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Nernst equation and equilibrium constant

  1. Jun 19, 2013 #1
    The equation ΔG=-nFEcell, I understand.

    I also understand that, at unity activities of all species, Q=1 so R*T*ln(Q)=R*T*ln(1)=0. And therefore Ecell=E°cell and ΔG=ΔG°.

    However, surely we cannot then write that ΔG°=-nFE°cell=-R*T*ln(K), since ΔG° is measured at unity activity when Q=1, rather than at K? In other words, ΔG°=-nFE°cell requires that the species be at unity activity (i.e. Q=1) so that it can come from ΔG=-nFEcell. But surely if Q=1 and K≠Q (if K does equal Q then, as we discussed some time ago, nothing is measured because equilibrium has already been reached!), we cannot write ΔG=0, so we cannot write ΔG°=-R*T*ln(K). So we should not be able to make the connection -R*T*ln(K)=-nFE°cell as far as I can see, because ΔG°=-nFE°cell is only true when Q=1 and ΔG°=-R*T*ln(K) is only true when Q=K and thus ΔG=0, and we in general assume that K≠1 (so that we are not at equilibrium to start with, meaning there is some spontaneous reaction that enables us to measure potential).

    Separately: do the redox equilibria have the same standard electrode potential regardless of the direction (i.e. whether the forward reaction is the reduction or oxidation)? If not then I guess ΔG=-nFEcell must refer in particular to the equilibrium where the electrons are in the reactants of the forward reaction, since ΔG does depend on direction; if Ecell and E°cell do depend on the direction then of course this is fine.
  2. jcsd
  3. Jun 20, 2013 #2


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    You must be careful about the interpretation: When the system is in equilibrium, i.e. all activities fulfill the mass action law with K, then Delta G=0 and the electrochemical cell will not show any voltage. However you can infer the value of the equilibrium constant from the voltage in a non-equilibrium situation where all activities are 1, as
    [itex] \Delta G=\Delta G^0-RT \ln K[/itex]. Once you know [itex]\Delta G^0[/itex] from a measurement where all activities are 1, you can calculate the value of K where Delta G=0.
  4. Jun 20, 2013 #3
    True. Thanks for this point. ΔG° is a constant value for a certain equilibrium under a certain set of thermodynamic conditions. Thus we can use it to go back to the equilibrium constant as you say.

    Can you help on this thread? https://www.physicsforums.com/showthread.php?t=697677
  5. Jun 21, 2013 #4


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    It was my understanding that the constant value of ΔG° = ΔG°f products - ΔG°f reactants in standard states.
    So in the quote above where ΔG° is a constant value. What are the certain set of thermodynamic
    conditions and certain equilibrium that ΔG° is derived from ?
    And also if at equilibrium ΔG = 0 when Q = K = 1
    ΔG = ΔG°+RT ln Q then ΔG° = -RT lnK. It looks like ΔG = ΔG° = 0
  6. Jun 22, 2013 #5
    In this case, it is derived from the measurement of E°(cell), which is identical to the measurement of E(cell), which we can actually measure, when all the species involved in the formal cell reaction for the electrode pair are at unity activity or fugacity meaning that Q=1 meaning that ±R*T*ln(Q)=0. E°(cell) can also be calculated from the standard electrode potentials (E°(cell)=E°(Right Electrode)-E°(Left Electrode) when it is written down; thus E°(cell) is dependent on which electrode you write on the right and which you write on the left).

    Unusual. Equilibrium's condition is ΔG = 0. Q=1's condition is that the activities and fugacities of all species are at unity. It's not likely that Q=1 coincides with equilibrium. But I'll bite:

    Of course. Q=1 automatically sets ΔG=ΔG°, as RTln(1) will just be 0. Since you've chosen to be at equilibrium at the same time as Q=1 (which, yes, means that K=1 as well) then ΔG=0 so of course ΔG°=0 as well.
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