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# Aqa chem 4/ chem 5 june 2016 thread

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1. i believe in 2014 past paper, it asked for the equations without giving you them in a table
(Original post by Suits101)
Strictly speaking, you don't need to know them.

In the exam they will tell you which equations to use, such as 'write the half equation at the hydrogen electrode in a hydrogen-oxygen fuel cell in acidic/alkaline conditions' or if they don't then you can use either one.

I say you don't need to learn them because you need to know what happens at the electrodes, e.g you know hydrogen is a reactant so you can deduce how to get to hydroxide ions (if alkaline) or hydrogen ions (if acidic) and you know oxygen is a reactant so you can deduce how to get to hydroxide ions (if alkaline) or hydrogen ions (if acidic) once again.
2. (Original post by chzm)
can someone explain to me why when writing out cell representations using the SHE you always put H2 | H+ on the left hand side even if it has an E value more positive than the other species?? e.g. i did a question where i had to write the cell representation for Fe2+ being reduced to Fe (s) using SHE but it had an E value of -1.63 (or something like that)

but surely SHE= 0V which is more positive than -1.63 so it should be written on the right?
I had noticed this and I have no idea.

I ALWAYS, since this question and as my notes say, put the SHE on left.
3. Can someone help me figure with oxygen reaction to use for hydrogen-oyygen in alkaline conditions for 3c and 3d please thanks http://filestore.aqa.org.uk/subjects...W-QP-JAN10.PDF
4. (Original post by chzm)
thank you !! how did you know m = -ΔS x = T
could they ever ask us to work out ΔH using a graph like this, and if so,how would u do it?
thanks once again!
compare the equation y = mx+c with ΔG = ΔH - TΔS. The graph given in the question has the gibbs free enrgy on the y axis so that's y and the temperature on the x axis so that is x. so now we need to know what is m and what is c. You should know that ΔH is a constant for a reaction so that will be c and so that means that m = -ΔS (The minus because the line is a negative gradient)

To calculate ΔH. We know that it's C in the y = mx + c. If you know your straight line equations then you should recall that the c value is the y-intercept. So to find ΔH you find where the line crosses the y axis and read of the Y axis.

Another way of looking at it, is looking at the equation ΔG = ΔH - TΔS. When the temperiature is 0 kelvin, ΔG = ΔH. So then you find where the temp is zero on the x axis and read the corresponding y value. This exactly the same as the y - intercept.
5. unit 5 cramming begins
6. (Original post by flylikeafly)
Can someone help me figure with oxygen reaction to use for hydrogen-oyygen in alkaline conditions for 3c and 3d please thanks http://filestore.aqa.org.uk/subjects...W-QP-JAN10.PDF
Firstly you have to select the alkaline half equations which are equations 2 and 4 in the table, which are:

O2 + 2H2O + 4e- ==> 4OH-
H2 + 2OH- ==> 2H2O + 2e-

The hydrogen electrode goes on the left and the oxygen electrode goes on the right. Remember, the species with the lowest oxidation states go on the outside while species with higher oxidation states go on the inside.

The oxidation state of hydrogen in H2 is 0, the oxidation state of hydrogen in OH- and H2O is +1, therefore H2 will be on the outside while OH- and H2O will go on the inside.

The oxidation state of oxygen in O2 is 0, the oxidation state of oxygen in H2O and OH- is -2, therefore O2 will be on the inside while H2O and OH- will be on the outside.

Therefore the conventional representation for an alkaline hydrogen-oxygen fuel cell is:

Pt(s) | H2(g) | OH-(aq) | H2O(l) || O2(g) | OH-(aq) | H2O(l) | Pt(s)
7. SHE is ALWAYS on the left hand side. All of the textbooks make this point quite prominent.
8. I have a c3 exam on tuesday, chem 5 wednesday, bio 5 Thursday & c4 on friday next week and I havent revised for most of them. C3 i just need to do past papers, chem 5 i literally just started revising on friday and im currently still doing chem 5, im on thermo and I still have transition metals to do. i wanted to do c3 papers tomorrow but now I cant bc i'll have to run chem into tomorrow. I need to re-revise bio 5, I havent done any past papers for that and theres a fricking 25 marker at the end....And c4 I still have integration left to do and I completely forgot vectors. And then i have s2 the finall exam on monday which I also havent revised for !! Worst thing is im in a private college and im paying 6k+ so if i flop my final year my mom will kill me. Im actually so stressed and its annoying bc the teachers at my college finished the content for everything so late !! 😭😭😭 im gonna do so bad im gonna cry
9. (Original post by Jpw1097)
Firstly you have to select the alkaline half equations which are equations 2 and 4 in the table, which are:

O2 + 2H2O + 4e- ==> 4OH-
H2 + 2OH- ==> 2H2O + 2e-

The hydrogen electrode goes on the left and the oxygen electrode goes on the right. Remember, the species with the lowest oxidation states go on the outside while species with higher oxidation states go on the inside.

The oxidation state of hydrogen in H2 is 0, the oxidation state of hydrogen in OH- and H2O is +1, therefore H2 will be on the outside while OH- and H2O will go on the inside.

The oxidation state of oxygen in O2 is 0, the oxidation state of oxygen in H2O and OH- is -2, therefore O2 will be on the inside while H2O and OH- will be on the outside.

Therefore the conventional representation for an alkaline hydrogen-oxygen fuel cell is:

Pt(s) | H2(g) | OH-(aq) | H2O(l) || O2(g) | OH-(aq) | H2O(l) | Pt(s)
Thank you sooo much
10. Does anyone know how to do this please???
11. (Original post by Chembio123)
Does anyone know how to do this please???
find dy/dx so find two the difference between two different values of y then divide it by the difference between the corresponding values of x
The units are the y units divided by the x units
Then delta S is derived from the formula of deltaG= deltaH-TdeltaS using the same as above
12. (Original post by Ohnis)
find dy/dx so find two the difference between two different values of y then divide it by the difference between the corresponding values of x
The units are the y units divided by the x units
Then delta S is derived from the formula of deltaG= deltaH-TdeltaS using the same as above

Excuse how awful I am at PHYSICS, but I found rise over run and got 0.096 where am I going wrong? Like I dunno what to do from there? Paper is June 12 Q2a
13. (Original post by Chembio123)
Excuse how awful I am at PHYSICS, but I found rise over run and got 0.096 where am I going wrong? Like I dunno what to do from there? Paper is June 12 Q2a
As an example, use the y value at 700 and 460 to get 23.5-0/(700-460) = 0.0979
What AQA have done like the f**kers they are; they've made the line just under 24 so most people will mistake it for 24 when it's actually 23.5
Easy mistake to make, you'd have done it right otherwise, I did the same thing the first time I did that paper
14. (Original post by Ohnis)
As an example, use the y value at 700 and 460 to get 23.5-0/(700-460) = 0.0979
What AQA have done like the f**kers they are; they've made the line just under 24 so most people will mistake it for 24 when it's actually 23.5
Easy mistake to make, you'd have done it right otherwise, I did the same thing the first time I did that paper
Ah thanks😎😎
15. What does this mean? The writing in the boxes?

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16. (Original post by Ohnis)
As an example, use the y value at 700 and 460 to get 23.5-0/(700-460) = 0.0979
What AQA have done like the f**kers they are; they've made the line just under 24 so most people will mistake it for 24 when it's actually 23.5
Easy mistake to make, you'd have done it right otherwise, I did the same thing the first time I did that paper
When I did this paper, I did it on A5. This makes the graph harder to read but when you did this paper (I assume on A4) was it easy to see it as 23.5 instead of 24? I made that mistake.

(Original post by Bloom77)
What does this mean? The writing in the boxes?

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I can't explain it to you sorry, but in terms of knowing it for the exam you shouldn't worry about it!
17. (Original post by Suits101)
When I did this paper, I did it on A5. This makes the graph harder to read but when you did this paper (I assume on A4) was it easy to see it as 23.5 instead of 24? I made that mistake.
It's barely noticeable, you can just about see that it doesn't touch the 24 so the best option is to presume that the line is halfway between 23 and 24; it's a horrid question
18. (Original post by Ohnis)
It's barely noticeable, you can just about see that it doesn't touch the 24 so the best option is to presume that the line is halfway between 23 and 24; it's a horrid question
I can barely make it out on A5 so I assume A4 would be a little bit better. I liked the question because it required a little thought, but hated this accuracy aspect! Thanks!
19. Is the general consensus that CHEM5 will be an OK-ish paper or do you reckon AQA will pull a horrible one out of the bag?

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20. (Original post by 26december)
Is the general consensus that CHEM5 will be an OK-ish paper or do you reckon AQA will pull a horrible one out of the bag?

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With maths, C3 was horrible but C4 was a standard paper but still difficult

I expect the same with CHEM5.

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