# Balancing using oxidation numbers

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?

• I'd rather go with half-reactions, but if you insist, this way may work out as well.We don't touch the redox part anymore; now what elements are not balanced yet? Is the reaction balanced in S, for example? What can we do about it? Jun 13 '16 at 8:01
• @Ivan Neretin I prefer half reactions as well, however teacher is marking it based on our ability to do it without it. Also what your saying is my method is correct, and all that's left is to balance the S O and H's? Jun 13 '16 at 8:12
• Yeah, that's about right. Jun 13 '16 at 8:21

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) + MnO2 (s) + 2H2SO4 (ac) --> K2SO4 (ac) + MnSO4 (ac) + Cl2 (g) + 2H2O

### 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) + MnO2 (s) + H2SO4 (ac) --> K2SO4 (ac) + 2MnSO4 (ac) + 2Cl2 (g) + H2O

and considering this as valid just left to adjust the rest of the elements and finally you'll have:
4KCl (ac) + 2MnO2 (s) + 4H2SO4 (ac) --> 2K2SO4 (ac) + 2MnSO4 (ac) + 2Cl2 (g) + 4H2O Wich is also a valid adjust

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