Is Q in the Nernst equation [Reduction]/[Oxidation] or [Oxidation]/[Reduction]?

The Nernst equation is given by :


$$E\_{\text{cell}} = E\_0 – \frac{RT}{nF}\ln{Q}$$


I have seen ${Q}$ given as both ${[\text{oxidation}]/[\text{reduction}]}$ and ${[\text{reduction}]/[\text{oxidation}]}$. When does each form apply?


Also, for the following cell:
$${\ce{Cu}|\ce{Cu2+}(\pu{0.001 M})|\ce{Cu2+}(\pu{1 M})|\ce{Cu}}$$ 


${[\text{oxidation}]}$ is 0.001, as the half reaction would be $\ce{Cu -> 2e- + Cu2+}$ because $\ce{Cu}$ is losing 2 electrons and becoming oxidised, right? 




It depends on the sign before the log term. The way you have written it is reduction/oxidation e.g.


$$E\_{\text{Ox/Red}} = E^0\_{\text{Ox/Red}} - \frac{RT}{nF}\ln\left(\frac{a\_{\text{Red}}}{a\_{\text{Ox}}}\right)$$


where *a* is the activity. Note the ratio is the opposite way round *red/ox* . (If you write *Q* the opposite way up then there is a positive sign before the log)


A derivation is given below for the Daniel cell: $$\ce{Zn(s) + Cu^{2+}(aq)-> Zn^{2+} (aq) + Cu(s)}$$


At the anode (where oxidation occurs) the left half cell is $$\text{Red}\rightarrow \text{Ox}^{n+}+ne^{-}$$
and at the cathode $$ \text{Ox}+ne^-\rightarrow \text{Red}^{n+}$$


The free energy $\Delta G\_{\text{Rev}}$ is determined from electrochemical potentials for the species.


$$\Delta G\_{\text{rev}} = \mu\_{\ce{Zn^{2+}}} + \mu\_{\ce{Cu}} -\mu\_{\ce{Cu^{2+}}} - \mu\_{\ce{Zn}}$$
$$ \Delta G\_{\text{rev}} = \Delta G^0\_{\text{rev}} +RT\ln\left(\frac{a\_{\ce{Zn^{2+}}}}{a\_{\ce{Cu^{2+}}}}\right)$$


where the last equation is obtained from definition of chemical potential in terms of activities, $\mu = \mu^0+\ln a$. (The activity of each metal is 1) If the reaction is carried out reversibly then the work done is equal to charge time potential difference through which the charge is moved. Therefore $$ \Delta G\_{\text{rev}} =-nF\Delta \phi$$ In a reversible reaction we change potential difference to *E* the 'electromotive force'(emf). Thus we can write 


$$-2FE=\Delta G^0\_{\text{rev}}+RT\ln\left(\frac{a\_{\ce{Zn^{2+}}}}{a\_{\ce{Cu^{2+}}}}\right)$$ 


For standard conditions the activities $ a\_{\ce{Zn^{2+}}}=a\_{E}=1$ and then we can write $$ E=E^0-\frac{RT}{2F}\ln\left(\frac{a\_{\ce{Zn^{2+}}}}{a\_{\ce{Cu^{2+}}}}\right)$$ When there are *n* electrons exchanged and the reaction is $$\text{Ox}^{n^+}+ne^-=\text{Red}$$ then the Nernst equation is $$ E=E^0-\frac{RT}{nF}\ln\left(\frac{a\_{\text{Red}}}{a\_{\text{Ox}^{n+}}}\right)$$


 2