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OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD watch

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    (Original post by thad33)
    Does anyone have a nice way of explaining cathodes and anodes? I keep mixing them up. I know the direction that the electrons flow in but everywhere I look to help explain the terminals, they contradict each other.
    Cathode, Positive, Reduction
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    (Original post by VMD100)
    I would say from reading many of the conflicting responses about the F324 paper to say it was easy is not entirely true. If anything I believe many here would have scored higher if they sat the 2015 F324 as it had less of the... well... nonsense like the fats and stitches questions. My chem teacher looked at the 2016 F324 and said he would be quite surprised if the grade boundaries changed too much from last year. As far as a pattern I'm not really sure one can be drawn for the F325 grade boundaries.
    June 2015: A -79
    June 2014: A-74
    June 2013: A-75
    January 2013: A-70
    June 2012: A-76
    January 2012: A-77
    June 2011: A-72
    I agree, there were many sly questions in the f324 paper that I feel a lot, including myself, would have missed out marks on but I think the grade boundaries will be higher than last years but I doubt it'll be in the 50's for an A

    Surprised last years f325 was the highest marks for an A in this unit as I feel there were lots of easier papers than last years, strange
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    (Original post by tcameron)
    I agree, there were many sly questions in the f324 paper that I feel a lot, including myself, would have missed out marks on but I think the grade boundaries will be higher than last years but I doubt it'll be in the 50's for an A
    It won't be in the 50s, albeit a far nicer paper than last year's

    I think people left the exam hall feeling content since it was relatively straight forward in most aspects, i.e. the NMR question

    but lots of opportunities for small mistakes that filter the A/A* students out from the rest mean it wasn't as easy as most people say it was
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    (Original post by HediIsLord)
    It won't be in the 50s, albeit a far nicer paper than last year's

    I think people left the exam hall feeling content since it was relatively straight forward in most aspects, i.e. the NMR question

    but lots of opportunities for small mistakes that filter the A/A* students out from the rest mean it wasn't as easy as most people say it was
    hahah I guess I was one of the few who messed up my nmr question by not checking the number of proton environments - good thing it was only one mark
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    what is a conventional electrochemical cell? I always see it in questions but don't know what it actually is
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    (Original post by tcameron)
    what is a conventional electrochemical cell? I always see it in questions but don't know what it actually is
    Well the half-cells put together are technically an electrochemical cell but I am assuming it is referring to a battery?
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    (Original post by tcameron)
    what is a conventional electrochemical cell? I always see it in questions but don't know what it actually is
    It might be referring to a standard cell like battery which has all it's fuel/chemicals.
    In comparison to a fuel cell, like a hydrogen fuel cell that needs a constant supply of fuel.
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    Does anyone know a method to form balanced redox equations and balanced half-equations, like Question 8.a on this paper.

    Thanks
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  1. File Type: pdf 243458-question-paper-unit-f325-01-equilibria-energetics-and-elements.pdf (516.8 KB, 60 views)
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    I can't explain it in detail now but the trick with redox is to start with last step and work your way up for step 1 using stoichiometry to connect one step to the next via a common compound.

    Without looking at the question I'm pretty sure you'll have to do is calc moles of titrating solution. Usually chromate, manganate, or thiosulfate

    Then do the ratio method to calculate moles on the other product in the last equation. E.g. 2:1 ratio thiosulfate:iodine. So half mols from the paragraph above. Now you can move up a step as you've calculated the moles of, say, the iodine, to continue the example.

    If there is a volume change (commonly 25cm sample from 250cm total) then you times the moles by 10 to account for this.

    It gets pretty self-explanatory from there. Once you can do it fine, F325 final questions (usually redox calc) you'll be able to finish in under a minute.

    (Original post by greenorange)
    Does anyone know a method to form balanced redox equations and balanced half-equations, like Question 8.a on this paper.

    Thanks
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    (Original post by greenorange)
    Does anyone know a method to form balanced redox equations and balanced half-equations, like Question 8.a on this paper.

    Thanks
    Iron starts as Iron (III) Oxide Therefore its formula is likely Fe2O3 and when reacted with a solution containing hydroxide ions forms FeO4 2- Where the oxidation number of the iron is +6.
    So a change from +3 to +6 means it is being oxidised.

    To start with you have to balance atoms

    Fe2O3 + Cl2 + OH- ---> FeO4 2- + 2Cl-
    You need 2 iron atoms on the right hand side making it now
    Fe2O3 +OH- + Cl2 --> 2FeO4 2- + 2Cl-
    Now there are too many oxygens on the right hand side so you have to increase the number of OH-
    Fe2O3 + 5OH- +Cl2 --> 2FeO4 2- +2Cl-
    As there are more hydrogens on the left than right you add water molecules to the right hand side. However each water molecule is H2O so you have to double the amount of OH-
    Fe2O3 +10OH- +Cl2 --> 2FeO4 2- +2Cl- + 5H2O

    Now all the atoms are balanced you have to balance charges.
    The charge of the left hand side is 10- and of the right hand side is 6-
    If you increase the number of Cl- to 6Cl- you will have balanced charges. Then change the number of Cl2 so it balances. Overall you will get
    Fe2O3 + 10OH- + 3Cl2 --> 2FeO4 2- + 6Cl- +5H2O
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    Can anyone explain to me how to work out the lattice enthalpy for a enthalpy change of solution question?
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    (Original post by ShannonD_1697)
    Can anyone explain to me how to work out the lattice enthalpy for a enthalpy change of solution question?
    Typically the lattice enthalpy and enthalpy of solution will be on the same side and the enthalpies of hydration on another.
    The sum of the left = the right
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    (Original post by ShannonD_1697)
    Can anyone explain to me how to work out the lattice enthalpy for a enthalpy change of solution question?
    You have to draw a Born-Haber cycle with gaseous ions at the top, being hydrated, one at a time (X+(g) --> X+(aq)), after hydration of cation and anion you have the dissolved ions.
    Draw from the gaseous ions down to the ionic solid (this is lattice enthalpy) and draw a final arrow from the ionic solid to the aqueous ions. That's how to draw an energy cycle, you can work out missing values if you have all the energy changes bar 1
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    Formation = Sum of everything else
    Therefore
    Formation = Lattice + X
    Therefore
    Lattice = Formation - X

    X is the sum of all the known values (excluding formation obviously)

    Lattice is always negative (hence down arrow). Don't forget to times things like atomisation (and affinity and ionisation if no second affinity/ionisation value provided) by 2 if necessary.

    E.g. MgCl2. You'd times atomisation Cl by 2. Along with the affinity if you're only given first affinity (because they're needs to be two for each ion)

    -

    If you're wondering why it's formation = sum of rest, look at the cycle and you'll see that formation arrow goes on way whilst the other arrows all go in another direction. So different routes for the same value
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    (Original post by Rust Cohle)
    Formation = Sum of everything else
    Therefore
    Formation = Lattice + X
    Therefore
    Lattice = Formation - X

    X is the sum of all the known values (excluding formation obviously)

    Lattice is always negative (hence down arrow). Don't forget to times things like atomisation (and affinity and ionisation if no second affinity/ionisation value provided) by 2 if necessary.

    E.g. MgCl2. You'd times atomisation Cl by 2. Along with the affinity if you're only given first affinity (because they're needs to be two for each ion)

    -

    If you're wondering why it's formation = sum of rest, look at the cycle and you'll see that formation arrow goes on way whilst the other arrows all go in another direction. So different routes for the same value

    Thank-you this makes sense it's always something i forget to do I realise afterwards
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    (Original post by k.russell)
    You have to draw a Born-Haber cycle with gaseous ions at the top, being hydrated, one at a time (X+(g) --> X+(aq)), after hydration of cation and anion you have the dissolved ions.
    Draw from the gaseous ions down to the ionic solid (this is lattice enthalpy) and draw a final arrow from the ionic solid to the aqueous ions. That's how to draw an energy cycle, you can work out missing values if you have all the energy changes bar 1
    Thank-you
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    (Original post by VMD100)
    Typically the lattice enthalpy and enthalpy of solution will be on the same side and the enthalpies of hydration on another.
    The sum of the left = the right
    Thank-you
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    https://goo.gl/mGFVGf

    Share and share alike
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    how to do f325 faster?
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    (Original post by lai812matthew)
    how to do f325 faster?
    Practice.
 
 
 
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