Chemistry: Oxidation Numbers

CiaraBelle

I just did 2007 Q10c on oxidation numbers from the papers, and thought I did it right, but can't figure out how they got the answer they did from the marking scheme...

It's the balance of this equation using oxidation numbers:
Cl2 + SO3(2-) + H2O --> Cl(-) +SO4(2-) + H(+)


And I got:
[2]Cl2 + SO3(2-) + H2O --> [4]Cl(-) +SO4(2-) + [2]H(+)


But the marking scheme has:
Cl2 + SO3(2-) + H2O --> [2]Cl- +SO4(2-) + [2]H(+)


Sorry I can't write it any clearer that that but I would really appreciate if someone could take a minute to explain it to me! Really has me stuck!

zam

CiaraBelle wrote: »

I just did 2007 Q10c on oxidation numbers from the papers, and thought I did it right, but can't figure out how they got the answer they did from the marking scheme...

It's the balance of this equation using oxidation numbers:
Cl2 + SO3(2-) + H2O --> Cl(-) +SO4(2-) + H(+)


And I got:
[2]Cl2 + SO3(2-) + H2O --> [4]Cl(-) +SO4(2-) + [2]H(+)


But the marking scheme has:
Cl2 + SO3(2-) + H2O --> [2]Cl- +SO4(2-) + [2]H(+)


Sorry I can't write it any clearer that that but I would really appreciate if someone could take a minute to explain it to me! Really has me stuck!

Ok well say something is like Cl2 and it loses one electron, it loses one electron TWICE (because it's Cl2) so therefore it loses 2

CiaraBelle

Ok, that makes sense, but how come this one from 2008:

I2 + S2O3(2-) --> I(-) +S4O6(2-)

works out as:

I2 + [2]S2O3(2-) --> [2]I(-) +S4O6(2-) (from the marking scheme)

Sorry if I'm missing something obvious here, but should the I2 also be gaining 2 electrons in this case... ? I'm confused

jumpguy

CiaraBelle wrote: »

I just did 2007 Q10c on oxidation numbers from the papers, and thought I did it right, but can't figure out how they got the answer they did from the marking scheme...

It's the balance of this equation using oxidation numbers:
Cl2 + SO3(2-) + H2O --> Cl(-) +SO4(2-) + H(+)


And I got:
[2]Cl2 + SO3(2-) + H2O --> [4]Cl(-) +SO4(2-) + [2]H(+)


But the marking scheme has:
Cl2 + SO3(2-) + H2O --> [2]Cl- +SO4(2-) + [2]H(+)


Sorry I can't write it any clearer that that but I would really appreciate if someone could take a minute to explain it to me! Really has me stuck!

The question is...

"Hence or otherwise balance the second equation. (6)"

It does not say "balance using oxidation numbers". However, I'm not 100% sure, if that's the case, the question is very, very poorly phrased altogether. Balancing using oxidation numbers I got the same answer as you did, but by balancing only by inspection I got the same answer as the marking scheme.

Best of luck.

CiaraBelle

I'll just stick with the oxidation numbers I suppose and hope that it's a clearer question if it comes up this year! Thanks for your help

rainbowtrout

CiaraBelle wrote: »

I just did 2007 Q10c on oxidation numbers from the papers, and thought I did it right, but can't figure out how they got the answer they did from the marking scheme...

It's the balance of this equation using oxidation numbers:
Cl2 + SO3(2-) + H2O --> Cl(-) +SO4(2-) + H(+)


And I got:
[2]Cl2 + SO3(2-) + H2O --> [4]Cl(-) +SO4(2-) + [2]H(+)


But the marking scheme has:
Cl2 + SO3(2-) + H2O --> [2]Cl- +SO4(2-) + [2]H(+)


Sorry I can't write it any clearer that that but I would really appreciate if someone could take a minute to explain it to me! Really has me stuck!

You need to look at what is being oxidised and reduced every time you do one of these questions.

1. The Cl2 has an O.N. of zero
2. The Cl- has an O.N. of -1
3. Chlorine is being reduced, gaining 1 electron
4. In the SO3- ion S has an O.N of +4 (as the three oxygens = -6)
5. In the SO4(2-)ion the S has an O.N of +6
6. Sulphur is oxidised losing 2 electrons
7. Oxidation and Reduction have to be equal so if S is losing 2 electrons overall they both have to be gained somewhere, you only have one Cl on the right so by putting a 2 in front of it you double the number of Cl ions and each can gain an electron meaning the number of electrons lost and gained is now equal.
8. At this point once the number of electrons lost and gained is balanced you can then balance the rest of the equation as normal if needs be.

Your original answer is wrong because 2Cl2 still has an O.N. of zero while 4Cl- has an O.N. of 4(-1) meaning four electrons are gained. As you have not inspected the sulphur in the equation you have not seen that sulphur is only losing 2 electrons, so it is not possible for chlorine to gain four electrons when only 2 are lost.

CiaraBelle wrote: »

Ok, that makes sense, but how come this one from 2008:

I2 + S2O3(2-) --> I(-) +S4O6(2-)

works out as:

I2 + [2]S2O3(2-) --> [2]I(-) +S4O6(2-) (from the marking scheme)

Sorry if I'm missing something obvious here, but should the I2 also be gaining 2 electrons in this case... ? I'm confused

Something similar here

I2 = ON of zero
I- = ON -1
S on LHS = ON +2
S on RHS = +2.5

So Iodine is gaining one electron and sulphur is losing half an electron. The sulphur needs to be doubled to lose a whole electron hence doubling it on the LHS. The electrons are now balanced so proceed with balancing the rest of the equation as normal. The only thing in this case that needs to be balanced is the iodines so put a 2 in front of I- on the RHS.

jumpguy wrote: »

The question is...

"Hence or otherwise balance the second equation. (6)"

It does not say "balance using oxidation numbers". However, I'm not 100% sure, if that's the case, the question is very, very poorly phrased altogether. Balancing using oxidation numbers I got the same answer as you did, but by balancing only by inspection I got the same answer as the marking scheme.

Best of luck.

It's an oxidation number question on a higher level paper. It is assumed you know that you have to balance it with oxidation numbers. You are not going to get a balance the equation type question on a HL paper which only requires you to stick in a two in front of one of the compounds.

jumpguy

rainbowtrout wrote: »

It's an oxidation number question on a higher level paper. It is assumed you know that you have to balance it with oxidation numbers. You are not going to get a balance the equation type question on a HL paper which only requires you to stick in a two in front of one of the compounds.

Ahh, thanks alot, it turns out myself and the OP were making the same mistake! I knew the question couldn't be that badly phrased...

theowen

jumpguy wrote: »

The question is...

"Hence or otherwise balance the second equation. (6)"

It does not say "balance using oxidation numbers". However, I'm not 100% sure, if that's the case, the question is very, very poorly phrased altogether. Balancing using oxidation numbers I got the same answer as you did, but by balancing only by inspection I got the same answer as the marking scheme.

Best of luck.

Would you ever not use oxidation numbers. Recipe for disaster if you don't:P