Why is the theoretical value of the electrode potential different to the experimental value for Cu and Al for a galvanic cell by using Nernst equation

I've construct a galvanic cell with the Al strip putting in the Aluminum nitrate solution and Cu strip putting in the CuSO4 solution. KNO3 is used as the salt bridge in this experiment. The question is about how does changing the concentration of the copper sulfate solution affect the electrode potential in this galvanic cell. I use 5 different variables of the concentration of the copper sulfate, which are 1.5M, 1M, 0.5M 0.25M, and 0M which is distilled water.


I choose using 1M of concentration of CuSO4 as the example to calculate the theoretical value of the electrode potential by using Nernst equation. The following picture is my working process and my data table. I'm just wondering about do I do something wrong in my work. If I didn't make mistakes in my working process, why is the experimental value so differ from the theoretical value? When the copper sulfate is being replaced by water, why does it still have a voltage? When I replaced both copper sulfate and aluminium nitrate to water, why does it have a voltage? [![enter image description here](https://i.stack.imgur.com/GWb3i.jpg)](https://i.stack.imgur.com/GWb3i.jpg)




Don't use an aluminium electrode. In the presence of air, aluminium is always covered by a thin, continuous, colorless, insulating and waterproof layer of alumina (Aluminium oxide $\ce{Al2O3}$). If some accident happens that rubs and removes part of this this layer, the metallic surface will be immediately reoxidized by air. So the metallic electrode is never in contact with the aqueous solution.


If the aqueous solution is acidic, this alumina layer is partially dissolved, and aluminium metal is in contact with water. It will immediately react with water according to :$$\ce{2 Al + 3 H2O -> 2 Al(OH)3 + 3 H2}$$ so that the potential taken by this electrode follows Nernst's law for the $\ce{H2/H^+}$ redox half-cell. The two electrodes are then : $$\ce{H2 -> 2 H+ + 2 e^-}$$ $$\ce{Cu^{2+} + 2 e- -> Cu}$$ And Aluminium does not play a role in this cell. The potential of the Hydrogen electrode depends only on the pH of the solution in contact with the aluminum plate.


If the pH of the solution is $5$, the $\ce{H2/H^+}$ electrode has a potential $$\pu{E} = - (0.059/2)\times5 = - 0.15 \mathrm{V}$$Coupled with the $1$ molar $\ce{CuSO4}$ electrode, it produces cell voltage of $0.15 + 0.34 = 0.49 $V. This is nearly exactly what you found in your measurements with $1$ M $\ce{CuSO4}$. Your pH was not exactly $5$.


 2