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Hi could someone show me these equations as a balanced ones:

(Hot)NaOH+Cl2---->NaClO3+Nacl+H2O

(Cold) NaOH+Cl2---->NaClO+Nacl+H2O
(edited 10 years ago)
Original post by Ratboy
Hi could someone show me these equations as a balanced ones:

(Hot)NaOH+Cl2---->NaClO3+Nacl+H2O

(Cold) NaOH+Cl2---->NaClO+Nacl+H2O


The first is

equation is changing one chlorine atom into a chlorate(V) ion and the other chlorine atom into a chloride ion.

i.e.

Cl(0) --> Cl(V) + 5e
Cl(0) + 1e --> Cl(-1)

To balance the electrons you must multiply the second equation by 5

Cl(0) --> Cl(V) + 5e
5Cl(0) + 5e --> 5Cl(-1)

Hence 6 chlorine atoms are needed on the LHS, i.e. 3 chlorine molecules ....

can you get there now?
Reply 2
sorry but even more confused now!!
Reply 3
These are redox reactions. There are two commonly used approaches to balancing redox reactions - using half reactions, and using oxidation numbers. I bet you are expected to use one of them - do you know which one?
Original post by Ratboy
Hi could someone show me these equations as a balanced ones:

(Hot)NaOH+Cl2---->NaClO3+Nacl+H2O

(Cold) NaOH+Cl2---->NaClO+Nacl+H2O


HOT: 6NaOH + 3Cl2 5NaCl + NaClO3 + 3H2O

COLD: Cl2 + 2NaOH →NaCl + NaOCl + H2O
Reply 5
Original post by Plantagenet Crown
HOT: 6NaOH + 3Cl2 5NaCl + NaClO3 + 3H2O

COLD: Cl2 + 2NaOH →NaCl + NaOCl + H2O


I thought posting final solutions without a real need is against forum rules.
Original post by Borek
I thought posting final solutions without a real need is against forum rules.


Maybe, but he asked 2 days ago and is still 'confused' so I though I might as well give the equations. I dont know why people have trouble with this though, it took me 2 seconds of google searching to come across them.

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