₂S + Br₂ -----> HBr + S

Step 1: Write the equation in the ionic form and do not balance it.

H⁺+S^2-+ Br₂ -----> H⁺+Br^-+S

Step 2: Ignore common ions on both side

S^2- + Br₂ -----> Br^- + S

Step 3: Write the O.N of elements at the top now

S^2- + Br₂⁰ -----> Br^- + S⁰

Step 4: Find the O.N change of any element by:

For example, O.N change of S = [0 – (--2)] X 1 = +2

O.N change of Br = [–1 – 0] X 2 = –2

  Balancing change in oxidation number:

We must be clear that increase in oxidation number is called the oxidation and decrease in oxidation number is the reduction. The molecule in which the oxidation  number of an element increases is the reducing agent or reductant and whose oxidation number decreases is the oxidising agent  or oxidant. Consider the example of:

NH₃ + O₂  ------> NO + H₂O

Now to balance the change in oxidation number,

1. Multiply the molecule of oxdising agent (O₂ on the reactant side), with the change in oxidation number in the oxidation process.

 i.e., O₂ becomes 5O₂

 2. Multiply the molecule of reducing agent (NH₃  on the reactant side), with the change in oxidation number in the reduction p;rocess.

i.e.,   NH₃ becomes 4NH₃

3. The reaction is written including point 1 and 2 together as:

4 NH₃ + 5 O₂  ------> NO + H₂O

4. Now keep the reactant side unaltered and make the number of atoms of different elements (whose O.N has changed) on the product side equal to the number of atoms of the same elements in the reactant side by multiplying suitable integer. 

5. Thus the reaction with balanced O.N change is:

4 NH₃ + 5 O₂  ------> 4 NO + 6 H₂O