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

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#501

(Original post by

Whoops!

I meant like if you get 37 out of 40, this will be 48 out of 60.

But if you get 38 out of 40, this will be 54 out of 60.

Sorry. :/

**senz72**)Whoops!

I meant like if you get 37 out of 40, this will be 48 out of 60.

But if you get 38 out of 40, this will be 54 out of 60.

Sorry. :/

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#502

What should we revise from units 1,2 and 4 apart from organic chemistry, since unit 5 is synoptic.

Thanks...

Thanks...

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#503

(Original post by

What should we revise from units 1,2 and 4 apart from organic chemistry, since unit 5 is synoptic.

Thanks...

**YAH**)What should we revise from units 1,2 and 4 apart from organic chemistry, since unit 5 is synoptic.

Thanks...

- All the basic things from the unit 1 exam like writing ionic equations, calculating concentration & moles, enthalpy, types of bonding etc.

- The bit on atomic structure would be useful to know since it links in with the transition metal topic and also a little bit with benzene. This is stuff like knowing about the spdf orbitals and how to write out the electron configuration for things.

- All of the various tests for different organic functional groups as these can come up in those questions where you have to identify the substance. Eg. Sodium and alcohol making bubbles of H2, silver nitrate test with halogens and all the tests to see if an alcohol has been oxidised etc.

- Also how entropy values and equilibrium constants can be used to predict the extent of reactions etc. because this ties in to the Ecell values bit.

There's probably lots of other stuff you should know but I'd say these things should definitely be remembered.

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Revision isn't going too badly, but it could be better. How's yours going?

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#506

(Original post by

Hey AS01, glad your back...!

Revision isn't going too badly, but it could be better. How's yours going?

**David Tennant**)Hey AS01, glad your back...!

Revision isn't going too badly, but it could be better. How's yours going?

How you finding chem? It took me quite long to get hang of electrochemistry lol

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(Original post by

same here. Its alright.

How you finding chem? It took me quite long to get hang of electrochemistry lol

**AS01**)same here. Its alright.

How you finding chem? It took me quite long to get hang of electrochemistry lol

Transition metals-I haven't done them yet.

The only seemingly hard part of the specification is the electrochemical chapter which I still haven't got the hang of,

But we have like 78 days untill the exam so for the meanwhile, lol

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#508

(Original post by

The Benzene and nitrogen chemistry is not too bad and neither is the chemical synthesis part...Though there is a lot to remember.

Transition metals-I haven't done them yet.

The only seemingly hard part of the specification is the electrochemical chapter which I still haven't got the hang of,

But we have like 78 days untill the exam so for the meanwhile, lol

**David Tennant**)The Benzene and nitrogen chemistry is not too bad and neither is the chemical synthesis part...Though there is a lot to remember.

Transition metals-I haven't done them yet.

The only seemingly hard part of the specification is the electrochemical chapter which I still haven't got the hang of,

But we have like 78 days untill the exam so for the meanwhile, lol

78 days! I think that will be enough

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#509

(Original post by

that cathode and anode thing confuses me. So that is still left. I think I like organics of unit 5 than TM and electrochemistry.

78 days! I think that will be enough

**AS01**)that cathode and anode thing confuses me. So that is still left. I think I like organics of unit 5 than TM and electrochemistry.

78 days! I think that will be enough

78 days sounds like ages o.O

I love organics, I always have really.

I've learnt the transition metal colours, but I need to learn some of the explanations as to why they're coloured in solution and I need to practice a couple of equations, but otherwise, it's just stuff to learn-learn-learn.

Goodluck! Nice to see you back, I like learning from the guys on this thread. We'll pull through guys

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#510

(Original post by

Here are the first set of equations needed.

Half equation for Manganese being reduced:

MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

Half equation for Iron being oxidised:

Fe2+ → Fe3+ + e-

Full equation for manganate oxidising iron:

MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+

The First step now is to calculate the number of moles used to react with the Iron.

Moles of MnO4- in 1st titration: n=cv/1000, n=(0.02 x 18.2)/1000 = 3.64x10^-4 moles.

From the balanced equation we can see this is 5x less than the moles of Iron present, so the moles of iron are:

5 x 3.64x10^-4 = 1.82x10^-3 moles

Now before the second titration, zinc is added so it oxidises the iron(III) to iron(II):

Zn + 2Fe3+ → Zn2+ + 2Fe2+

So now in the second equation, all of the Iron is Iron(II).

So now we do the same process with this second titration:

Moles of MnO4- = (0.02 x 25.3)/1000 = 5.06x10^-4 moles

Moles of Fe(II) = 5 x 5.06x10^-4 = 2.53x10^-3

Now this total amount of Fe(II) includes the moles of Fe(III) in it, so to find the moles of Fe(III) we need to take away the first answer for moles of iron:

Moles of Iron(III) = (2.53x10^-3) - (1.82x10^-3 moles) = 7.1x10^-4 moles

Now from the Iron and Zinc equation you can see that the number of moles of Zinc are half that of Iron(III):

Moles of zinc = (7.1x10^-4 moles) / 2 = 3.55x10^-4 moles

Now you can just use Mass = Moles x RMM (RMM of Zn is 65.4):

3.55x10^-4 x 65.4 = 0.023217g

So your answer for mass of Zinc is 0.0232g

**GeorgeL3**)Here are the first set of equations needed.

Half equation for Manganese being reduced:

MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

Half equation for Iron being oxidised:

Fe2+ → Fe3+ + e-

Full equation for manganate oxidising iron:

MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+

The First step now is to calculate the number of moles used to react with the Iron.

Moles of MnO4- in 1st titration: n=cv/1000, n=(0.02 x 18.2)/1000 = 3.64x10^-4 moles.

From the balanced equation we can see this is 5x less than the moles of Iron present, so the moles of iron are:

5 x 3.64x10^-4 = 1.82x10^-3 moles

Now before the second titration, zinc is added so it oxidises the iron(III) to iron(II):

Zn + 2Fe3+ → Zn2+ + 2Fe2+

So now in the second equation, all of the Iron is Iron(II).

So now we do the same process with this second titration:

Moles of MnO4- = (0.02 x 25.3)/1000 = 5.06x10^-4 moles

Moles of Fe(II) = 5 x 5.06x10^-4 = 2.53x10^-3

Now this total amount of Fe(II) includes the moles of Fe(III) in it, so to find the moles of Fe(III) we need to take away the first answer for moles of iron:

Moles of Iron(III) = (2.53x10^-3) - (1.82x10^-3 moles) = 7.1x10^-4 moles

Now from the Iron and Zinc equation you can see that the number of moles of Zinc are half that of Iron(III):

Moles of zinc = (7.1x10^-4 moles) / 2 = 3.55x10^-4 moles

Now you can just use Mass = Moles x RMM (RMM of Zn is 65.4):

3.55x10^-4 x 65.4 = 0.023217g

So your answer for mass of Zinc is 0.0232g

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#511

I have 60/60 for the PSAs and got 108/120 in the Jan exam, so I suppose not that worried about this one, but I do want an A*, and right now I'm having trouble memorising a lot of the titration stuff in the redox part of the syllabus ;( I've always sucked at titrationy-acidy-iodiney-stuff

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#512

(Original post by

Yeah the electrodes are GCSE recap for me; it's something I need to remind myself of. I selected my firm and insurance universities for Biochemistry so it just feels like I really have grades to aim for, even though I've known my firm f or ages.

78 days sounds like ages o.O

I love organics, I always have really.

I've learnt the transition metal colours, but I need to learn some of the explanations as to why they're coloured in solution and I need to practice a couple of equations, but otherwise, it's just stuff to learn-learn-learn.

Goodluck! Nice to see you back, I like learning from the guys on this thread. We'll pull through guys

**Mollymod**)Yeah the electrodes are GCSE recap for me; it's something I need to remind myself of. I selected my firm and insurance universities for Biochemistry so it just feels like I really have grades to aim for, even though I've known my firm f or ages.

78 days sounds like ages o.O

I love organics, I always have really.

I've learnt the transition metal colours, but I need to learn some of the explanations as to why they're coloured in solution and I need to practice a couple of equations, but otherwise, it's just stuff to learn-learn-learn.

Goodluck! Nice to see you back, I like learning from the guys on this thread. We'll pull through guys

That's good!So what grades do you need to get?

I like organics too eventhough its memorising reagents and all its easier if you understand it. Same! I have to clear some doubts on TM and redox then hopefully I will be alright on that

I hope we all get the grades we want on August

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**Hey guys, just a little shout out to anyone whose resitting the Unit 4 Edexcel exam this June...The thread AS01 made is at following link:**

http://www.thestudentroom.co.uk/showthread.php?t=2308998&p=42071 563#post42071563

http://www.thestudentroom.co.uk/showthread.php?t=2308998&p=42071 563#post42071563

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#514

I always get confused about which half-equation to switch to get the overall equation & standard electrode potential of the cell.

I first thought I saw somewhere that you take the higher value of electrode potential and subtract the smaller one which doesn't sound right as that would make all reactions thermodynamically spontaneous.

Will they always give us in the question... what the reactants are & products ? Or give us a full-equation in the exam If so... that would answer my question

Thanks in advance

I first thought I saw somewhere that you take the higher value of electrode potential and subtract the smaller one which doesn't sound right as that would make all reactions thermodynamically spontaneous.

Will they always give us in the question... what the reactants are & products ? Or give us a full-equation in the exam If so... that would answer my question

Thanks in advance

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#515

(Original post by

I always get confused about which half-equation to switch to get the overall equation & standard electrode potential of the cell.

I first thought I saw somewhere that you take the higher value of electrode potential and subtract the smaller one which doesn't sound right as that would make all reactions thermodynamically spontaneous.

Will they always give us in the question... what the reactants are & products ? Or give us a full-equation in the exam If so... that would answer my question

Thanks in advance

**posthumus**)I always get confused about which half-equation to switch to get the overall equation & standard electrode potential of the cell.

I first thought I saw somewhere that you take the higher value of electrode potential and subtract the smaller one which doesn't sound right as that would make all reactions thermodynamically spontaneous.

Will they always give us in the question... what the reactants are & products ? Or give us a full-equation in the exam If so... that would answer my question

Thanks in advance

1. Write the two half equations so that they both show a reduction happening, look up the Ecell values and make sure the equation with the more negative value is on top.

2. Now apply LORA: The reactant (left) on the bottom equation will oxidise the product (right) of the top equation.

3. Write out the full equation, the top, most negative one will be the one which gets switched round.

As well as this, if you're given half cell notation then it's just Right minus Left.

Hope that helped.

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#516

(Original post by

That's good!So what grades do you need to get?

I like organics too eventhough its memorising reagents and all its easier if you understand it. Same! I have to clear some doubts on TM and redox then hopefully I will be alright on that

I hope we all get the grades we want on August

**AS01**)That's good!So what grades do you need to get?

I like organics too eventhough its memorising reagents and all its easier if you understand it. Same! I have to clear some doubts on TM and redox then hopefully I will be alright on that

I hope we all get the grades we want on August

And yeah, it's a lot of memorisation but I like memorising things, it's almost fun when I test myself, although it's not really proving if I'm good at Chemistry or not, it's just a memory game really :P

Yeah, TM's are a bit iffy but thankfully it should be easy to sort out once I fully understand the theory. We've got quite a while left which is good

I've just started looking at a couple of Unit 1 things for my retake which is 23rd of May. What grades do you need?

**posthumus**)

I always get confused about which half-equation to switch to get the overall equation & standard electrode potential of the cell.

I first thought I saw somewhere that you take the higher value of electrode potential and subtract the smaller one which doesn't sound right as that would make all reactions thermodynamically spontaneous.

Will they always give us in the question... what the reactants are & products ? Or give us a full-equation in the exam If so... that would answer my question

Thanks in advance

(Original post by

I have exactly the same problem-

I use Rhs-Lhs, which works sometimes...But I get the answer quite a lot of the time wrong because if you always make the more negative value negative then you will always get a positive value.

They're must be a better way to do it!

**David Tennant**)I have exactly the same problem-

I use Rhs-Lhs, which works sometimes...But I get the answer quite a lot of the time wrong because if you always make the more negative value negative then you will always get a positive value.

They're must be a better way to do it!

Forward Cell + Reversed Cell = your overall Electrode potential

So, if you've got +0.16, and +0.78 electrode potentials, the

**most**positive one is +0.78 which means it

**will**be a reduction reaction. It

**wants**to go that way, so you reverse the +0.16 which makes it -0.16.

+0.78 -0.16(because +/-) makes a - overall = +0.62 which makes it feasible overall. They probably will give you the half equations (ie, the reduction potentials) and you'd be able to write the reduction half equation, and the oxidation half equation, which are simultaneously happening. From that, you'd be able to combine the equations and make a full one.

"Use the following data to predict whether the dichromate ions in acidified potassium dichromate will react with Cl- in HCl"

Cr2O72- + 14H+ + 6e- ----> 2Cr3+ +7H2O (Reduction potential: +1.33)

1/2Cl2 + e- ------> Cl- (Reduction potential: +1.36)

So the chlorine one will be reduced. It is more positive than the chromate half equation.

So you reverse its reduction potential; -1.36 instead of +1.36

You therefore reverse the equation too, so it's 6Cl- ----> 3Cl2 +6e-

**Remember when you are going to combine the equations, you multiply the chlorine one by 6 because the ratio's 1:6, so we need to cancel out the electrons.**which is what I've done above.

So combined, it's Cr2O72- + 14H+ +6Cl- ----> 2Cr2+ +7H2O + 3Cl2

Sorry for the lack of subscript superscript changing!

So E cell is: +1.33

**-**1.36 (as forward cell + reversed cell is the formula and these ones are + -, so it's minus overall) because the sign has been changed.

E Cell is -0.03 which is negative, so it's

**not feasible**

Sorry that was quite longwinded to answer a simple question, hope it helps

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**posthumus**)

I always get confused about which half-equation to switch to get the overall equation & standard electrode potential of the cell.

I first thought I saw somewhere that you take the higher value of electrode potential and subtract the smaller one which doesn't sound right as that would make all reactions thermodynamically spontaneous.

Will they always give us in the question... what the reactants are & products ? Or give us a full-equation in the exam If so... that would answer my question

Thanks in advance

I use Rhs-Lhs, which works sometimes...But I get the answer quite a lot of the time wrong because if you always make the more negative value negative then you will always get a positive value.

They're must be a better way to do it!

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#518

(Original post by

Not sure if they do it in all schools but we use the LORA rule to find out which way round the half equations should go. it stands for Left Oxidises Right Above:

1. Write the two half equations so that they both show a reduction happening, look up the Ecell values and make sure the equation with the more negative value is on top.

2. Now apply LORA: The reactant (left) on the bottom equation will oxidise the product (right) of the top equation.

3. Write out the full equation, the top, most negative one will be the one which gets switched round.

As well as this, if you're given half cell notation then it's just Right minus Left.

Hope that helped.

**GeorgeL3**)Not sure if they do it in all schools but we use the LORA rule to find out which way round the half equations should go. it stands for Left Oxidises Right Above:

1. Write the two half equations so that they both show a reduction happening, look up the Ecell values and make sure the equation with the more negative value is on top.

2. Now apply LORA: The reactant (left) on the bottom equation will oxidise the product (right) of the top equation.

3. Write out the full equation, the top, most negative one will be the one which gets switched round.

As well as this, if you're given half cell notation then it's just Right minus Left.

Hope that helped.

**David Tennant**)

I have exactly the same problem-

I use Rhs-Lhs, which works sometimes...But I get the answer quite a lot of the time wrong because if you always make the more negative value negative then you will always get a positive value.

They're must be a better way to do it!

I don't know really I hope that is the case though ...

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#519

(Original post by

Hey thanks for the reply ! I've seen something like this in the George Facer book... but to me it doesn't seem to always work or make sense Because I've gotten that the general idea is you are switching the half-equation with the most negative electrode potential - but if that is the case then all electrode cell potentials would be positive! (thus all reactions are thermodynamically spontaneous). I must be missing something

I was thinking it might just be what they give you in the question? Since its reversible, one way would be negative (not feasible) & another positive (feasible). So they will ask you to find the electrode potential of the cell where they have stated what way they want you to consider the overall equation ?

I don't know really I hope that is the case though ...

**posthumus**)Hey thanks for the reply ! I've seen something like this in the George Facer book... but to me it doesn't seem to always work or make sense Because I've gotten that the general idea is you are switching the half-equation with the most negative electrode potential - but if that is the case then all electrode cell potentials would be positive! (thus all reactions are thermodynamically spontaneous). I must be missing something

I was thinking it might just be what they give you in the question? Since its reversible, one way would be negative (not feasible) & another positive (feasible). So they will ask you to find the electrode potential of the cell where they have stated what way they want you to consider the overall equation ?

I don't know really I hope that is the case though ...

*will*

occur with the reactants you've got.

The answer should only be negative if for example they ask you to work out whether a specific reaction will occur or give you a full equation to work out the value for.

In that case you have to look at which species is being oxidised/reduced and decide from there which needs the sign swapping.

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#520

(Original post by

Hey thanks for the reply ! I've seen something like this in the George Facer book... but to me it doesn't seem to always work or make sense Because I've gotten that the general idea is you are switching the half-equation with the most negative electrode potential - but if that is the case then all electrode cell potentials would be positive! (thus all reactions are thermodynamically spontaneous). I must be missing something

I don't know really I hope that is the case though ...

**posthumus**)Hey thanks for the reply ! I've seen something like this in the George Facer book... but to me it doesn't seem to always work or make sense Because I've gotten that the general idea is you are switching the half-equation with the most negative electrode potential - but if that is the case then all electrode cell potentials would be positive! (thus all reactions are thermodynamically spontaneous). I must be missing something

I don't know really I hope that is the case though ...

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