For a school project, I am testing how concentration affects the voltage output of three metals. So I am building voltaic cells with 0.1 M copper nitrate and 0.1 M, 0.01 M, and 0.001 M of zinc, aluminum and nickel nitrates as the electrolyte solutions. For the cells with zinc and copper, I found that the voltage produced was generally fairly close to their theoretical value of 1.10 V with the lowest voltage produced being 0.95. However, for aluminum and nickel, I have found that they are very far off from their theoretical values of 2.00 and 0.57 respectively as I have only reached a maximum voltage of 0.50 for aluminum and 0.20 for nickel.

I am using a porous cup and have made sure all of the electrodes are thoroughly sanded before each trial. I am fairly confident the issue is not related to the concentrations, or to the porous cup. At this point, I just have no idea why it isn't working. I read a similar question on here suggesting that the issue might be that aluminum does not easily dissolve in NO3 and I was wondering if the same was true for nickel.

  • $\begingroup$ Check the resistance of your porous cup. What is it made of? Internal resistance of the cell can reduce the potential. $\endgroup$ – M. Farooq Jan 10 at 20:41
  • $\begingroup$ Are you also familiar with Nernst equation? $\endgroup$ – M. Farooq Jan 10 at 20:42
  • $\begingroup$ Yep I think I have a pretty good understanding of the Nernst equation. Would the resistance have a different effect depending on the electrolyte? Like could the resistance explain why zinc matched its theoretical value while the others didn’t? $\endgroup$ – John Jan 10 at 22:06
  • $\begingroup$ Since the concentration of the electrolyte is changing, I feel the resistance of the internal cell is also changing. I am talking about the resistance of the solution. Can you bring the electrodes closer and see if the potential increases. $\endgroup$ – M. Farooq Jan 10 at 22:49

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$$\ce{5 Al + 3 HgCl2 -> 3 Al(Hg) + 2 Al^{3+} + 6 Cl^-}$$ 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 : $$\ce{2 Al(Hg) + 6 H2O -> 2 Al(OH)3 + 3 H2 + 2Hg}$$ or$$\ce{2Al + 6 H2O -> 2 Al(OH)3 + 3 H2}$$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 : $$\ce{H2 -> 2 H^+ + 2 e-}$$ As a consequence, this "aluminum" 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).

  • $\begingroup$ Thank you that is very helpful. In regard to the nickel electrodes would I be correct in assuming that nickel, like aluminum, also forms this instantaneous oxide layer? Also, would zinc then take longer to form this oxidation layer or is there some other explanation as to why its value matched the theoretical? $\endgroup$ – John Jan 11 at 1:26
  • $\begingroup$ @John Purely by the standard redox potential for $\ce{Ni(s)/Ni^2+(aq)}$, nickel should be quite reactive. The fact that it is not and is used in corrosion resistent alloys or metal platings says something. $\ce{ZnO}$ is probably not so compact and is quite soluble in acidic solutions. Note also that some standard redox potentials are more or less just theoretical equivalents of related standard reaction Gibbs energy and cannot be ( or just with difficulties ) directly measured, at least in simple scenarios. I have once seen redox potential for C/CH4 and would like to see its measurement. $\endgroup$ – Poutnik Jan 11 at 7:31
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    $\begingroup$ The operation is forbidden by law? I'm surprised to hear that it's illegal to mix two chemicals. I would expect that it's legal to mix them but then the law would be very picky about how you dispose of the mercury compounds - including the mercury dissolved in the aluminium, if it does that. $\endgroup$ – user253751 Jan 11 at 16:25
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    $\begingroup$ @User253751 In my country you cannot use mercury in chemistry lectures and in the laboratory. $\endgroup$ – Maurice Jan 11 at 19:34
  • $\begingroup$ @Maurice - Mine also. There used to be a demonstration of a simple electric motor consisting of a star shaped sheet of metal rotating above a dish of mercury, between the poles of a horseshoe magnet. The points of the star dipped in the mercury. The current flowed from the battery to the centre of the star to the tips, via the mercury, and back to the battery. The star rotated and successive points dipped in the mercury. And out again, making a little spark. Not enough to hurt the students much, but sufficient to hurt the teacher during his or her career. $\endgroup$ – Michael Harvey Jan 11 at 19:45

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