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This question is on the same topic as this one but more complex. So, let's consider the cell with $\ce{Zn}$ and $\ce{Cu}$ electrodes inside the $\ce{NaCl}$ solution. On the $\ce{Zn}$ electrode we have $$\ce{Zn + 2Cl^- -> ZnCl2 + 2e^-},$$ and on the copper one $$\ce{H3+O + e^- -> 1/2 H2 + H2O}.$$

For both reactions we can find standard electrode potentials, but they are given for normal conditions, and $1~\mathrm{M}$ concentration of $\ce{Zn^{2+}}$ and $\ce{H3+O}$. To calculate the EMF for non-$1~\mathrm{M}$ concentration we use the Nernst equation. In this case we have something like $$E=E_0 - \frac{\mathcal{R}T}{n\mathcal{F}}\ln{\frac{[\ce{Zn^{2+}}]}{[\ce{H3+O}]}}.$$ But really in the initial $\ce{NaCl}$ solution we do not have any $\ce{Zn}$ ions (so their concentration is very close to zero and unknown to us). Of course, in few microseconds some $\ce{Zn}$ from electrode will be dissolved creating some finite potential. But in the experiment when putting $\ce{Zn}$ electrode into the same solution we always observe the approximately same EMF value, $\approx 0.8~\mathrm{V}$. How to predict theoretically this value?

Furthermore, my question is how to calculate the EMF of such cell as precisely as possible for different experimental conditions, e.g. different temperature?

I have made an experiment showing that the EMF of a 'cell' with $\ce{Zn}$ and $\ce{Cu}$ electrodes into $\ce{NaCl}$ solution is decreasing when heating the $\ce{NaCl}$ solution.

It's strange, let me explain why. The Nernst equation tells us that EMF is increasing with $T$ when $[\ce{Zn^{2+}}]$ is less than $[\ce{H3+O}]$ (because $\log$ is negative), and otherwise EMF is decreasing with $T$ ($\log$ is positive). But in my opinion the concentration of $\ce{Zn}$ ions is always less than $[\ce{H3+O}]$ because we do not have $\ce{Zn}$ in the initial solution. So then logarithm is always negative so EMF is growing with $T$.
In this case, why do I observe the EMF decreasing for $\ce{Zn-Cu}$ pair?

And the last question. When we talk about $\ce{Zn-Cu}$ pair we assume that zinc dissolves; and the hydrogen gas is created on the copper electrode. But why do I have the EMF when putting into a $\ce{NaCl}$ solution the $\ce{Al}$ ($\ce{Fe, Cr, etc}$) electrodes instead of $\ce{Cu}$, and remain instant the $\ce{Zn}$ electrode? I am not sure that hydrogen is created in such pairs: all these metals have a negative standard electrode potential. Measured EMFs of all these ($\ce{Zn-Al, Zn-Fe, Zn-Cr}$) pairs are significantly different, so the second metal is important in these cases. What reactions are going on in such pairs?

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  • $\begingroup$ Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the help center. Do you want to know more about this site, please take the tour. $\endgroup$ – Martin - マーチン Jan 8 '15 at 4:45
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I suspect that the reason you see a decay in the cell potential of $\ce{Zn/1M~ NaCl/Cu}$ with increasing temperature is due to increased corrosion of zinc. In $\ce{NaCl}$ solutions the anodic corrosion reaction is dissolution of $\ce{Zn}$ as $\ce{Zn^{2+}}$ and the cathodic reaction is reduction of dissolved oxygen to form an insoluble zinc hydroxide (Ichiro Suzuki, Corrosion Science 1985,25 (11), 1029-1034). The insoluble hydroxide passivates the metal surface. So, you have two effects which would contribute to EMF decay: 1) increased $\ce{Zn^{2+}}$ concentration and 2) passivation of the electrode.

One last thing which may or may not be helpful to you - if you are planning on utilizing the EMF from this cell to drive any kind of current you must account for the hydrogen overpotential on the copper electrode. It takes a lot of extra voltage to make hydrogen on pretty much any metal except platinum. You may find that you have a cell that looks good at open circuit (i.e. high potential with no current flow) only to see it crumble to nothing when you try connecting it to a load.

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  • $\begingroup$ Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the help center. Do you want to know more about this site, please take the tour. $\endgroup$ – Martin - マーチン Jan 8 '15 at 4:45
  • $\begingroup$ Thank you, now I understand there are two parallel reactions on the cathode: reduction of oxygen to a zinc hydroxide, and reduction of hydrogen ions to gaseous hydrogen. So, now the question is how to estimate the emf decay theoretically. Is the zinc ions' concentration linear increasing with T or more complicated? $\endgroup$ – Martino Jan 8 '15 at 11:09
  • $\begingroup$ Thanks for the edit to clean up my original answer. The concentration of zinc ions would be determined by the corrosion rate of the zinc electrode at a given temperature. In the event that the zinc electrode becomes passivated, you might expect to observe parabolic kinetics. On the other hand, if the cathodic corrosion reaction were primarily shuttled to the copper electrode, there may be little passivation for either electrode and you might find linear corrosion kinetics. I would suggest that you do some digging in the literature on zinc corrosion kinetics. $\endgroup$ – Qubit1028 Jan 8 '15 at 11:46
  • $\begingroup$ Thank you. But if the cathodic corrosion reaction is primarily shuttled to the copper electrode, we will have some EMF due to it. The standard electrode potential for $$O_2+2H_2O+4e^−\rightarrow 4 OH^−$$ is +0.401. So due to the zinc corrosion we must have EMF of 0.40+0.76=1.16 V at the normal conditions and 1M concentrations? $\endgroup$ – Martino Jan 8 '15 at 14:11
  • $\begingroup$ The Nernst equation for the corrosion cell is: $$E_{cell}=E_{cell}^{o}-\frac{nF}{RT}ln\left ( \frac{\left [Zn^{2+} \right ]^{2}\left [ OH^{-} \right ]^{4}}{\left [ O_{2} \right ]} \right )$$ The cell potential you gave would be correct for standard state conditions, but the solubility of oxygen in water at room temperature is only about 10 ppm. The other species are also present at concentrations much lower than standard state. As the temperature is increased, the solubility of oxygen would decrease and the cell potential would be shifted in the negative direction. $\endgroup$ – Qubit1028 Jan 9 '15 at 0:17

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