We created a voltaic pile battery and tested voltage with a multimeter. Each cell is comprised of a penny and a nickel with wet paper in between (vinegar + salt solution). Then you simply connect these cells serially.

The penny end is supposed to become positive and the nickel end is supposed to become negative.

However, we see positive voltage readings on the multimeter by connecting the the negative probe to the penny and the positive probe to the nickel. I've verified that the probes are of correct orientation. Am I misunderstanding something about negative electrodes, i.e instead of being the end of a battery where there is a buildup of negative charge, are they instead the end where negative charge "wants" to go to?


2 Answers 2


Each cell in this voltaic pile consists of two USA coins (a nickel and a cent) plus an electrolye solution of vinegar and table salt. The coins are separated by wet paper that is saturated in the electrolyte solution. The wet paper prevents the coin electrodes from touching, which would short out the cell, and is otherwise unimportant: it is not serving as a salt bridge because there is only one electrolyte solution.

The nickel coin is cupronickel, i.e., 25% nickel and 75% copper. The cent (aka “penny”) coin, if minted after 1982, is a zinc planchet (99.2% zinc and 0.8% copper), plated with copper. The overall composition is 97.5% zinc and 2.5% copper. Provided the copper plating completely prevents the electrolyte from contacting the underlying zinc planchet, the penny will simply behave as pure copper would. But, if the underlying zinc planchet is exposed to the electrolyte, then the penny will behave as a zinc electrode. The damaged cent will also slowly dissolve in the vinegar, which is typically a 5% solution of acetic acid (a weak acid) in water.

The figure below shows a schematic version of one of the voltaic pile cells, assuming the nickel and cent coins act as pure nickel and copper metals, respectively. The standard electrode potentials of possible reductions are also shown. The most favored reduction would be the one with the most positive standard reduction potential. However, the electrolyte is far from standard state for any of the reductions shown in the figure, so the standard electrode potentials are not definitive: the Nernst equation would be necessary.

Coins voltaic pile

Figure attribution: I drew it myself using Keynote on my iMac.

Also note that the top reduction will not be happening because the concentrations of dissolved oxygen gas and hydrogen ion are too low (this is not a fuel cell). The next reduction, of \$Cu^{2+} \$ to Cu(s) will not happen because there are no copper ions in the electrolyte and none will be produced during operation: copper metal will not be oxidized. So \$H^+ \$ will be reduced, at the copper cathode, to hydrogen gas. At the anode, nickel metal will be oxidized to \$Ni^{2+} \$. The potential would be +0.236 V under standard state conditions. In the actual cell, the voltage would be whatever the DMM reads. The energy source is the oxidation of the nickel.

If the cent’s copper plating is breached, and the zinc planchet is exposed to the electrolyte, then the zinc would be oxidized to \$Zn^{2+} \$ and what was the cathode would become the anode, and vice versa.

So the nickel electrode would be the cathode and \$H^+ \$ would be reduced at it to hydrogen gas. The potential would be -0.762 V under standard state conditions and with the DMM’s reference (black) lead still on the nickel electrode. Swapping the DMM’s leads, so that the reference lead was on the zinc anode, i.e., the damaged cent coin, would give a DMM reading of +0.762 V under standard state conditions. Again, the actual cell voltage would be whatever the DMM reads. The voltage is higher because the energy source is oxidation of zinc rather than nickel.


The standard electrode potential is the following:

  • nickel (Ni) -0,25 V
  • copper (Cu) +0,16 V

I think Wikipedia provides the answer when it says: "Since the electrode potentials are conventionally defined as reduction potentials, the sign of the potential for the metal electrode being oxidized must be reversed when calculating the overall cell potential."

  • \$\begingroup\$ So, if I understand what you're saying, since the nickel is the metal being oxidized in this case, we indicate the nickel as being positive? And this explains why the positive multimeter probe touching the nickel produces positive voltage readings? \$\endgroup\$
    – JohnQ
    Feb 12, 2017 at 19:47
  • \$\begingroup\$ Yes, see Cell voltage \$\endgroup\$
    – skvery
    Feb 12, 2017 at 19:56
  • \$\begingroup\$ Thanks for your help so far. Really appreciate it. That wiki article seems to support the issue I'm seeing with the anode being the metal that's paired with copper. In that article, it's zinc and copper with zinc being the element that's oxidized. "The anode is the electrode where oxidation (loss of electrons) takes place." In my case, I used a penny with a nickel coin, which as I understand is nickel-plated copper. I observed the nickel end being the cathode, however. \$\endgroup\$
    – JohnQ
    Feb 13, 2017 at 0:36

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