Calculating cell-potentials

I have measured the cell potential between a number of metals e.g. Ni, Pb, Al, Cu and Zn by constructing a galvanic cell and then used a voltmeter to measure the potential. 


Now, if I want to calculate the theoretical potential (if I have understood this correctly) I need to look at a standard reduction potential table and compare the two elements in question. Take Al and Cu for example. The element that will act as the anode where oxidation will occur will be the element with the more negative cell potential. In this case Al because it has a value of -1,66 V compared to +0,34 V for Cu. Then since it is a reduction table I need to flip the reaction of Al to show an oxidation with a value of +1,66 V. 


The full cell potential will be 1,66 V + 0,34 V = 2,0 V


I have used this same logic and gotten results for Ni and Pb (0,13 V), Al and Ni (1,4 V) etc. 


Now my teacher asks me how come you haven't gotten any negative theoretical values and to check my work again? But surely I can't get any negative cell potential values since this is a galvanic cell or am I mistaken? 


She asked the same thing with the real measured potentials with the voltmeter and of course here I got a few negative readings, but isn’t that simply because I placed the positive and negative ends of the voltmeter on the opposite electrodes? So I made all negative measurements positive as if I HAD placed them correctly… 


Am I wrong here, should there be any negative values theoretically? 




As I understood your problem with signs of the potentials, I think it's better if you write your redox equations, starting with half-reactions. Like @Kian Stevens said, consider your Galvanic cell is as standard battery (albeit a battery is a set of galvanic cells). So, to have a electric current, the redox reaction should be spontaneous $\Delta G^o \le 0$). That means $ E\_{cell}^0 \gt 0$. Let's consider, your galvanic cell with $\ce {Al}$ and $\ce {Cu}$:


Write both reduction half-reactions:



> 
> For $\ce{Cu}$: $\ce {Cu^{2+} + 2e- <=> Cu} \pu \space\space {E\_{\ce{Cu^{2+}/Cu}}^0 = \pu{+0.34 V}}\tag{1}$
> 
> 
> For Al: $\ce {Al^{3+} + 3e- <=> Al} \pu \space\space {E\_{\ce{Al^{3+}/Al}}^0 = \pu{-1.66 V}}\tag{2}$
> 
> 
> 


Now you may see why you needed to switch the equation (2). If you switched the half-reaction of $\ce {Al}$, the full redox equation would have a positive $E\_{\text{cell}}^0$ value.



> 
> When you reversed the equation (2): $\ce {Al <=> Al^{3+} + 3e-} {E\_{\ce{Al^{3+}/Al}}^0 = \pu{+1.66V}}$ $\tag{3}$
> 
> 
> Thus, total redox equation would be:
> $\ce {3Cu^{2+} + 2Al -> 3Cu + 2Al^{3+}} {E\_{\text{cell}}^0 = \pu{+2.00 V}}$
> 
> 
> 


Note that followings:


(A) this would have been your voltmeter reading if you have used the standard conditions. Actual cell potentials can be calculated by using the Nernst equation to compare your findings, if you know the solution concentrations of half-cells ($\ce {[Cu^{2+}]}$ & $\ce {[Al^{3+}]}$), and the temperature of which you performed the experiment.


(B) In equation (2), $E^0$ of $\ce {Al}$ have more negative value compared to that of $\ce {Cu}$. That means, $\ce {Cu}$ has a greater electronegativity (liking to keep electrons or pulling electron, comparatively speaking) than $\ce {Al}$, thus $\ce {Cu}$ tends to steal electrons from $\ce {Al}$ or in other words, $\ce {Cu}$ tends to oxidize $\ce {Al}$. Hence, this is an one way reaction (favoring one direction of the reaction; see total redox equation).


For more info, read the [Wikipedia article on Galvanic cell](https://en.wikipedia.org/wiki/Galvanic_cell).


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