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    We weren't really taught this well so now I've come to revise it and I'm failing miserably.
    I'm trying to do this worksheet:
    Name:  uploadfromtaptalk1365089064843.jpg
Views: 62
Size:  87.5 KB

    so far I've got:
    1) zero; as it can be used to compare other reactions to see if they are a reduction or oxidation

    2) Zn (s) + Cd^2+ (aq) -> Zn^2+ (aq) + Cd (s)

    3) -0.4--0.76=0.36v (is this right?)

    4) I'm stuck. Is it just -0.34--0.26? It asks about reducing Cu^2- to copper but the table shows values for Cu^2+ :dontknow:

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    (Original post by Squirb)
    We weren't really taught this well so now I've come to revise it and I'm failing miserably.
    I'm trying to do this worksheet:
    Name:  uploadfromtaptalk1365089064843.jpg
Views: 62
Size:  87.5 KB

    so far I've got:
    1) zero; as it can be used to compare other reactions to see if they are a reduction or oxidation
    correct

    2) Zn (s) + Cd^2+ (aq) -> Zn^2+ (aq) + Cd (s)
    correct

    3) -0.4--0.76=0.36v (is this right?)
    correct

    4) I'm stuck. Is it just -0.34--0.26? It asks about reducing Cu^2- to copper but the table shows values for Cu^2+ :dontknow:

    Posted from TSR Mobile
    If the reaction were feasible vanadium(II) would be the oxidised state and Cu(II) the reduced state.

    E = E(red)-E(ox) = +0.34 -- 0.26 = +0.60 V

    This is positive and greater then 0.3V, so the reaction is spontaneous (feasible) and goes to completion...
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    (Original post by charco)
    If the reaction were feasible vanadium(II) would be the oxidised state and Cu(II) the reduced state.

    E = E(red)-E(ox) = +0.34 -- 0.26 = +0.60 V

    This is positive and greater then 0.3V, so the reaction is spontaneous (feasible) and goes to completion...
    Thank you very much! I think I was just over-complicating the last one

    Posted from TSR Mobile
 
 
 
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