Edexcel Chemistry A2 Unit 5 ~ Wednesday 19th June 2013 (Now Closed)

Urgh. Just done the Jan13 paper and got an E.



It's the memorising things that's letting me down; I can't remember any of the reactions, reagents, conditions etc, both organic and inorganic.

Time to plaster posters all over the house, methinks.

aha okay so far I guess, thanks for asking I have one tomorrow as well in fact !

The only retakes I'm dreading is for Chemistry really How's unit 2? & pardon me if I've forgotten... are you doing unit 4 as well ?

I did the jan 13 for real and when I saw the copper question, I was stumped. I took some educated guesses, got my script back and got the full 6 marks for that question. I lost a couple of marks here and there. Got a B which was disappointing but chem5 is such a rigorous paper. You need to be really on the ball for the top marks. I just could never hack it.

Yeah I know, even Wikipedia says a freshly exposed surface has a reddish-orange color.
Ah well, it's just copper and chromium then can ask us stuff like that for so as long as we learn them both it'll be fine.

How do you know if the catalyst with benzene is going to be AlCl3 or FeCl3? I've seen in mark schemes where they allow one and reject the other and I'm not sure how you determine which one you use?

FeCl₃ is a catalyst for substituting a halogen onto the benzene ring, it polarises the diatomic chlorine molecule to make one atom delta positive so it will go to attack the benzene ring. (Note, it doesn't have to be Cl).
eg. FeBr₃ + Br₂ → FeBr₃.Br^δ--Br^δ+

AlCl₃ on the other hand is used as a catalyst for the Friedel-Crafts reaction. It will react with a halogenoalkane to form a carbocation which then substitutes onto the benzene ring.
eg. CH₃CH₂Br + AlCl₃ → CH₃CH₂⁺ + AlCl₃Br ^-

Thank you. Could you use AlCl3 for both?

No it I'm pretty sure it has to be FeX₃ for halogenation and AlX₃ for Friedel-Crafts. (Where X= a halogen, usually Cl or Br).

Someone please help me? How is the ans -0.13?? The one that gets oxidised is on the left hand side right? And the equations for calculating electrode potential is E=right hand-left hand. So this makes chlorine the one on the left hand resulting in 1.23-(-1.36) giving 2.59.

The awkward moment where the teacher basically calls you stupid >.<

... Then gets in a muddle when trying to "help" someone answer a question.


Anyway, I don't think I'll be going back after that. I think independent working is going to work better anyway. The teacher is pretty ****e tbh, walk out of most lessons more confused than at the start!

Currently making a card 'game' (such fun!) to help me memorize the aromatic and nitrogen reactions, reagents and conditions

I'm pretty sure that the 'Right - Left' rule only works if both equations are showing reduction whereas currently one is showing oxidation and one reduction.
If they were both reduction the values would be +1.23 & +1.36, meaning that when you do 'Reduced - Oxidised' it would be 1.23 - 1.36 = -0.13

You should check the overall equation to see 'what direction' each half equation is going in and therefore which sign the values should have.

I never learn anything either at school... that's why tomorrow I'm going to catch a cold

That sounds cool, hehe Please share with us when you done ??

Oh wow, that's an unfortunate coincidence eh?

Will do, if it turns out ok and I don't turn it into mini paper aeroplanes out of frustration and/or boredom

So you mean to say when both the equations are showing reduction only thats when we can use this formula?
And as far as i remember i've read (in the cgp guide) that we change the sign of the more negative one and then use the equation, is it right? So i presumed that since the signs are already changed, i dont need to change them n directly applied the formula. And after ur reply m guessing thats wrong..so i just need to directly put in the values n calc the emf of the cell, when both are reduction values?

Basically there are loads of ways to work out Ecell values but I find that a lot of them are subject to certain conditions which can get confusing when trying to remember what should be a simple calculation.
I find that looking at the question in a more intuitive way and thinking about it will always give you the correct result.

For example, if given an overall reaction, look up the two half equations that are involved.
Since the values in the data book always show reduction, the half equation in the overall reaction which is going in the same direction as the half equation by itself is the one being reduced, so you leave it's Ecell value as it is.
The half equation in the reaction which is going the other way round in the overall equation will be the one getting oxidised, so you have to flip the sign round for Ecell.
All you need to do then is just add the two values together to get the correct answer.

So for example:
An overall equation is: 2Fe³⁺+ 2I^- → 2Fe²⁺ + I₂
The two reduction half cells as given in the data book are:
Fe³⁺,Fe²⁺|Pt +0.77
I₂ ,2I^-|Pt +0.54

These represent the half equations:
Fe³⁺+ e^- → Fe²⁺
I₂ +2e^- → 2I^-

Comparing these to the overall equation, the Iron half equation is 'the right way round' since it is being reduced but the Iodine half equation is 'the wrong way round' since it should be getting oxidised, this means we need to change it's value to -0.54

Now that we've done the necessary changing, you just need to add the two values together:
0.77 + (-0.54) = +0.23 V

Anyone got notes for unit 4 and unit 5 that they can upload pleaaaaaase