# 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:

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

From this I determined the following two half-reactions:

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

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

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

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

• It is an arithmetical problem, watch out over the mathematical operations you did. May 23, 2016 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. May 23, 2016 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}