Balancing redox equations

Use oxidation states to balance the following equation:
Cl₂ + NaOH → NaClO₃ + NaCl + H₂O

I know that first you need to find which elements have been oxidised and reduced, one of which I think is Na. But in the products, there are three different species containing Na which is confusing me.

A step by step guide really would be appreciated.

Cheers

Reply 1

cptbigt

Laura373

Use oxidation states to balance the following equation:
Cl₂ + NaOH → NaClO₃ + NaCl + H₂O

I know that first you need to find which elements have been oxidised and reduced, which I think is Na. But in the products, there are three different species containing Na which is confusing me.

A step by step guide really would be appreciated.

Cheers

Ok you have NaOH - the oxidation state here is 1+ for sodium, you also made NaCl - the oxidation state here is still 1+ so you can ignore NaCl as nothing has changed :smile:

In NaClO3, oxygen gives a total of -6 oxidation state, chlorine is in its (V) state which also means sodium is +1 still. So it hasn't changed at all throughout the reaction.

On the otherhand you have gone from Cl2 to 1. NaCl then 2. NaClO3

In NaCl Cl is -1 state, its gone from 0 (all elements are 0) to -1 so it has been reduced (gained electrons). But in NaClO3, its now +5 state so its gone from 0 to +5 and hence been oxidised (lost electrons) :smile:

Now you know what is being reduced/oxidised you just need to balance it.

wrong post nvm

Laura373

The chlorine has undergone reduction from the 0 oxidation state to the -1 oxidation state in NaCl. It has also been oxidised from the 0 oxidation state to the +1 oxidation state in NaClO as the oxygen is -2 and the sodium is +1, so the chlorine must be +1 to get the overall neutral charge.

Forget about the sodium and just concentrate on the chlorine undergoing disproportionation (oxidation and reduction simultaneously).

Reply 4

cptbigt

Danielisew

Its sodium chlorate (V) not NaClO :smile:

Reply 5

Laura373

Thank you!!!

(might be back in a minute though... :tongue:

)

Reply 6

Laura373

Haha, told you I'd be back.
Cl₂ → NaClO₃ + 5e^-

Cl₂ + e^- → NaCl
5Cl₂ + 5e^- → 5NaCl

Then I put the two together and get a completely wrong equation.
Where am I going wrong?

Reply 7

cptbigt

Laura373

Haha, told you I'd be back.
Cl₂ → NaClO₃ + 5e^-

Cl₂ + e^- → NaCl
5Cl₂ + 5e^- → 5NaCl

Then I put the two together and get a completely wrong equation.
Where am I going wrong?

They are not balanced.

Try doing:

Cl2 -> 2ClO3^- + 10e^-

and

5Cl2 + 10e^- -> 10Cl^-

Then put them together into:

6Cl2 + 12NaOH -> 2NaClO3 + 10NaCl + 6H2O

I think that should be right.

Reply 8

Laura373

The answer in the book is 3Cl₂ + 6NaOH → NaClO₃ + 5NaCl + 3H₂O.
So basically, half of that.
I thought it was only the electrons you balanced?
I also don't understand how my half equations weren't balanced but yours were. Balanced with what?

Thanks for this help btw! :biggrin:

Reply 9

cptbigt

Laura373

Yeah you could easily just half it to make it simpler.

And if you look at it, you said:

Cl2 → NaClO3 + 5e-

You have to balance the electrons between the two half equations but before that you have to make sure the actual half equation is balanced. Your going from 2 chlorines to just 1 on the products side. You need to balance that first.

Each chlorine looses 5 electrons each, you start with 2 so you need 10 electrons lost and 2NaClO3 to balance the chlorines :smile:

You made the same mistake with the second one too:

5Cl2 + 5e- → 5NaCl

You have 10 chlorines going to just 5, so you need 10NaCl made and 10 electrons being gained as you 5Cl2 = 10 chlorines.

Hope it helps!

Reply 10

Laura373