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

Poll: How pumped up are you for this exam?-(warning)-(bad jokes arene this poll!)
"Titanium-I'm not going to corrode (even at high temperatures)" (A*) (22)
16.67%
"Benzene's my middle name, give me the paper in a week and I'll ace it!" (A) (27)
20.45%
"Yeah, I'm fairly electrophillic (positively charged) about the exam" (B) (27)
20.45%
"I'm in the middle of the salt bridge, but I will pass-eventually" (C) (21)
15.91%
"I'm feeling rather electroNegative about this exam" (D) (18)
13.64%
"Benzene, what's that?" (E) (6)
4.55%
"Chemistry, what's that?" (F) (11)
8.33%
This discussion is closed.
F1's Finest
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#1041
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#1041
(Original post by Gnome :))
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.
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.
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Gnome :)
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#1042
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#1042
(Original post by posthumus)
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 ?
Ah, good luck!
I haven't done any past papers for unit 2 yet, but it's definitely making so much more sense this time round. I could never understand any of the molecular shapes or organics last year, but compared to U5 it seems so much easier! And I'm not doing U4 (thankfully!)

(Original post by James A)
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.
Well done Are you aiming for an A/A* this time round? I'd quite literally skip around the room if I got a B in this unit haha!

Need 124 UMS from 2+5... Hoping that's realistic, with a little... err... a LOT of cramming. 50ish UMS in the paper I just did, so...
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Mollymod
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#1043
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#1043
(Original post by GeorgeL3)
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.
Yeah Edexcel are trolls on this kind of thing
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LeaX
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#1044
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#1044
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?
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GeorgeL3
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#1045
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#1045
(Original post by LeaX)
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.Brδ--Brδ+

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 -
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AT95
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#1046
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#1046
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.
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LeaX
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#1047
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#1047
(Original post by GeorgeL3)
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.Brδ--Brδ+

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?
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GeorgeL3
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#1048
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#1048
(Original post by LeaX)
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).
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LeaX
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#1049
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#1049
(Original post by GeorgeL3)
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).
Ohh damn. Just when you think you've mastered a topic there's tiny details you've missed out.
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GeorgeL3
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#1050
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#1050
(Original post by AT95)
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.
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.
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Gnome :)
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#1051
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#1051
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
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posthumus
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#1052
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#1052
(Original post by Gnome :))
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 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 ??
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AT95
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#1053
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(Original post by GeorgeL3)
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.
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?
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Gnome :)
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#1054
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#1054
(Original post by posthumus)
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
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posthumus
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#1055
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#1055
(Original post by Gnome :))
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
ahaha

& well good luck ! I really need some inspiration right now
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GeorgeL3
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#1056
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#1056
(Original post by AT95)
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
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AT95
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#1057
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#1057
(Original post by GeorgeL3)
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
Omg you are a lifeeee saverrr..thankk youu SOO muchhh

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KhanR
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#1058
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#1058
Anyone got notes for unit 4 and unit 5 that they can upload pleaaaaaase
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GeorgeL3
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#1059
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#1059
(Original post by KhanR)
Anyone got notes for unit 4 and unit 5 that they can upload pleaaaaaase
I have uploaded notes for unit 4 on page 52 and notes for unit 5 on page 50 of this thread if you want them.
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1234lime
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#1060
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#1060
Hey everyone. On page 174 of the edexcel A2 textbook it mentions that a transition metal with complete d-orbitals will be tetrahedral, and one with incomplete d-orbitals will be square planar. It uses cobalt as an example of a t.m. ion forming a tetrahedral complex- but the electronic configuration of co2+ is 3d7? It also has a diagram showing nickel as both tetrahedral and square planar... ;_; I've had a good look around trying to find an explanation for this, but no luck. Does this mean that there isn't really a pattern?
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