To equalize the Oxidation Numbers of Cl and Na, 3 atoms of Cl and 6 atoms of Na are required. Thus,
3Cl2 + 6NaOH -> 5NaCl + NaClO3 + xH2O
By inspection x = 3
∴ 3Cl2 + 6NaOH -> 5NaCl + NaClO3 + 3H2O
I think this is correct, but if I have done something wrong please tell me so, as I'm new to this topic.
Thanks in advance.
No idea what you mean by "equalize the ON", specially in the context of sodium - ON for sodium doesn't change, and Na+ is only a spectator.
You have Cl(0) on the left and Cl(+1) and Cl(-1) on the right. This is disproportionation - teh same element becomes both oxidized and reduced at the same time. Can you tell how many atoms of Cl get oxidized if 1 atom gets reduced?