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

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++Cl2 Balance the number of Cl first. MnO-4+2Cl-→Mn2++C l2

MnO-4+10Cl-→2M n2 + + 5C l2

Balance the number of oxygen by adding H2O 2MnO-4+10Cl-→2M n2 + + 5Cl2+8H2O

Balance the number of H by adding H + 2MnO-4+10Cl-+16H+→2M n2 + + 5Cl2+8H2O

Now, we know that there are 2 H + in H2SO4, so, we can get 2MnO-4+10Cl-+8H2SO4→2M n2 + + 5Cl2+8H2O

Put S O2-4 with M n2 + 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+5C l 2 + 8H20

The final answer is 2KMn O4 + 10KCl+8 H2 S O4 →6 K2 S O4 + 2MnS O4 + 5C l2 + 8 H2 O