Redox Reaction Equation with elusive oxygen
First of all you need to write equations with real ions and molecules that do exist in solution. Your oxidation agent here is $\ce{BrO3-}$. In neutral media you can use $\ce{H+}$, $\ce{OH-}$ and $\ce{H2O}$. Your half-reaction can be made from this scheme:

$$\ce{BrO3- -&gt; Br-}$$

There is no oxygen in the right part so we need to add $\ce{H2O}$ there, we need 3 molecules.

$$\ce{BrO3- -&gt; Br- + 3H2O}$$

Now there are no H atoms in the left part and 6 H atoms in the right part. Fix this by adding 6 protons, $\ce{H+}$, in the left part:

$$\ce{BrO3- + 6H+ -&gt; Br- + 3H2O}$$

Charge of left part is +5 and charge of right part is -1. Now equalize charges in left and right parts by adding 6 electrons to the left part.

$$\ce{BrO3- + 6H+ + 6e- -&gt; Br- + 3H2O}$$

Half-reaction is ready.