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    (Original post by Simonon)
    The emf of the reaction "VO2+(aq) + 2H+(aq) + e– → VO2+(aq) + H2O(l)" is less than the one for MnO4- so V2+ makes V3+ which makes VO(2+) which makes VO2(+). The MnO4- must be reduced.

    VO2+ +5 + -4 = +1

    And the equation should be pretty straight forward once you know the final species because you just balance the oxygen and water etc.

    Hope that helped because that first part is always hard to explain!
    But it says it was added to a solution containing V^{2+}, then wouldn't you have to use the first equation making the species V^{3+}?
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    (Original post by brittanna)
    But it says it was added to a solution containing V^{2+}, then wouldn't you have to use the first equation making the species V^{3+}?
    Yeah because the MnO4- species has a higher emf it is reduced and the other is oxidised so the equation is reversed. Bare in mind that they're all written as reductions in the electrochemical series.
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    (Original post by Jesss9)
    Same I'm doing chem 2, 4 and 5 but I'm worried I'm leaving chem 5 too late
    Ouch that's a lot of chem exams! I'm planning to start after my next exam which is tommorow though!
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    (Original post by Simonon)
    Yeah because the MnO4- species has a higher emf it is reduced and the other is oxidised so the equation is reversed. Bare in mind that they're all written as reductions in the electrochemical series.
    Sorry, I still don't get it .

    The way I see it is as you add it into a solution containing V^{2+} ions, the two equations must be

    V^{2+} \rightarrow V^{3+}+e^- and

    MnO_4^{-}+8H^++5e^- \rightarrow Mn^{2+}+4H_2O

    which would the produce an overall equation of

    5V^{2+}+MnO_4^{-}+8H^+\rightarrow 5V^{3+}+Mn^{2+}+4H_2O

    This then gives V^{3+} as the final Vanadium species and so I can't see where the VO_2^{+} is coming from.
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    (Original post by brittanna)
    Sorry, I still don't get it .

    The way I see it is as you add it into a solution containing V^{2+} ions, the two equations must be

    V^{2+} \rightarrow V^{3+}+e^- and

    MnO_4^{-}+8H^++5e^- \rightarrow Mn^{2+}+4H_2O

    which would the produce an overall equation of

    5V^{2+}+MnO_4^{-}+8H^+\rightarrow 5V^{3+}+Mn^{2+}+4H_2O

    This then gives V^{3+} as the final Vanadium species and so I can't see where the VO_2^{+} is coming from.
    In the same table V3+ can be oxidised to VO2+ because the emf is still less than that of MnO4- and you carry on until you can no longer make another product.
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    (Original post by Simonon)
    In the same table V3+ can be oxidised to VO2+ because the emf is still less than that of MnO4- and you carry on until you can no longer make another product.
    Thank you!
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    From the Chem5 jan 12 paper, question 8.e:

    Why doesn't the diaminoethane form a co-ordinate bond with the aluminium ion, yet it does with the cobalt ion in the following question?
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    (Original post by jimmy_95)
    From the Chem5 jan 12 paper, question 8.e:

    Why doesn't the diaminoethane form a co-ordinate bond with the aluminium ion, yet it does with the cobalt ion in the following question?
    It can. It told you diaminoethane can act as a ligand or a base. It asked for the equation where it acts as a base.
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    Little bit of help, for question 8e)ii), Jan 2012, it says there are various ways to balance the oxidation of the diaminoethane complex.

    Is this one?

    4[Co(en)3]^2+ + O2 --> 4[Co(en)3]^3+ +2O^2-

    That's what I did, just want to know if it's OK.
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    (Original post by Chris-69)
    It can. It told you diaminoethane can act as a ligand or a base. It asked for the equation where it acts as a base.
    The question DOES NOT specify whether the diaminoethane acts as a base or a ligand, nor does it in the following question. It just states the fact that it could.
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    (Original post by jimmy_95)
    The question DOES NOT specify whether the diaminoethane acts as a base or a ligand, nor does it in the following question. It just states the fact that it could.
    Fair point. I think you were suppose to make the assumption they wanted the reaction where it acted as a base since the reaction was with an acid and also that it asked for the appearance of the product. I suppose 'colourless solution' could have been the appearance for the substitution reaction but, since they provide you with an acid, I'd again assume they're asking for an acid-base reaction.

    You know it can act as a ligand, what would be the point of mentioning it can act as a base if that wasn't the point of the quesiton?
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    (Original post by jimmy_95)
    From the Chem5 jan 12 paper, question 8.e:

    Why doesn't the diaminoethane form a co-ordinate bond with the aluminium ion, yet it does with the cobalt ion in the following question?
    The important thing in this question is where it says LIKE AMMONIA so you treat it as you would ammonia acting as a base.

    Posted from TSR Mobile
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    (Original post by Chris-69)
    Fair point. I think you were suppose to make the assumption they wanted the reaction where it acted as a base since the reaction was with an acid and also that it asked for the appearance of the product. I suppose 'colourless solution' could have been the appearance for the substitution reaction but, since they provide you with an acid, I'd again assume they're asking for an acid-base reaction.

    You know it can act as a ligand, what would be the point of mentioning it can act as a base if that wasn't the point of the quesiton?
    I see what you mean, I put it down to to a little negligence on AQA's part.

    As for your equation in 8e)ii), I also gave the same answer when I did the paper earlier and only awarded myself a single mark. You should gain the mark for having the correct Co(III) species but i'm not sure whether oxygen can be shown to be reduced like that. I remember in a similar question in another past paper where it said you need to show H20 or H+ in the equation when O2 is being reduced. But i'm not entirely sure if you'd lose the mark for definite. But in future just add the H20 or H+ to guarantee the mark.
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    (Original post by Chewy29)
    The important thing in this question is where it says LIKE AMMONIA so you treat it as you would ammonia acting as a base.

    Posted from TSR Mobile
    That makes sense, thanks.
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    (Original post by jimmy_95)
    I see what you mean, I put it down to to a little negligence on AQA's part.

    As for your equation in 8e)ii), I also gave the same answer when I did the paper earlier and only awarded myself a single mark. You should gain the mark for having the correct Co(III) species but i'm not sure whether oxygen can be shown to be reduced like that. I remember in a similar question in another past paper where it said you need to show H20 or H+ in the equation when O2 is being reduced. But i'm not entirely sure if you'd lose the mark for definite. But in future just add the H20 or H+ to guarantee the mark.
    Oh right, thanks. Will keep that in mind.
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    what exactly is the cisplatin structure? it seems that the aqa nt book mixes it up with trans and im not sure which it is?
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    (Original post by gingerandice)
    what exactly is the cisplatin structure? it seems that the aqa nt book mixes it up with trans and im not sure which it is?
    cisplatin is square planar structure. It involves two chloride ions and two ammonia molecules attached to a platinum ion via coordinate bonding.

    Looks like this: http://upload.wikimedia.org/wikipedi...splatin-2D.png
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    What are the reactions we need to know for Hydrogen Peroxide?
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    why when drawing the conventional representation we dont include water or h+ ions?
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    when working out the temperature at which a reaction is feasible does it matter if the reaction is endothermic or exothermuc? is it always the region higher than the temperature that uve calculated for which the reaction is feasible for?
 
 
 
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