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Electrochemisty Concept- Help Plz

  1. May 1, 2005 #1
    If E(cell) = E(cathode)-E(anode)

    and

    E(cell) = (RT/nF)ln K

    Then shouldnt E(cathode)-E(anode)=(RT/nF)ln k ?

    I have a question asking to find the voltage of the cell and I am given a equilibrium constant.... which means i should use the second equation with "k" in it, and i get the right answer. But why does this value differ from the first equation's E(cell)?

    I hope this explains my question, Thx!
     
  2. jcsd
  3. May 1, 2005 #2
    i thought E(cell) = E(anode) + E(cathode)
     
  4. May 1, 2005 #3
    Welcome to the wonderful world of electrochemistry, perhaps the most confusing area of analytical chemistry. First off the cell potential can be expressed as:

    Ecell =E0red + E0ox

    or

    Ecell =E0red - E0red

    As to Ecell=RT/nF ln(K) This is incorrect. Think about it where are the standard redox potentials ? This is actually part of the nernst equation and used to calculate half cell effective potentials (i.e. concentration dependence).

    for:

    ox + ne- -> red

    E=E0+RT/nF ln(K) ... K=[ox]/[red]

    So then:

    Ecell=Ered - Eox
     
    Last edited: May 1, 2005
  5. May 1, 2005 #4
    Hmm... my text explicitly says Ecell0=(RT/nF)(ln k). The description is "Relating the standard emf of the cell to the equilibrium constant". This is why i thought it should equal the emf calculated from Ered+Eox. Where are the standard redox potentials? I dont know...seems like they should be there. I suppose i understand less than i thought :confused: Thanks for trying to help though.....
     
  6. May 2, 2005 #5
    Sorry, some confusion here on my part. What your talking about is the standard emf, that is when all activities for products and reactants are 1. This then leads to E0=RT/nF ln(K) for a give redox reaction. The value of E0red - E0red should give the same result. Edit: Edit: Are both reduction potentials using the same reference electrode.
     
    Last edited: May 2, 2005
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