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Chemistry Electrode potentials question

  1. Jul 15, 2013 #1
    Am I right to think that the potential established at an electrode immediately when placed in solution is given by

    [tex]E = E° - R \cdot T \cdot log_e(Q_{surface})[/tex]

    where the potential E of the electrode results from the standard electrode potential (E°) of the reaction and the reaction quotient Qsurface at the electrode surface.

    Then, once equilibrium is reached (usually fairly quickly for potential determining equilibria), we get a steady value of potential established, which is given by

    [tex]E_{eq} = E° - R \cdot T \cdot log_e(K)[/tex]

    Noting also that, once equilibrium conditions are established, the value of activities or concentrations at the electrode surface will be the same as their values in the bulk solution.

    If the conditions of the system are held constant at a certain reaction quotient Qconstant instead of allowed to go to equilibrium, then the potential established is always

    [tex]E_{constant} = E° - R \cdot T \cdot log_e(Q_{constant})[/tex]

    Is this a correct understanding? And if this potential will be established on its own, is it correct to say that the only need for another electrode to join the first one is so that the difference in potentials can be measured (i.e. Ecell=E(Electrode 1)-E(Electrode 2))?
  2. jcsd
  3. Jul 16, 2013 #2
    Small modification to all my equations above. Instead of

    [tex]- R \cdot T [/tex]

    I should have

    [tex]- \frac{R \cdot T}{F} [/tex]

    Where F is the Faraday constant.
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