# Why are potentials of cells with magnesium electrode always lower than expected?

I have prepared the 1M water solution of magnesium acetate. Then I placed $\ce{Mg}$ and $\ce{Cu}$ electrodes in it. I predict there such reactions take place: $$\ce{Mg-2e^- \rightarrow Mg^{2+}}$$ $$\ce{H_3^+O + e^- \rightarrow\frac12 H_2 + H_2O}$$ The gas bubbles emitting on both electrodes probably confirm my prediction (I did not check is it hydrogen or not). So I expect to obtain the potential difference about 2.3 V. But really I observe something like 1.62 V.

Then I changed the acetate solution to a water (simple water which we drink) and observed almost same value of 1.60 V. Adding there many crystals of $\ce{Mg}$ acetate (I added them till they could not dissolve more) almost did not change the voltage!

When I added much acid to the solution the potential difference became larger - up to 1.9 V.

So, I have two questions: 1) why the real voltage is so different from the expected value? 2) Why changing the $\ce{Mg^{2+}}$ concentration almost does not change the voltage?

You may say I have dirty magnesium electrode etc. But if you search e.g. experiments with potatoes (when one puts $\ce{Mg}$ and $\ce{Cu}$ strips in potato) you can observe that everywhere it is mentioned that such $\ce{Mg-Cu}$ cell gives 1.7 V.

I have several ideas why can it happen: a)some story with overpotentials, b)oxide film on magnesium surface. What is the real reason?

• Note that mhchem does exist here. Here's more info. – M.A.R. Jan 31 '15 at 21:22

At least one significant reason is that the voltage you "expect" is probably for a standard state of one mole per liter of $\ce{H+}$, i.e. at pH 0. Neither the magnesium acetate nor the plain (tap?) water you used for the experiments come close to this pH. Also, the standard state for $\ce{H2}$ would be one bar of atmospheric pressure. Probably this condition was not achieved either.

You can calculate the expected potential at any pH using the Nernst Equation.

The half-cell reactions in your system are

• $\ce{Mg^0 -> +2e- +Mg^{+2}}$ (this has a potential of 2.372 V not 2.7 V as a mistakenly wrote in a prev. version of this answer)
• $\ce{2H^+ +2e- -> H2}$

The total reaction is thus $\ce{Mg^0 +2H+ -> Mg^{+2} + H2}$

So applying the Nernst equation to a one molar magnesium solution at pH 7 (i.e. $\ce{H+}$ concentrations of $10^{-7}$ mol/L) would give:

• $E_\text{cell} = E^{\ominus}_\text{cell} - \frac{RT}{zF} \ln Q$
• $E_\text{cell} = 2.4 \text{ V} - \frac{RT}{zF} \ln \frac{1}{(10^{-7})^2 }$
• $E_\text{cell} = 2.4 \text{ V} - 0.41 \text{ V} = 1.9 \text{ V}$

Obviously, this correction gives you a predicted voltage that is still higher than what you measure closer what you measured. But you reported using acid and getting a higher potential, so it's clear that the pH effect matters for your system. But something else must be going on too. Some other possibilities for other minor effects that tweak the voltage a bit more:

• Hydrogen partial pressure is unaccounted for
• overpotential
• your cell is not open but is conducting current
• dirty electrodes etc.
• other unaccounted for electrochemical reactions (copper oxidation, etc.)
• ""At least one significant reason is that the voltage you "expect" is probably for a standard state of one mole per liter of H+, i.e. at pH 0. "" Normal potentials are always for a solution of the same metal ion 1 molar and neutral, if not otherwise stated. – Georg Feb 4 '15 at 16:45
• Georg, please have a look at the wikipedia page for the standard hydrogen electrode. The SHE is the standard that is usually assumed in electrochemistry, unless otherwise stated. It is approximated in the laboratory by using 1 mole per liter of strong acid -- i.e., pH 0. – Curt F. Feb 4 '15 at 16:56
• Also, if you see this table of standard reduction potentials, you will see that Mg is listed at the potential stated in the OP, 2.7 V, and that preface to the table says "standard electrode potentials are given in the table below in volts relative to the standard hydrogen electrode". – Curt F. Feb 4 '15 at 17:09
• You are not correct about 2.7V: 2.7V is potential for $Mg-1e^-\rightarrow Mg^+$ but here we expect $Mg-2e^-\rightarrow Mg^{2+}$ with standard potential 2.3V. So the expected voltage seems to be ~2.3-0.4=1.9 what looks closer to the truth :) Thank you for your answer! – Martino Feb 6 '15 at 4:47
• Thanks for the comment Martino. I mis-read the wikipedia table and then assumed (without checking) that the number matched the OP! I'll edit my answer to eliminate the mistake. – Curt F. Feb 6 '15 at 4:51