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    xo_brookie's Avatar
    xo_brookie Posts: 1, Reputation: 1
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    #1

    May 20, 2009, 04:14 PM
    Predicting the product
    Ag+CuSO4
    Perito's Avatar
    Perito Posts: 3,139, Reputation: 150
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    #2

    May 20, 2009, 08:17 PM

    You can check out the standard potentials for the half reactions



    and



    to see if the reaction has any potential to make it go.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #3

    May 21, 2009, 08:22 AM

    Hummm, I don't about potentials, perhaps you'll tell me Perito.

    What I have to say here is that a more reactive metal will 'displace' (or take the place if you understand better that way) another less reactive metal from a solution of that less reactive metal salt. So, if Ag is more reactive than Cu, then, a reaction will occur.
    Perito's Avatar
    Perito Posts: 3,139, Reputation: 150
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    #4

    May 21, 2009, 08:50 AM

    from Applications of Standard Electrode Potentials

    Cu2+(aq) + 2e- --> Cu(s) 0.34 V

    from Foundations of Electronics - Standard Potentials

    Ag+(aq) + e- --> Ag(s) +.80 V

    Reversing the silver half-reaction (since the reaction is written Ag -> Ag+) gives -0.80v. Adding that to the +.34v gives -.46v. This indicates a lack of driving force (Gibbs free energy) for the reaction to occur as stated (Ag+CuSO4). In fact the reverse reaction, Cu + Ag+ = Cu+2 + Ag, is commonly done in quantitative analysis chemistry laboratories.

    In other words, there will probably be little or no reaction.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #5

    May 21, 2009, 08:54 AM

    Oh OK! I got it! Thanks Perito. I forgot that that formed part of electrolysis too! I remember now, though I doubt I was told of the terms 'electrode potential' in the chapter. :)

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