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_{cell}, 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 E

_{cell}=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=-nFE

_{cell}. 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=-nFE

_{cell}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 E

_{cell}and E°

_{cell}do depend on the direction then of course this is fine.