Balancing using oxidation numbers

Question

For the reaction

KCl + MnO2 + H2SO4 -> K2SO4 + MnSO4 + Cl2 + H2O

I have Mn going under oxidation from +4 to +2 I have Cl going under reduction from -1 to 0

Then after using my teachers method o multiplying the oxidation and reduction atoms I have

4KCl + 1MnO2 + H2SO4 -> K2SO4 + 1MnSO4 + 2Cl2 + H2O

However I'm not sure I'm on the right track, but if I am, what are the next steps to completely balance the equation?

Answer 1

It is really easy to solve by the half reaction method, here is my answer:

Half reaction method

Global eq: KCl + MnO2 + H2SO4 -> K2SO4 + MnSO4 + Cl2 + H2O
Cathode: 2Cl^-   --> Cl₂ + 2e^-
Anode: MnO₂ + 4H⁺ + 2e^- --> Mn⁺² + 2H₂O

So the final balanced eq is this one:
2KCl _(ac) + MnO_{2 (s)} + 2H₂SO_{4 (ac)} --> K₂SO_{4 (ac)} + MnSO_{4 (ac)} + Cl_{2 (g)} + 2H₂O

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Variations method

Without half-reaction method will be like this: You have one Mn⁺⁴ reduced to Mn⁺² That is 2e x 1 atom (ion indeed) so this is a net variation of 2 On the other hand you have Cl^- wich oxidizes to form a Cl₂ molecule so this is 1e x 2 atoms wich also is a net variation of 2 Now you use these numbers as the coefficient of the products that contain these atoms and you'll have

KCl _(ac) + MnO_{2 (s)} + H₂SO_{4 (ac)} --> K₂SO_{4 (ac)} + 2MnSO_{4 (ac)} + 2Cl_{2 (g)} + H₂O

and considering this as valid just left to adjust the rest of the elements and finally you'll have:
4KCl _(ac) + 2MnO_{2 (s)} + 4H₂SO_{4 (ac)} --> 2K₂SO_{4 (ac)} + 2MnSO_{4 (ac)} + 2Cl_{2 (g)} + 4H₂O Wich is also a valid adjust

I always recommend using half-reaction method or undetermined coefficient method instead. Good luck!!