[Steenrod]

Here is a question from my school paper:


Balance the following by ion electron method:

[itex]Cl_2+OH^{-}\rightarrow ClO_3^{-}+Cl^{-}+H_2O[/itex]

For some reason I am not being able to do it by this method.(I have to do it and formally write it up so that I get maximum credit)

I was in a dilemma, so, I tried changing the equation into this:

[itex]Cl_2+KOH\rightarrow KClO_3+KCl+H_2O[/itex] and by the oxidation number method, I got this :

[itex]3Cl_2+6KOH\rightarrow KClO_3+5KCl+3H_2O[/itex]

Can anyone please write up a solution by the ion-electron method so that I read it and understand where I am going wrong?
Thanks a ton.

[Borek]

Can you write skeletal reduction and oxidation reactions? Or at least list what is being oxidized (and to what) and what is being reduced (and to what)?

[Steenrod]

This is what I tried:

Reduction half reaction:

[itex]Cl_2\rightarrow ClO_3^{-}+Cl^{-}[/itex] and the
reduction half reaction:[itex]OH^{-}\rightarrow H_2O[/itex]

The first one can be balanced by
[itex]Cl_2+6OH^{-}\rightarrow ClO_3^{-}+Cl^{-}+3H_2O+4e[/itex]
I don't understand where I am going wrong.

[Borek]

Reduction half reaction:

[itex]Cl_2\rightarrow ClO_3^{-}+Cl^{-}[/itex]

This is not reduction. You have both reduction and oxidation here. Assign oxidation numbers to Cl in all three compounds.

[Steenrod]

I give up.

[Borek]

Before even trying?

What is oxidation number of Cl in Cl₂?

[Steenrod]

0 ?

I am in trouble.I am surely going to fail my chemistry exam.

[Borek]

Don't give up too early.

Yes, zero.

This one is also easy - what is the oxidation number of chlorine in Cl^-?

This one can be a little bit harder - what is the oxidation number of chlorine in ClO₃^-?

[Steenrod]

I used the oxidation number method to do the balancing.

The oxidation number of [itex]Cl^{-}[/itex] is -1 and in [itex]ClO_3^{-}[/itex] is +5.

[Borek]

OK.

Now, you have chlorine that is oxidized to +5, and chlorine that is reduced to -1. THESE ARE TWO SEPARATE HALF REACTIONS. Can you write separately oxidation half reaction and reduction half reaction (just as a skeletal ones, don't worry about balancing yet)?

[Steenrod]

Here are they:

[tex]Cl_2\rightarrow ClO_3^{-}[/tex]
and [itex]Cl_2\rightarrow Cl^{-}[/itex].Sorry for making you shout.

[Borek]

I wasn't shouting, I just wanted to make it clear what is important here 

[tex]Cl_2\rightarrow ClO_3^{-}[/tex]
[tex]Cl_2\rightarrow Cl^{-}[/tex]

That's correct. Now, that you have both half reactions in skeletal form, can you balance each one separately?

[Steenrod]

Like this?

[itex]Cl_2+12OH^{-} \rightarrow 2ClO_3^{-}+6H_2O+10e\dots \boxed{1}[/itex] and [itex]Cl_2+2e\rightarrow 2Cl^{-}\dots \boxed{2}[/itex]

Multiplying (2) by 5 and adding, we get
[itex]6Cl_2+12OH^{-}\rightarrow 2ClO_3^{-}+6H_2O+10Cl^{-}[/itex].

[Borek]

Almost there - just note you can still divide everything by 2.

Wasn't that hard 

[Steenrod]

Yeah, I noticed that.

To be precise, I was making a fundamental error(besides being careless not to balance Cl).That led to all that confusion.

Thanks for being patient.