Verified answer

I think this would work

2IO3- + 12H+ + 10e- ----------> I2 + 6H2O as one half-reaction and

HSO3- + H2O ----> SO4-2 + 3H+ + 2e- Multiplying the bottom equation by 5 to balance the electrons would give a final equation after they are added as:

2IO3- + 5HSO3- -----> 5SO4-2 + I2 + 3H+ + H2O