# Balancing the redox disproportionation of chlorine by half-reactions

I am having some difficulty balancing the following reaction using the half-reaction method:

$$\ce{Cl2(g) -> Cl^{-}(aq) + ClO3^{-}(aq)}$$

My attempt at the problem:

By determining oxidation numbers it is possible to see what is being oxidised and what is being reduced. I have indicated oxidation numbers below in square brackets:

$$\ce{Cl2  ->Cl^{-}}\ [-1] \ce{+ ClO3^{-}}\ [\ce{Cl} = +5, \ce{O} = -2]$$

From this I determined the following two half-reactions:

Reduction: $\ce{Cl2(g) + 2e- -> 2Cl-}$

Oxidation: $\ce{Cl2(g) + 3H2O (l) -> 2 ClO3- + 6H+ + 4e-}$

and a net ionic equation of: $\ce{6Cl2 + 6H2O(l) -> 8Cl- + 4ClO3- + 12H+}$

The answer to this problem is however given as $$\ce{2 Cl2 (g) + 6 H2O (l) -> 2 ClO3- (aq) + 12 H+ + 10 Cl- (aq)}$$

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

Update:

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

• It is an arithmetical problem, watch out over the mathematical operations you did. – Another.Chemist May 23 '16 at 20:24
• Your oxidation half-reaction (and, consequently, your tentative answer) are not balanced in oxygen. You should have known something's amiss when you saw that $4e^-$ part. We oxidize from 0 to +5, that is, by 5 steps; the number of electrons should be a multiple of 5. – Ivan Neretin May 23 '16 at 20:24

\begin{align}&\overset{\pm 0}{\ce{Cl2}}/\overset{\mathrm{+V}}{\ce{Cl}}\ce{O3-}\\[0.9em] \ce{Cl2 \phantom{\ce{{} + 6 H2O }} &-> 2 ClO3- + 10 e-}\tag{Ox1}\\ \ce{Cl2 \phantom{\ce{{} + 6 H2O }} &-> 2 ClO3- + 10 e- + 12 H+}\tag{Ox2}\\ \ce{Cl2 + 6 H2O &-> 2 ClO3- + 10 e- + 12 H+}\tag{Ox3} \end{align}
\begin{align}\ce{6 Cl2 + 6 H2O &-> 2 ClO3 + 10 Cl- + 12 H+}\tag{Redox1}\\ \ce{3 Cl2 + 3 H2O &-> ClO3 + 5 Cl- + 6 H+}\tag{Redox2}\end{align}