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

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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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.
• 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? – Martino Jan 8 '15 at 14:11
• 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. – Qubit1028 Jan 9 '15 at 0:17