AQA A2 CHEM4 & CHEM5 June 2015 [official discussion thread]

Hey could someone please help me with Jan 11 queston 7bii) (Chem 5)

http://filestore.aqa.org.uk/subjects/AQA-CHEM5-W-QP-JAN11.PDF

I understood that the reduction of hydrogen peroxide to hydroxide ions is H202 + 2e- --> 2OH- but I'm not sure how to do the following bit

Reply 781

randlemcmurphy

11

Original post by adorablegirl1202

Hey could someone please help me with Jan 11 queston 7bii) (Chem 5)

http://filestore.aqa.org.uk/subjects/AQA-CHEM5-W-QP-JAN11.PDF

I understood that the reduction of hydrogen peroxide to hydroxide ions is H202 + 2e- --> 2OH- but I'm not sure how to do the following bit

Do you know what the "chromium (III) complex" is?

Reply 782

_Caz_

10

Original post by randlemcmurphy

Good question, I have thought about that same point as well. Hopefully it will be cleared up later on!

I hate buffer calculations :frown:

Reply 783

LookIts_Liam

10

Original post by adorablegirl1202

Hey could someone please help me with Jan 11 queston 7bii) (Chem 5)

http://filestore.aqa.org.uk/subjects/AQA-CHEM5-W-QP-JAN11.PDF

I understood that the reduction of hydrogen peroxide to hydroxide ions is H202 + 2e- --> 2OH- but I'm not sure how to do the following bit

Okay I've just done it. The ''chromium complex'' from reaction 3 is [Cr(OH)6]3- and its just knowledge that the final solution is yellow (Im fairly sure it is anyway).

As for the equation, you now know the half equation of the H2O2, but you have to work out the ration in which it reacts with the complex. Well, the chromium goes from [Cr(OH)6]3- to CrO42- so in the first part it has an oxidation state of 3+ and in the second it has an oxidation state of 6+, so it is oxidised losing 3 electrons.

To get rid of electrons on both sides you'd need to multiply the peroxide equation by 3 and the chromium one by 2, to get 6e- on each side, therefore they cancel, then just stick the equations together and add any water needed to balance etc. Quote me if you need anything clearing up!

Reply 784

adorablegirl1202

2

Original post by LookIts_Liam

Okay I've just done it. The ''chromium complex'' from reaction 3 is [Cr(OH)6]3- and its just knowledge that the final solution is yellow (Im fairly sure it is anyway).

As for the equation, you now know the half equation of the H2O2, but you have to work out the ration in which it reacts with the complex. Well, the chromium goes from [Cr(OH)6]3- to CrO42- so in the first part it has an oxidation state of 3+ and in the second it has an oxidation state of 6+, so it is oxidised losing 3 electrons.

To get rid of electrons on both sides you'd need to multiply the peroxide equation by 3 and the chromium one by 2, to get 6e- on each side, therefore they cancel, then just stick the equations together and add any water needed to balance etc. Quote me if you need anything clearing up!

If I understand correctly,
the half equation for the chromium one is [Cr(0H)6]3- ---> Cr207^2- + 3e+?
And then you just put it in one like AS? but is that correct though? Because it's not balanced and I'm a bit confused how to balance it

Reply 785

randlemcmurphy

11

Original post by adorablegirl1202

If I understand correctly,
the half equation for the chromium one is [Cr(0H)6]3- ---> Cr207^2- + 3e+?
And then you just put it in one like AS? but is that correct though? Because it's not balanced and I'm a bit confused how to balance it

It goes to CrO₄^2-. To balance oxygens add water, to balance hydrogens add H+, then to balance the overall charges on either side add electrons.

Reply 786

LookIts_Liam

10

Original post by adorablegirl1202

If I understand correctly,
the half equation for the chromium one is [Cr(0H)6]3- ---> Cr207^2- + 3e+?
And then you just put it in one like AS? but is that correct though? Because it's not balanced and I'm a bit confused how to balance it

no no, because it says in the question that the final solution is crO42-..but yes it is 3 electrons. Well, if you write out both half equations, you know that to form a full equation you need to make the electrons cancel on both sides. Meaning you have to multiply the Chromium equation by 2 and the peroxide one by 3 to get 6 electrons on both side which cancel.

H2O2 + 2e- ---> 2OH-
[Cr(OH)6]3- ---> CrO42- + 3e- + 2H2O + 2H+

So you'd get something like:

2[Cr(OH)6]3- + 3H2O2 ---> 2CrO42- + 6OH- + 4H+ + 4H20

But the OH- and H+ will form H2O so your final equation will be:

2[Cr(OH)6]3- + 3H2O2 ---> 2CrO42- + 2OH- + 8H20

Reply 787

adorablegirl1202

2

Original post by randlemcmurphy

It goes to CrO₄^2-. To balance oxygens add water, to balance hydrogens add H+, then to balance the overall charges on either side add electrons.

Original post by LookIts_Liam

no no, because it says in the question that the final solution is crO42-..but yes it is 3 electrons. Well, if you write out both half equations, you know that to form a full equation you need to make the electrons cancel on both sides. Meaning you have to multiply the Chromium equation by 2 and the peroxide one by 3 to get 6 electrons on both side which cancel.

H2O2 + 2e- ---> 2OH-
[Cr(OH)6]3- ---> CrO42- + 3e- + 2H2O + 2H+

So you'd get something like:

2[Cr(OH)6]3- + 3H2O2 ---> 2CrO42- + 6OH- + 4H+ + 4H20

But the OH- and H+ will form H2O so your final equation will be:

2[Cr(OH)6]3- + 3H2O2 ---> 2CrO42- + 2OH- + 8H20

Thanks so much! I completely get it now!!!

Reply 788

sleepisheaven

5

Original post by LookIts_Liam

no no, because it says in the question that the final solution is crO42-..but yes it is 3 electrons. Well, if you write out both half equations, you know that to form a full equation you need to make the electrons cancel on both sides. Meaning you have to multiply the Chromium equation by 2 and the peroxide one by 3 to get 6 electrons on both side which cancel.

H2O2 + 2e- ---> 2OH-
[Cr(OH)6]3- ---> CrO42- + 3e- + 2H2O + 2H+

So you'd get something like:

2[Cr(OH)6]3- + 3H2O2 ---> 2CrO42- + 6OH- + 4H+ + 4H20

But the OH- and H+ will form H2O so your final equation will be:

2[Cr(OH)6]3- + 3H2O2 ---> 2CrO42- + 2OH- + 8H20

Would you have to cancel the H2O's at the end so you have no H2O's on the left and 5 on the right. (-3H2O from both sides?)

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Reply 789

Hudl

16

Original post by sleepisheaven

Would you have to cancel the H2O's at the end so you have no H2O's on the left and 5 on the right. (-3H2O from both sides?)

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Yes you should!

Reply 790

sleepisheaven

5

Original post by Hudl

Yes you should!

I just realised I read it wrong. On the left is H2O2 not H2O so you don't need to do what I said :P

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Reply 791

djhodnett

9

on question 5e(ii) of june 11 it asks to draw an overall equation for the cell being recharged, but could someone please explain how they got that answer from the mark scheme because it make no sense!

Reply 792

l1lvink

11

Does anyone have a complete list of all the 'test-tube' experiments you can do to find out what is inside for CHEM4?

Reply 793

Emilyb1006

Does this help? :smile:

https://www.youtube.com/watch?v=cfA4HHj4tzg

Reply 794

LookIts_Liam

10

So...are we allowed to discuss the EMPA now?

Reply 795

Absent Agent

21

Original post by LookIts_Liam

So...are we allowed to discuss the EMPA now?

We don't need to ask for special permission to discuss it. It's impossible that any college would take the EMPA tomorrow which is the deadline because that's the deadline for submission. Having said that, there is no point discussing that nonsense. For example why containers of samples in colorimetry should have the same dimensions?
I know that those that want to study or have a strong interest in chemistry answered this question because they went and searched for it. If I'm expected to be questioned from beyond the scope of the specification then I certainly don't need to take exams so that a handful of people would tell me that I'm wrong

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Reply 796

Absent Agent

21

I will upload the mark scheme of the EMPA hopefully on the 25th of May, if not then on the 26th.

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Reply 797

chem@uni

9

Anyone have a nice table or something of all the colours of the transitions metals/reactions we need to know?

Reply 798

jasskaurr11

Original post by chem@uni

Anyone have a nice table or something of all the colours of the transitions metals/reactions we need to know?

There's a table in the blue CGP revision book, if you haven't got it let me know and I'll post it :smile:

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Reply 799

AJeeeee

6

Jan 2010 Question 7 b)

It says that C which is CH3CH2CH2CH2CH3 has 3 peaks in its H nmr spectra but i only see two peaks of 1 quatret and 4 triplets. Where does the third peak come from? So confused, pls help :frown: