# 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?

• You're right in saying you flip the reaction, but you don't flip the sign; it will still be a negative potential. Because of that, 0.34 - (-1.66) = 2V, since potential difference is finding the difference (by subtraction). It's the same value but you will have a negative in there. Think of it this way: oxidation is loss of electrons, so electrons are being released into the cell, meaning that cell will become negative. – user60221 Apr 8 '18 at 17:34
• Oh I see! So just to make sure I got it, in the case of Zn with Cu, Zn (-0,76V) will be the anode and Cu (+0,34V) the cathode. However I don't switch the sign, simply calculate E-cell as E-cathode - E-anode? – J.Se Apr 8 '18 at 18:12
• This still gives me the same potentials, but do You have a guess on what my teacher could mean with there being a problem that I haven't gotten any negative E-cell values? I shouldn't right? – J.Se Apr 8 '18 at 18:19
• I believe it could be that I wrote as I told you: 1,66 + 0,34 = 2,0 V and that she is hinting on why I havent used any negative values as you explained I should.. Could you help me with one last thing? If I created a table with all combinations of electrodes and the experimental values that they gave, how would I go about in ranking their reduction potentials from best to worst? – J.Se Apr 8 '18 at 18:30
• Your teacher probably does mean that. In a galvanic cell, you'll have positive and negative cells due to oxidation and reduction happening at each cell. Think of it like a standard battery, you have a positive and negative end due to the same reason. To calculate the cell potential, it's always the positive voltage - the negative voltage; subtracting a negative number just adds it instead, hence why you reached the same value either way. By the way, what do you mean by 'best to worst'? – user60221 Apr 8 '18 at 19:11

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).