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AQA - Unit 5 - Energetics, Redox and Inorganic Chemistry watch

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    (Original post by choc1234)
    theres barely any synoptic in unit 5 chem ! its not like bio lol
    i dont think there was any in the jan11 paper
    i probs wont revise any synoptic
    Yeah, I think it's just AS periodicity and energetics that could come up. They usually have all the synoptic stuff in Chem 4.
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    anyone got the jan 11 paper please?
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    does anyone have answers to the NT book?
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    hey guys, good luck for the exam on friday
    just have a little question to ask, if anyone could help, i would be grateful!

    in redox reactions, when do you know when to flip one of the half equations around, when combining? because in some cases one is flipped, in other cases they are combined like a normal redox equation, and i'm confused as to how this comes about... thanks!
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    (Original post by Rapidfire8)
    hey guys, good luck for the exam on friday
    just have a little question to ask, if anyone could help, i would be grateful!

    in redox reactions, when do you know when to flip one of the half equations around, when combining? because in some cases one is flipped, in other cases they are combined like a normal redox equation, and i'm confused as to how this comes about... thanks!
    you only reverse the reactions when looking at spontaneous cell reactions , meaning , to see if they are feasible the question will usuallly specify by saying spontaneous or feeasible.
    you dont flip them aroound when your looking for a overall reaction
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    (Original post by Willyg11)
    6+ for Cr2O72-

    3+ for 2Cr3+ but counts as +6 due to the '2'

    don't understand your last question
    sorry where does 6e- come from i knw its silly que. but this is the only bit :confused:
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    (Original post by jimmy303)
    It's used when there isn't a solid involved.

    E.g. Zn2+(aq) + 2e- ---> Zn(s)

    doesn't need platinum as it has a solid (Zn) which acts as the electrode so it becomes:

    Zn(s)|Zn2+(aq) ||

    However,

    Fe3+(aq) + e- ---> Fe2+(aq)

    there is no solid involved here so a platinum electrode is needed to transfer electrons, and the cell becomes:

    Pt | Fe2+(aq), Fe3+(aq) ||

    So if you see a half cell equation with no solid involved, chuck in platinum

    It can be left or right depending on if the half cell without the solid is more negative or more positive, and can appear on both if neither half cell involves a solid as it does with the hydrogen fuel cell:

    Pt|H2(g)|OH-(aq), H2O(l) || H2O(l) | O2(g), OH-(aq)|Pt

    Hope this helped
    I thought this "Pt|H2(g)|OH-(aq), H2O(l) || H2O(l) | O2(g), OH-(aq)|Pt" should be.. Pt|H2(g)|OH-(aq), H2O(l) || O2(g) | H20(l), OH-(aq)|Pt
    Because the O2 is in a different phase to the water and hydroxide...
    Or maybe im wrong
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    (Original post by ben10)
    does anyone have answers to the NT book?
    there you go http://www.thomas-reddington.com/chemistry-answers.html
    unit 4 n 5 enjoy
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    (Original post by ben10)
    you only reverse the reactions when looking at spontaneous cell reactions , meaning , to see if they are feasible the question will usuallly specify by saying spontaneous or feeasible.
    you dont flip them aroound when your looking for a overall reaction
    thanks man
    so just judging by the question, i should be able to know?
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    (Original post by Rapidfire8)
    thanks man
    so just judging by the question, i should be able to know?
    yes . it wil say either
    overall reaction ( where you dont flip)


    or find the spontanoues cell reaction/ is this reaction feasible / wil a particular species oxidise/reduce somthing ( this is where u flip)
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    Is everyone just memorizing the tons of equations from chapter 15/16? Cos I don't know what else to do about them. :p:
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    (Original post by al_habib)
    sorry where does 6e- come from i knw its silly que. but this is the only bit :confused:
    Start with: Cr2O72- -> 2Cr3+ (Cr's balanced)

    Balance oxygens by adding 7 H20 to right

    Balance Hydrogens by adding 14 H+ to the left

    The charge on the left is +12, the charge on the right is +6, so to make it +6 on both sides, add 6e- to the left, giving:

    Cr2O72- + 14H+ + 6e- -> 2Cr3+ +7H2O

    Hope this helps
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    can some one explain to me how to find the charge outside the bracket, {Cr(NH3)6}3+ ?

    for {Cr(OH)6}3- i guess cr is 3+ and OH is -6 :. 3-6=-3 am i right?
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    (Original post by Sparkly-Star)
    Is everyone just memorizing the tons of equations from chapter 15/16? Cos I don't know what else to do about them. :p:
    I was thinking that, do we really have to remember every single one?
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    (Original post by Willyg11)
    Start with: Cr2O72- -> 2Cr3+ (Cr's balanced)

    Balance oxygens by adding 7 H20 to right

    Balance Hydrogens by adding 14 H+ to the left

    The charge on the left is +12, the charge on the right is +6, so to make it +6 on both sides, add 6e- to the left, giving:

    Cr2O72- + 14H+ + 6e- -> 2Cr3+ +7H2O

    Hope this helps
    nice one
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    (Original post by Rapidfire8)
    I was thinking that, do we really have to remember every single one?
    I guess so. So many to remember. :cry2:
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    Any tips on how to achieve an a* in this unit!!
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    (Original post by al_habib)
    can some one explain to me how to find the charge outside the bracket, {Cr(NH3)6}3+ ?

    for {Cr(OH)6}3- i guess cr is 3+ and OH is -6 :. 3-6=-3 am i right?
    Yes, each OH has a charge of -1......

    NH3 doesn't have a charge, Cr has charge of 3+ on this occsion but can have 0, +2, +3 and +6.
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    Do we need to know specific details of the fuel cell? I just read over it. =/
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    (Original post by Ebony_x)
    I thought this "Pt|H2(g)|OH-(aq), H2O(l) || H2O(l) | O2(g), OH-(aq)|Pt" should be.. Pt|H2(g)|OH-(aq), H2O(l) || O2(g) | H20(l), OH-(aq)|Pt
    Because the O2 is in a different phase to the water and hydroxide...
    Or maybe im wrong
    I thought the same thing but let it slide since how I wrote it is how it was stated in the Jan 10 mark scheme.

    I've looked into it and I can't find a single mention of the conventional representation of a hydrogen fuel cell anywhere in any text book or online source :confused: Don't know if it's an AQA mess up or we're missing something.
 
 
 
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