When I originally learned the Nernst equation, it was just conceptual without any hands on experience. I am now a chemistry teacher, and wanted to provide some hands on experience with it for my students. We did a lab where we set up a galvanic cell as follows:

Anode: Aluminum electrode in 1 M AlCl3 solution Cathode: Copper electrode in 1 M CuSO4 solution

Then, we varied the concentration of each solution. The overall reaction should be: 2 Al + 3 Cu+2 --> 2 Al+3 + 3 Cu

So, we expected that diluting the CuSO4 solution would decrease the cell potential, while diluting the AlCl3 solution would increase the cell potential.

When we tried it, diluting CuSO4 decreased the cell potential as expected. However, diluting AlCl3 also decreased the cell potential! In fact, it seemed to decrease the cell potential MORE than diluting the CuSO4 did. All 10 labs groups independently got this same result, and their measured voltage values were all remarkably consistent, so I have ruled out procedural flukes. It is clear that there is something about this reaction I do not understand. What am I missing?

Here is some of our data in case it helps:

1 M AlCl3 and 1 M CuSO4: 0.83 V

0.01 M AlCl3 and 1 M CuSO4: 0.66 V

0.0001 M AlCl3 and 1 M CuSO4: 0.54 V

1 M AlCl3 and 0.01 M CuSO4: 0.78 V

1 M AlCl3 and 0.0001 M CuSO4: 0.73 V

  • $\begingroup$ I have always thought teachers verify the intended experiments themselves before deploying them in the lab class, unless overtaking already verified ones. Unexpected troubles confuse students a lot, if the teacher does not have answers for their questions. But the pro is you at least try to get answers a posteriori. $\endgroup$
    – Poutnik
    Feb 4, 2023 at 11:06

2 Answers 2


Aluminium electrode is not a good choice, as there are several side effects fighting against theoretical expectations.

If $\ce{Al/Al^3+}$ had been well behaving electrochemical system, $\ce{Al}$ would have created hydrogen bubbles in contact with the solution, with its strongly negative potential $E^\circ = \pu{-1.66 V}$.

With aluminium salt more and more diluted, it converges to water and metallic aluminium becomes more inert, with decreasing cell voltage.


The overall reaction is not $\ce{2 Al + 3 Cu^{2+} -> 3 Cu + 2 Al^{3+}}$ as expected, because, in contact with air, aluminium plates are always covered by a thin, continuous and colorless layer of aluminium oxide $\ce{Al2O3}$ (or alumina) that prevents any aluminum atom to get in touch and react with the aqueous solution it is dipped in. If a scratch happens on this outer surface, the $\ce{O2}$ molecules from air will immediately react and regenerate this protecting alumina layer. So in your $\ce{Al/Cu}$ cell, the metallic $\ce{Al}$ atom does not react : the overall reaction is due to water, as will be explained later. So please never try to make and measure an electrolytic cell with Aluminium plates in a classroom.

You may remove this alumina layer by dipping any aluminium plate into a mercury chloride solution for a couple of seconds. The $\ce{Hg^{2+}}$ ions have the strange property of going through the aluminium oxide layer, and more or less destroy it. The aluminum metal surface is transformed into an amalgam, which reacts immediately with air. You could see white alumina filaments growing at a visible rate on the unprotected metal. It stops when all aluminium amalgam has been oxidized. It is a surprising phenomenon, which was popular in classrooms when mercury compounds were allowed.

To go back to your cell, what happens can be described so : first a few molecules water are crossing the alumina layer through pores or by diffusion and react with aluminium metal according to : $$\ce{2 Al + 6 H2O -> 2 Al(OH)3 + 3 H2}$$ This reaction happens as soon as the aluminium plate is dipped into water, and stops quickly. So a rather small amount (some nanomoles ?) of $\ce{H2}$ is adsorbed on the metallic surface under the oxide layer. It does not produce any bubbles. Later on, when the plate is coupled with another electrode, this adsorbed hydrogen produces the anodic half-equation $$\ce{H2 -> 2 H+ + 2 e-}$$ $$ \ce{E = E° + \frac{0.059 V}{2} log\frac{[H+]}{p(H2)}= \frac{0.059 V}{2} log\frac{[H+]}{p(H2)}}$$ But its potential is difficult to calculate with Nernst law, because both the pressure $\ce{p(H2)}$ is unknown and must be rather small; and the concentration [$\ce{H^+}$] must also be small (with an unknown pH < $5$). On the other electrode, made of copper, there's no problem: the half-reaction is as expected $$\ce{Cu^{2+} + 2 e^- -> Cu}$$ Its potential can be calculated by Nernst law.

So the cell is not $\ce{Al/Cu}$ but $\ce{H2/Cu}$ and the overall cell tension is not easy to calculate in a class room by ordinary students. So please don't make students measure and discuss an aluminium-copper cell. Replace the aluminium by another metal like zinc $\ce{Zn}$ or lead $\ce{Pb}$. This is an old chemistry high school teacher's advice !

However, let me analyze some of your results. First let's compare the cells with $1$ M $\ce{AlCl3}$ solution, when they are coupled with different copper solutions. When both $\ce{Cu}$ and $\ce{Al}$ solutions are $1$ M, the overall tension is $0.83$ V. As the $1$ M copper solution (from Tables) is at $\ce{E_{Cu, 1M} = +0.34}$ V, it means that the redox potential of the $1$ M $\ce{Al}$ electrode is $$\ce{E_{Al, 1M} = 0.83 \mathrm{V} - 0.34 \mathrm{V} = + 0.49 \mathrm{V} = \ce{E_{H_2,nearly zero M}}}$$ Doing the same calculation with $0.01$ M and $0.0001$ M $\ce{Cu^{2+}}$ solutions and their respective overall tensions $0.78$ V and $0.73$ V, one gets the following $\ce{Cu}$ redox potential $$\ce{E_{Cu, 0.01M}} = 0.78 \mathrm{V} - 0.49 \mathrm{V} = + 0.29 \mathrm{V}$$ $$\ce{E_{Cu, 0.0001M}} = 0.73 \mathrm{V} - 0.49 \mathrm{V} = + 0.24 \mathrm{V}$$ As a consequence, the three Cu potentials derived from measurements show a difference ($0.05$ V), which is nearly exactly what is derived from Nernst law : the theoretical difference should be $0.059$ V. It shows that the copper electrode behaves logically.

Similar calculations made on the aluminium electrode can hardly be carried out, for lack of knowledge about $p$H and pressure of $\ce{H2}$ on the electrode.


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