[Kaleyrvt]

Troubles with this particular problem:

I need to Balance the following equations by the half-cell method. Show both half-cell
reactions and identify them as oxidation or reduction.

Cl2 (g)       +     OH-     <--->       Cl-      +     ClO3^-        +        H2O(l)

so far I have: Cl2 --> Cl-
                   Cl2  + 2e-  --> 2Cl-
                 
                  Cl2   --> ClO3-
                  Cl2  +  3H2O  --> 2ClO3^-  +  6H+   + 6e-

2 e-'s on the Left and 6 e-'s on the right therefore I now have to multiply the first equation by 3.
=> 3Cl2 + 6e- --> 6 Cl-

Is this right so far??? I am now unsure what to do next??? What do I do with the OH-? Please help as I have no clue where to go from  here!!
Thanks a bunch 

[chiralic]

If the solution is alkaline, the OH- is used. By each oxygen in excess in a side of an equation one add a H₂O in same side and 2OH-, in the other side. If the hydrogen is without equaling after this, adds a OH- by each hydrogen in excess in same side and a H₂O in the other side. If exist in excess hydrogen and oxygen in the same side of the schematic equation, you write a OH- in the other side by each pair in excess of H and O.

I hope that this info help you and others members of this list....

Chiralic

[DevaDevil]

That works well indeed, Chiralic.
The reaction to perchlorate isn't balanced by the way, Kaleyrvt, look at the oxygen atoms and electrons.

In this case however the half-reaction to perchlorate could also be written with OH^- in this form:

Cl₂ + 12 OH^- --> 2 ClO₃^- + 6 H₂O + 10e^-

Then you immediately have hydroxide and water on the right sides

[Borek]

http://www.chembuddy.com/?left=balancing-stoichiometry&right=half-reactions-method