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    (Original post by Zakir)
    WOW :eek:

    thanks alot lol

    its starting to make a bit more sense...

    so when the equilibrium shifts to the right, does the emf ALWAYS increase?

    and when it shifts to the left, does the emf always decrease??

    thanks
    I checked all the questions I came across in the booklet of exam questions on http://www.a-levelchemistry.co.uk/AQ...ria%20home.htm

    And the short answer is yes.

    But don't get caught out on this. Whilst it's true that when the reaction is written out in full and the equilibrium shifts to the right an increase in emf is the result, the standard way of writing half cells is as a reduction reaction so if you've not had the opportuntiy to swap it around it may be the opposite.

    E.g.

    Mg2+(aq) + 2e– --->Mg(s) –2.37
    Fe2+(aq) + 2e– ----->Fe(s) –0.44

    Deduce how the e.m.f. of the cell Mg(s)|Mg2+(aq)||Fe2+(aq)|Fe(s) changes when the concentration of Mg2+ is decreased. Explain your answer.

    The magnesium half cell is more negative so would be swapped to

    Mg(s) --> Mg2+(aq) + 2e-

    Thus you can see a decrease in Mg2+ would cause the equilibrium to shift right and increase the emf.

    BUT, the question hasn't given you the opportunity to swap the reaction around, so the markscheme says that the equilibrium shifts to the left, increasing emf as it is referring to the original equation Mg2+(aq) + 2e– --->Mg(s).

    That contradicts my explanation that the arrow points towards equilibrium, since with Mg2+(aq) + 2e– --->Mg(s) the equilibrium is far on the left, but that's because the half equation hasn't been swapped.

    Therefore I would suggest that if such a question came up, convert the half reaction to what's actually happening and refer to the old half equation to ensure you get the mark since sometimes markschemes are kind, sometimes not.

    So it would be:

    In the reaction Mg(s) --> Mg2+(aq) + 2e-
    a decrease in Mg2+ would cause the equilibrium to shift right
    (or shift left in the original Mg2+(aq) + 2e– --->Mg(s) reaction)
    Thus emf becomes more positive.

    For an added point you could mention that magnesium's electrode potential would become more negative (deduced from the fact more electrons are given off when the equilibrium shifts) since this is what really causes emf to be more positive.

    Emf = Epositive - Enegative

    Wow, yeah....

    My advice if you didn't quite get that (neither do I tbh) is to go through the exam booklet and do the questions based on a change in concentration so that you get your head around it.

    P.S. I'm really sorry for all the long replies lol, it's just that this is my form of revision and I don't feel like I've done it justice unless I've explored every nook and cranny
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    (Original post by jimmy303)
    x
    If you don't get at 100% in this paper, I'll be shocked. lol

    I think you should definitely take a chemistry based degree at university.

    Good luck with everything
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    (Original post by T-Toe)
    What was wrong with chem4?
    Just heard it was quite a bit harder than January, which itself wasn't straightfoward, guess we'll find out soon enough
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    (Original post by Y.D)
    Just heard it was quite a bit harder than January, which itself wasn't straightfoward, guess we'll find out soon enough
    I though January's paper was harder.

    Yep, I'm bricking it :afraid:
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    hey, could anyone help me with this question, i know the answer i just dont know why lol:

    Write a half-equation for the formation of hydrogen and NH2 minus ions from ammonia.
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    For periodicity questions, if there's a question where you add an acid or base to a metal oxide, can you just do H+ or OH-?
    For example, sodium oxide + nitric acid, you could write Na2O + 2H+ > 2Na+ + H20?
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    (Original post by flowersandtrees)
    For periodicity questions, if there's a question where you add an acid or base to a metal oxide, can you just do H+ or OH-?
    For example, sodium oxide + nitric acid, you could write Na2O + 2H+ > 2Na+ + H20?
    Not really, because it's not Na+ that is produced, it would be NaCl or NaNO3 etc.
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    (Original post by Seasick Steve)
    Not really, because it's not Na+ that is produced, it would be NaCl or NaNO3 etc.
    That's what I thought, but then in some of the mark schemes in the nelson thornes book it says you can?
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    guys how do you identify reducing agent and write half equation for this reaction

    5S2O + Br2 + 6H2O ? 2BrO + 12H+ + 10SO

    its a bit hard coz you dont know the E values which makes life harder, any help would be appreciated.
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    (Original post by ben10)
    is it possible to revise chem5 in a week?
    I am learning BIO5 and CHM5 in a week, impossible, but I'm trying!
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    Can anyone help me on 3bi of the specimen paper pls (redox equilibria question)? It seems pretty basic but I'm doing something wrong...

    Thanks.. and i'll defo +rep
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    (Original post by englishman129)
    Can anyone help me on 3bi of the specimen paper pls (redox equilibria question)? It seems pretty basic but I'm doing something wrong...

    Thanks.. and i'll defo +rep
    This one caught me out for ages too. I could have sworn it was a typo error on the mark scheme

    You start with MnO4- and V2+.

    The electrode potential of a V2+ half cell is more negative so you'd assume MnO4- would oxidise V2+ into V3+ and that would be the end of the story.

    However, if you look down the table, there is also a half cell equation for V3+.
    The electrode potential is more negative than the electrode potential of MnO4- again so V3+ is oxidised by MnO4- to VO(2+)

    Looking down the table further there's another half cell equation for VO(2+). This is also more negative than MnO4- so VO(2+) is oxidised by MnO4- into VO2(+)

    There are no more vanadium species between VO2(+) and MnO4- so this is the final species.

    Hope this helps!

    Edit: Just realised a key aspect of this question is the phrase 'excess of potassium manganate'. Thus any products of the first reaction can be further oxidised as there is still MnO4- to react with.
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    (Original post by jimmy303)
    This one caught me out for ages too. I could have sworn it was a typo error on the mark scheme

    You start with MnO4- and V2+.

    The electrode potential of a V2+ half cell is more negative so you'd assume MnO4- would oxidise V2+ into V3+ and that would be the end of the story.

    However, if you look down the table, there is also a half cell equation for V3+.
    The electrode potential is more negative than the electrode potential of MnO4- again so V3+ is oxidised by MnO4- to VO(2+)

    Looking down the table further there's another half cell equation for VO(2+). This is also more negative than MnO4- so VO(2+) is oxidised by MnO4- into VO2(+)

    There are no more vanadium species between VO2(+) and MnO4- so this is the final species.

    Hope this helps!

    Edit: Just realised a key aspect of this question is the phrase 'excess of potassium manganate'. Thus any products of the first reaction can be further oxidised as there is still MnO4- to react with.
    Ohhhh right i seeee! That defiantly caught me out. I get it now!

    Thanks a lot!
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    Out of interest - how did everyone score in the spec paper? I got 80/100 ish just now..
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    (Original post by englishman129)
    Out of interest - how did everyone score in the spec paper? I got 80/100 ish just now..
    i did crap and got 73
    on the jan 11 i got 94
    on the june 10 i got 91
    and jan 10 i never did in one whole go
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    (Original post by alicetalbot)
    I am learning BIO5 and CHM5 in a week, impossible, but I'm trying!
    the same, hopeful it will work out coz it makes you go crazy esp synoptic in bio
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    (Original post by Sheldon)
    i did crap and got 73
    on the jan 11 i got 94
    on the june 10 i got 91
    and jan 10 i never did in one whole go
    a* thats cool, how many past papers have you done.
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    hey guys is it possible for me to get an A star in chemistry
    i got 104 UMS in chem 4
    assuming i got an A in my EMPA ?
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    (Original post by al_habib)
    a* thats cool, how many past papers have you done.
    Not that many really a few old style ones, but tehy are cluttered with unit 4 and 2 questions in them
    Just been going over notes.
    If any one wants anything explained please ask it would help me as well
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    Hey just wandering why isn't EDTA poisonous? It is used as an antidote to metal poisoning. But i don't understand why it wouldn't substitute the nitrogens and oxygen in haemoglobin as they are unidentate ligands, so substituting for EDTA (a multidentate) would have an increase in entropy??
 
 
 
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