[taylor2264]

I have been trying to figure this out all day I had a test earlier today and I know I missed it and any help will be greatly appreciated.

Cl2 + I- => IO3- + 2Cl-

    3H2O + I-=> IO3- + 6H+ + 6e-
3 (2e- +2Cl-=> Cl2)

3H2O + I- => IO3- + 6H+ + 6e-
6e- +6Cl- => 3Cl2                       cancel electrons and add reaction

3H2O + 6Cl- + I- => 6H+ + 3Cl2 + IO3-   

This is where I'm lost the charges are not equal, now I know for basic solutions you add OH- for every H+ and I even tried that but I still can not get my charges to equal.

[UG]

3 (2e^- +2Cl^- Cl₂)

This half-equation is incorrect, the electrons are on the wrong side, also, it is Cl₂ being reduced to 2Cl^- not the other way round. Once these are fixed you should get a balanced equation. 

[taylor2264]

But for reduction the electrons go on the left side right?

[UG]

Yes. 
It would be Cl₂ + 2e^-  2Cl^-

[taylor2264]

Thank You So Much!! so the answer would be
3H2O + I- + 3Cl2 => IO3- + 6Cl- + 6H+

[UG]

Yes! 
Just to be sure, it is IO₃^- and not IO^3- right?