Here is example 4b:

KMnO_{4}+ KCl + H_{2}SO_{4}---> K_{2}SO_{4}+ MnSO_{4}+ Cl_{2}+ H_{2}O

1) identify the atoms being oxidized and reduced:

Mn goes from +7 to +2; it is reduced

Cl goes from -1 to zero; it is oxidized

2) write the two half-reactions:

Mn^{7+}+ 5e¯ ---> Mn^{2+}

2Cl¯ ---> Cl_{2}+ 2e¯

3) multiply the first reaction by 2 and the second by 5, and add them to get:

2Mn^{7+}+ 10Cl¯ ---> 2Mn^{2+}+ 5Cl_{2}

4) since this a complete redox reaction and the number of electrons are equal on both sides, we can now use these coefficients in the molecular equation to balance it:

2KMnO5) balancing the other atoms can be done easily now by 'trial and error' method:_{4}+ 10KCl + H_{2}SO_{4}---> K_{2}SO_{4}+ 2MnSO_{4}+ 5Cl_{2}+ H_{2}O

This is left to the reader.