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Chemistry unit 6 June 2016

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Original post by samb1234
The mistake you made was the - sign. If you are doing it that way, i.e. reduced species -oxidised species, you do NOT change the sign of the ecell values, i.e. in this case you would have X--0.14


But why we don't change it, as in what they have put, it is the value for the reduction of Sn and it's oxidation should be the opposite of the equation with a change to the sign ?

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Original post by PlayerBB
But why we don't change it, as in what they have put, it is the value for the reduction of Sn and it's oxidation should be the opposite of the equation with a change to the sign ?

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There are different ways of approaching it. Either you flip the sign and treat it as a normal sum and add the two numbers together (my personal preference) or you can remember the formula Ecell = Ereduction -Eoxidation (which is doing exactly the same thing, Eoxidation is the reduction potential for the oxidised species so by subtracting it you are flipping the sign, just as you would if you did the first method)
Original post by samb1234
There are different ways of approaching it. Either you flip the sign and treat it as a normal sum and add the two numbers together (my personal preference) or you can remember the formula Ecell = Ereduction -Eoxidation (which is doing exactly the same thing, Eoxidation is the reduction potential for the oxidised species so by subtracting it you are flipping the sign, just as you would if you did the first method)


Ahh, sorry for asking, but if the question gave us that the reaction is feasible, and gave us one of the potentials, do we also do this or directly, go by Ecell= Ered - Eoxi ?

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Original post by PlayerBB
Ahh, sorry for asking, but if the question gave us that the reaction is feasible, and gave us one of the potentials, do we also do this or directly, go by Ecell= Ered - Eoxi ?

Posted from TSR Mobile


It's the same process, the only difference with feasibility questions is you have to do what it tells you to do. Normally you would say the more negative one is the one which would be oxidised, but for feasibility you have to assume it happens as it tells you in the question and do the maths as if they are right and then comment after if it is feasible
Original post by samb1234
It's the same process, the only difference with feasibility questions is you have to do what it tells you to do. Normally you would say the more negative one is the one which would be oxidised, but for feasibility you have to assume it happens as it tells you in the question and do the maths as if they are right and then comment after if it is feasible


Ahhh, thank you SO MUCH!!!

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Original post by Adorable98
http://qualifications.pearson.com/content/dam/pdf/A%20Level/Chemistry/2013/Exam%20materials/6CH08_01_que_20110119.pdf

Q2(ii)

The answer is:
The concentration of iodine is proportional to the titre

Why's that?:redface:


Anyone ?? :frown:
Original post by Adorable98
Anyone ?? :frown:


CS2O32−VS2O32−=CI2VI2  ⟺  VS2O32−=CI2VI2CS2O32−⇒Vs∝CI2\displaystyle \mathrm{C_{S_{2}O_{3}^{2-}}V_{S_{2}O_{3}^{2-}}=C_{I_{2}}V_{I_{2}}} \iff \mathrm{V_{S_{2}O_{3}^{2-}}= \frac{C_{I_{2}}V_{I_{2}}}{C_{S_{2}O_{3}^{2-}}}} \Rightarrow \mathrm{V_{s} \propto C_{I_{2}} }

This is what I wrote on my mock.

Has anyone done the NMR question on June 2013? It was one of those questions you'd really appreciate so long it didn't come in your exam. :biggrin:
(edited 7 years ago)
Original post by Ayman!
CS2O32−VS2O32−=CI2VI2  ⟺  VS2O32−=CI2VI2CS2O32−⇒Vs∝CI2\displaystyle \mathrm{C_{S_{2}O_{3}^{2-}}V_{S_{2}O_{3}^{2-}}=C_{I_{2}}V_{I_{2}}} \iff \mathrm{V_{S_{2}O_{3}^{2-}}= \frac{C_{I_{2}}V_{I_{2}}}{C_{S_{2}O_{3}^{2-}}}} \Rightarrow \mathrm{V_{s} \propto C_{I_{2}} }

This is what I wrote on my mock.

Has anyone done the NMR question on June 2013? It was one of those questions you'd really appreciate so long it didn't come in your exam. :biggrin:


So is that a formula I should know?:redface:
Original post by Ayman!
CS2O32−VS2O32−=CI2VI2  ⟺  VS2O32−=CI2VI2CS2O32−⇒Vs∝CI2\displaystyle \mathrm{C_{S_{2}O_{3}^{2-}}V_{S_{2}O_{3}^{2-}}=C_{I_{2}}V_{I_{2}}} \iff \mathrm{V_{S_{2}O_{3}^{2-}}= \frac{C_{I_{2}}V_{I_{2}}}{C_{S_{2}O_{3}^{2-}}}} \Rightarrow \mathrm{V_{s} \propto C_{I_{2}} }

This is what I wrote on my mock.

Has anyone done the NMR question on June 2013? It was one of those questions you'd really appreciate so long it didn't come in your exam. :biggrin:


That NMR question was honestly one of the more difficult ones, I couldn't even understand the spectrum
Original post by demotivated
That NMR question was honestly one of the more difficult ones, I couldn't even understand the spectrum


Heck yeah. IAL May 15 and Jan 16 also have decent ones but not nearly as difficult.

Original post by Adorable98
So is that a formula I should know?:redface:


I think so. I'm not sure how to show that they're proportional otherwise. We learnt the C1V1 = C2V2 formula in GCSEs so it kinda just stuck. :tongue:
Can anyone help me with an NMR question in Jan 2014 IAL https://52e1064b4dc7f9fd87a96faa7f7e5a9427a24547.googledrive.com/host/0B1ZiqBksUHNYNFpielJBNWdCRFU/January%202014%20(IAL)%20QP%20-%20Unit%206%20Edexcel%20Chemistry.pdf

Question 3 (h)

The answer for the displayed formula is

But I don't understand how to figure out that the methyl group is attached to a carbon from the spectrum? I would just draw 4 carbons in a straight chain
Original post by demotivated
Can anyone help me with an NMR question in Jan 2014 IAL https://52e1064b4dc7f9fd87a96faa7f7e5a9427a24547.googledrive.com/host/0B1ZiqBksUHNYNFpielJBNWdCRFU/January%202014%20(IAL)%20QP%20-%20Unit%206%20Edexcel%20Chemistry.pdf

Question 3 (h)

The answer for the displayed formula is

But I don't understand how to figure out that the methyl group is attached to a carbon from the spectrum? I would just draw 4 carbons in a straight chain


Check the spectrum. It shows that there are four different proton environments so based on the analysis of that spectrum, we can deduce a structure for the compound. If you see the spectrum, one of the environments shows around 8 small peaks and those number of peaks are actually telling us that the neighbouring carbon has 7 hydrogen around it which splits into 8(with reference to the n+1 rule).
I hope I'm correct lol I suck in this topic
Original post by Ayman!
Heck yeah. IAL May 15 and Jan 16 also have decent ones but not nearly as difficult.



I think so. I'm not sure how to show that they're proportional otherwise. We learnt the C1V1 = C2V2 formula in GCSEs so it kinda just stuck. :tongue:



Ayman, how are the moles of Iodine and Sodium Thiosulfate equal? Like did you assume this that they are equal?
Original post by sabahshahed294
Check the spectrum. It shows that there are four different proton environments so based on the analysis of that spectrum, we can deduce a structure for the compound. If you see the spectrum, one of the environments shows around 8 small peaks and those number of peaks are actually telling us that the neighbouring carbon has 7 hydrogen around it which splits into 8(with reference to the n+1 rule).


I think they are 9 peaks, but even then I don't see any carbon in the structure which has 8 hydrogen atoms attached to the carbons adjacent? It's also really inconvinient how we don't get the data booklet
Um, actually we should be provided with the data booklet because read what's written in page no 1.
http://qualifications.pearson.com/content/dam/pdf/A%20Level/Chemistry/2013/Teaching%20and%20learning%20materials/GCE_Chemistry_Data_booklet_Issue_2.pdf
Original post by Ayman!
Heck yeah. IAL May 15 and Jan 16 also have decent ones but not nearly as difficult.



I think so. I'm not sure how to show that they're proportional otherwise. We learnt the C1V1 = C2V2 formula in GCSEs so it kinda just stuck. :tongue:


I see, thank you!
Original post by Ayman!

Has anyone done the NMR question on June 2013? It was one of those questions you'd really appreciate so long it didn't come in your exam. :biggrin:


Yeah, It was tough lol but tbh I found the question about electrode potentials in IAL January 2015 much harder..........If I have done this paper(Jan 2015), I would come out subbing from that question :colondollar:
Original post by Adorable98
Anyone ?? :frown:


Also, in addition to what Ayman have wrote, Here as the volume of thiosulphate ions increase. then the concentration of Iodine increases as well that's because the concentration of iodine is directly proportional to the volume of thiosulphate ions

Which means, the more Sodium thiosulphate is added from the burette then it means there's more of the Iodine in the flask so the higher the concentration of Iodine
(edited 7 years ago)

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