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    http://pmt.physicsandmathstutor.com/...%20A-level.pdf -paper

    http://pmt.physicsandmathstutor.com/...%20A-level.pdf -ms

    I dont understand the fourth and fifth mark for 20c, if someone could kindly help explain it to me, id be very grateful
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    (Original post by hello654321)
    http://pmt.physicsandmathstutor.com/...%20A-level.pdf -paper

    http://pmt.physicsandmathstutor.com/...%20A-level.pdf -ms

    I dont understand the fourth and fifth mark for 20c, if someone could kindly help explain it to me, id be very grateful
    So the marks are for realising that you are in reducing conditions, so you could reduce V(V) to V(IV), then subsequently to V(III)

    To work out which oxidation state you end up with, the mark scheme wants you to go one oxidation state at a time.

    So first you work out the Cell potential for the first reduction, which turns out to be favourable. So you can get to V(IV) in these conditions.

    Next you want to see if Iodide can further reduce V(IV) to V(III).

    So mark 4 is for calculating the Cell potential for this second reduction.

    mark 5 is then for saying realising the meaning of your calculations, ie, Iodide is a good enough reducing agent to reduce V(V) to V(IV), but is not able to reduce V(IV) to V(III) so the reaction with results in a solution containing V(IV).
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    (Original post by MexicanKeith)
    So the marks are for realising that you are in reducing conditions, so you could reduce V(V) to V(IV), then subsequently to V(III)

    To work out which oxidation state you end up with, the mark scheme wants you to go one oxidation state at a time.

    So first you work out the Cell potential for the first reduction, which turns out to be favourable. So you can get to V(IV) in these conditions.

    Next you want to see if Iodide can further reduce V(IV) to V(III).

    So mark 4 is for calculating the Cell potential for this second reduction.

    mark 5 is then for saying realising the meaning of your calculations, ie, Iodide is a good enough reducing agent to reduce V(V) to V(IV), but is not able to reduce V(IV) to V(III) so the reaction with results in a solution containing V(IV).
    Thanks for the explanation! So just to be clear, you have to use all the equations with V to see what is feasible so you know how far I- can reduce??
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    (Original post by hello654321)
    Thanks for the explanation! So just to be clear, you have to use all the equations with V to see what is feasible so you know how far I- can reduce??
    Well the question asks what you will end up with, so you have to explore all the possibilities!
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    (Original post by MexicanKeith)
    Well the question asks what you will end up with, so you have to explore all the possibilities!
    Okay thank you so much !
 
 
 
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