Why isn't my electrochemical cell producing its potential voltage
Aluminum redox potential has nothing to do with the voltage of the cell you have build with it. Let me explain why, because it is not obvious.
Electrolytic cells made with aluminum anodes always yield rather low voltages. The reason is that usual pieces of aluminum are always covered by a thin, continuous and colorless layer of aluminum oxide $\ce{Al2O3}$. This oxide is an electrical insulator which makes a barrier between the metal and the solution. Aluminum ions cannot cross this barrier. There is no way of preventing this aluminum oxide formation. If you rub the aluminum surface to remove it, the pure metal so exposed will immediately react with air to remake this layer.
The only way to suppress this layer is to dip aluminum in a mercury chloride solution. Of course today this operation is forbidden by law. But this experiment was done before this legal decision. So let's see what was happening. Some mercury ions may cross the barrier and react with aluminum to produce an amalgam $\ce{Al(Hg)}$ plus some $\ce{Al^{3+}}$ ions getting in solution$$\ce{5 Al + 3 HgCl2 -&gt; 3 Al(Hg) + 2 Al^{3+} + 6 Cl^-}$$ The formula Al(Hg) is not sure, as it is an alloy. Maybe the ratio Al:Hg is different from $1:1$. It does not matter for the present study. The only important thing now is that the aluminum oxide layer does not perfectly cover the amalgamated metal any more, and is partly destroyed. Therefore aluminum atoms are suddenly in contact with water, and they quickly react with it according to :  $$\ce{2 Al(Hg) + 6 H2O -&gt; 2 Al(OH)3 + 3 H2 + 2Hg}$$ or$$\ce{2Al + 6 H2O -&gt;  2 Al(OH)3 + 3 H2}$$This shows that unprotected aluminum metal quickly reacts with water, producing $\ce{H2}$ bubbles, and these $\ce{H2}$ bubbles are responsible of the real anodic half cell according to :
$$\ce{H2  -&gt; 2 H^+ + 2 e-}$$ As a consequence, this &quot;aluminum&quot; electrode would have, at maybe $p$H$ 5$, a redox potential $\ce{E(H2/H+)}$ given by Nernst law : $$\ce{E(H_2/H^+) = E°(H_2/H^+) + 0.0296 V·log[H+]= 0 V + 0.0296 V·log 10^{-5} = - 0.148 V}$$ In your setup, this anode is coupled with an ordinary copper half-cell as cathode, whose standard redox potential is + $0.34$ V. The voltage of this cell is $$\ce{E(Cu/Cu^{2+}) - E(H2/H+) = 0.34 V -(-0.148 V) = 0.488 V}$$ This value is nearly the voltage that you have obtained in your cell ($0.50$ V).