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.
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.
