# How do we get NaClO3 in the reaction between Cl2 and NaOH?

From problem no. 4080 in a Russian example exam set:

$$\ce{6NaOH + 3Cl2 -> NaClO3 + 5NaCl + 3H2O}$$

How do we get that $\ce{NaClO3}$? I can't seem to guess. I found this explanation on YouTube, but there the products are different:

$$\ce{Cl2 + 2NaOH -> NaCl + H2O + NaClO}$$

In both reactions, there is twice as much sodium chloride as there is chlorine. How do the two additional oxygens connect to the chlorine?

P.S. A related question - thanks, Nilay Gosh.

• Why, this is a typical disproportionation reaction. Many non-metals would do something of this sort. Related: chemistry.stackexchange.com/questions/41182/… – Ivan Neretin May 23 '16 at 12:47
• @IvanNeretin - thank you! What I don't get is how we get a whole of 3 oxygens there, on the chlorine. I'll reread the basics about electrolysis and will try to wrap my head around this. – CowperKettle May 23 '16 at 13:09
• This is not related to electrolysis (chlorine that goes into the reaction may or may not be produced by electrolysis; that's not important). As for the particular oxidation state of chlorine, that's something to be memorized rather than deduced. – Ivan Neretin May 23 '16 at 13:16
• also see this:- chemistry.stackexchange.com/questions/50333/… – Nilay Ghosh May 24 '16 at 5:11
• If you substitute the spectator ion, the answer in the question Nilay linked is pretty much the solution to your problem. It depends on the reaction conditions. – Martin - マーチン May 24 '16 at 6:55

## 1 Answer

I find that when I'm faced when problems having to do with why the products are formed I make lewis structures.

if you make a lewis diagram of the $$\ce{NaClO3}$$

First off, you will see that a single Cl will be bonded to three oxygen molecules whom all carry full valance shells which were striped from the hydroxide ions in the reactants. (Remember that in solution the Na and OH will be around as free spectators.)

From the structure you will see the reason why there was a formation of the three waters. Since there were three oxygen stripped from the hydroxides and then recombination with more OH to make a reaction like this:

$$\ce{OH- + H+ -> H2O}$$

Also notice that on the products side for every molecule of NaClO3 made there will be a molecule for NaCl made. From there you can speculate that when the Cl-Cl bond breaks three oxygen atoms and a sodium atom will surround one of the Cl while another is tied to a Na molecule simultaneously.

• Thank you for the answer, and welcome to the SE! There, however, actually will be $\ce{5 NaCl}$ molecules per one sodium chlrorate – CowperKettle May 23 '16 at 13:24
• There are quite a few issues with charge balance here. In none of the examples hydroxide will act as a spectator. – Martin - マーチン May 24 '16 at 6:51