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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?

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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 $$ E=E^0-\frac{RT}{nF}\ln\left(\frac{a_{\text{Red}}}{a_{\text{Ox}^{n+}}}\right)$$

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  • $\begingroup$ I'm familiar with everything in your answer except that $\mu$ in the expression for $\Delta G$. Can you kindly tell what is? I would have expected it to be $\Delta G_{\text{rev}} = G_{\ce{Zn^{2+}}} + G_{\ce{Cu}} - G_{\ce{Cu^{2+}}} - G_{\ce{Zn}}$ instead (as $\Delta$ signifies a difference). Thanks! $\endgroup$ – Gaurang Tandon Mar 5 '18 at 16:32
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    $\begingroup$ $\mu$ is the chemical or the increase of free energy when one mole of a component is added to a large quantity of solution so that the overall composition is not affected. Mathematically for species $i$, $\displaystyle \mu_i=\left (\frac{\partial G}{\partial n_i} \right)_{T,P,n_j}$ which is also the partial molar free energy sometimes written as $\bar G$ $\endgroup$ – porphyrin Mar 5 '18 at 21:38

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