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E0 for Cu+ => Cu2+ + 1e- is -0,15V

  1. Jul 17, 2009 #1
    Given E[tex]^{0}[/tex]=0,52V for

    Cu[tex]^{+}[/tex](aq) + e[tex]^{-}[/tex] --> Cu(s)

    Calculate E[tex]^{0}[/tex] for the following reaction at 25ºC, 1atm.

    2Cu[tex]^{+}[/tex](aq) --> Cu[tex]^{2+}[/tex](aq) + Cu(s)

    Any ideas?
     
    Last edited by a moderator: Jul 17, 2009
  2. jcsd
  3. Jul 18, 2009 #2
    Re: Electrochemics

    Is that all the data you're given? Don't you need the E[tex]^{0}[/tex] for Cu+ => Cu2+?
     
  4. Jul 19, 2009 #3
    Re: Electrochemics

    While doing these exercises we can check the standart potential reduction table.

    E0 for Cu+ => Cu2+ + 1e- is -0,15V

    What would you do then?
     
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