How to accurately calculate the electromotive force for various conditions in a salt water battery?

Question

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?

Answer 1

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.