Balancing the redox disproportionation of chlorine by half-reactions
You have an error half-way through your oxidation half-reaction which probably multiplies itself. You correctly determined the oxidation states so it should be simple to spot it. But for reference, here they are:

$$\begin{align}&amp;\overset{\pm 0}{\ce{Cl2}}/\overset{\mathrm{+V}}{\ce{Cl}}\ce{O3-}\\[0.9em]
\ce{Cl2 \phantom{\ce{{} + 6 H2O }} &amp;-&gt; 2 ClO3- + 10 e-}\tag{Ox1}\\
\ce{Cl2 \phantom{\ce{{} + 6 H2O }} &amp;-&gt; 2 ClO3- + 10 e- + 12 H+}\tag{Ox2}\\
\ce{Cl2 + 6 H2O &amp;-&gt; 2 ClO3- + 10 e- + 12 H+}\tag{Ox3}
\end{align}$$

Thenceforth it should be clear to see that you need five reductions per oxidation, thus the redox equation is:

$$\begin{align}\ce{6 Cl2 + 6 H2O &amp;-&gt; 2 ClO3 + 10 Cl- + 12 H+}\tag{Redox1}\\
\ce{3 Cl2 + 3 H2O &amp;-&gt; ClO3 + 5 Cl- + 6 H+}\tag{Redox2}\end{align}$$

(The second version is the first but divided by two; you only have even coefficients.)