Advanced redox reactions

I have 2 advanced redox reactions due, but I've only learned how to work with simple redoxreactions.

The reactions are as follows:
S2O3(2-) + Cl2 --> SO4(2-) + Cl(-) (Beginning as a neutral environment)

NH3 + CLO(-) --> N2 + Cl(-)

Any help would be good and it would be a great help if the results were explained.

The S2O3 reaction is tricky, because one of the S atoms has an oxidation state of 0 and the other one an oxidation state of +4. You need to keep that in mind when determining total oxidation state change.

In SO4^2-, S has an oxidation state of +6. Therefore, one S atom changes by +6 and the other by +2, for a total of +8. Cl changes from 0 to -1, so you need 8 Cl atoms for 2 S atoms. You need to add water to the left to supply the extra O atoms, and then balance the H on the right.

S2O3^2- + 4Cl2 + 5H2O → 2SO4^2- + 8Cl^- + 10H^+

In the second reaction, N changes from -3 to 0, and Cl changes fro +1 to -1. You will need 3 Cl atoms per 2 N atoms. Once again, add water to balance the O and H.

2NH3 + 3ClO^- → N2 + 3Cl^- + 3H2O

To balance in neutral medium-separate oxidn. from redn,
Oxidation S2O3(2-) ---> SO4(2-) First balance S then O. To balance O add water to equal O and then double OH- ions on other side.Add electron on more +ve side to balance charges
S2O3(2-) + 10 OH- -----> 2 SO4(2-) + 5 H2O + 8 e-
Similarly
4 Cl2 + 2 e- ------> 2 Cl- Now multiply with suitable coefficients to cancel e- and add
S2O3(2-) + 4Cl2 + 10 OH- ---> 2 SO4(-) + 8 Cl- + 5H2O