Re: simple balanced equations

Panther wrote:
> hi,
>
> I would like the answers to these questions please. I've been given a ***
> to do for practise, and since I've no idea what the REAL answers are, I
> can't check my understanding/working out, so it's a bit useless without the
> answers as I won't know where I've gone wrong.

Here I would repeat what most responders have written,
say, for example, the reply of Bob.

> So please could you balance these equations by assigning oxidation no's and
> writing half equations for the oxidation and reduction process? Thanks.

>>From what you say above I suspect * (apologies if I'm wrong)
that even the concept of Oxidation Number is not clear to you.
There are at least three common methods that can be used
for balancing Red-Ox reactions. They have to obey (of course)
to atoms' and charge conservation, which are inescapable (chem.)
physical laws. Amongst those methods, the one that seem to me
you are going to use relies on the notion of Oxidation Number ON),
purportedly to make the task simpler. So you have first
to be familiar with ON and how to determine it for a given element
(atom type) in a given *chemical species* (atom, atomic
ion, molecule, molecular ion).

Even if ON cannot be reliably determined in *all* chemical
species, that isn't a problem with the species commonly encountered in
exercises written to exploit ON method,
in particular in the cases 1 - 5 you wrote below.

Unfortunately, learning to determine ON (by itself) previously entails,

in turn, the knowledge of the structure of the chemical species,
which is another (important) problem, although some shortcuts
and rules of thumb could help in most (easy) cases.

> 1. (S2O3)2- + I2 ---> (S4O6)2- + (I)-
>
> 2. Br2 + (S2O3)2- + H20 ---> (SO4)2- + (Br)- + (H)+
>
> 3. (Fe)3+ + (I)- ---> (Fe)2+ + I2
>
>
> 4. (IO3)- + (I)- + (H)+ -----> I2 + H2O
>
> 5. (CrO4)2- + (H)+ ---> Cr2O7(2-) + H2O

(*) If you knew how to use the ON, in the last case (No. 5),
you should have realized that this reaction is *not a Red-Ox
reaction*, so that ON is useless here. As a general help, I'll
choose that example. First balance Cr: write 2 (CrO4)2-.
Then you have
2 (CrO4)2- + (H)+ ---> Cr2O7(2-) + H2O
Step 2: now O and H must be balanced. O seems Ok, but,
to be sure, you will recheck that after having balanced H,
as a general rule. Now write 2 (H)+, so that you have
2 (CrO4)2- + 2 (H)+ ---> Cr2O7(2-) + H2O
Now atoms' conservation is achieved.
Step 3: charge conservation. The check is as follows:
2*(-2) + 2 (= -2) ---> -2 (Ok)

Notice that steps 2 and 3 may be executed in reverse order.
Wouldn't you try and figure out how?


> ---
> Thanks. This isn't homework -- the only person I'd be cheating
> is myself if I couldn't do it.

Best regards,
Angelo

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