Balance KMnO4 + KCl + H2SO4 → K2SO4 + MnSO4+Cl2+H2O Reaction by oxidation state change method?

Question:

Balance KMnO₄+KCl+H₂SO₄→K₂SO₄+MnSO₄+Cl₂+H₂O Reaction by oxidation state change method?

Solution:

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⁺+Cl₂

MnO⁻⁴+10Cl⁻→2Mn₂++5Cl₂

Balance the number of oxygens 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 2H⁺ 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