Balancing the redox disproportionation of chlorine by half-reactions I am having some difficulty balancing the following reaction using the half-reaction method:

$$\ce{Cl2(g) -&gt; Cl-(aq) + ClO3-(aq)}$$

My attempt at the problem:

By determining oxidation numbers it is possible to see what is being oxidised and what is being reduced:

$$\ce{\overset{0}{Cl}_2(g) -&gt; \overset{-1}{Cl}^-(aq) + \overset{+5}{Cl}\overset{-2}{O}_3^-(aq)}$$

From this I determined the following two half-reactions:

Reduction Half: $\ce{Cl2(g) + 2 e- -&gt; 2 Cl-(aq)}$

Oxidation Half: $\ce{Cl2(g) + 3 H2O (l) -&gt; 2 ClO3-(aq) + 6 H+ + 4 e-}$



Net ionic equation: $\ce{6 Cl2(g) + 6 H2O(l) -&gt; 8 Cl-(aq) + 4 ClO3-(aq) + 12 H+}$

The answer to this problem is however given as


  $$\ce{2 Cl2(g) + 6 H2O(l) -&gt; 2 ClO3-(aq) + 12 H+ + 10 Cl-(aq)}$$


I have tried this problem several times over and cannot work out how this answer was reached.



Update:

Silly mistake, oxidation reaction should be: $\ce{Cl2 + 6H2O -&gt; 2 ClO3- + 12 H+ + 10 e-}$ hence solving the mystery of why my answer is wrong.
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.)