Question

How to balance this by oxidation state change method? . KMnO4 + KCl + H2SO4 --> K2SO4 + MnSO4+Cl2+H2O

Answer 1

In this redox reaction, we must identify which is reduced and which is oxidized. Cl is oxidized from -1 to 0, Mn is reduced from +7 to +2. 

MnO−4+Cl−→Mn2++Cl2Balance the number of Cl first. MnO−4+2Cl−→Mn2++Cl2
MnO−4+10Cl−→2Mn2++5Cl2

Balance the number of oxygen by adding H2O 2MnO−4+10Cl−→2Mn2++5Cl2+8H2O

Balance the number of H by adding H+ 2MnO−4+10Cl−+16H+→2Mn2++5Cl2+8H2O

Now, we know that there are 2 H+ in H2SO4, so, we can get 2MnO−4+10Cl−+8H2SO4→2Mn2++5Cl2+8H2O

Put SO2−4 with Mn2+ 2MnO−4+10Cl−+8H2SO4→2MnSO4+5Cl2+8H2O

Add K in front of MnO−4 and Cl− and balance it on the right 2KMnO4+10KCl+8H2SO4→12K++2MnSO4+5Cl2 + 8H20

The final answer is 2KMnO4+10KCl+8H2SO4→6K2SO4+2MnSO4+5Cl2+8H2O