Consider a simple circuit that has a resister and a voltaic battery.

Here is what (I think) I know to be true:

  1. If I draw an arbitrary shape around the battery, the net charge enclosed in that battery is zero.

  2. At the anode, electrons are being generated from an oxidation event. Rather than accumulate on the anode electrode, the electrons from the anode electrode enter the wiring that is connected to the cathode of the battery. As electrons enter the wire (from the anode electrode) they can be thought of as "pushing" electrons throughout the wiring towards the cathode electrode of the battery, where a reduction event takes place. The anode is therefore a source of electrons, and the cathode is a sink of electrons.

Here is where I get confused.

Why does the anode hold a negative charge and the cathode hold a positive charge? If the moment electrons are generated at the anode electrode they are shuttled away from the electrode, why would this scenario result in any different charge than what is observed at the cathode electrode?

The only way I can think of mechanistically explaining this is that the anode does, in fact, accumulate some small portion of electrons even when the wire is present / circuit is closed (i.e. some small negative charge is statically present at the anode electrode). Conversely, no such accumulation occurs at the cathode electrode...perhaps, even a "baseline deficit" exists at the cathode.

Is this correct way of thinking about it?

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    \$\begingroup\$ For a shorted-out battery? Zero voltage? But with a parallel resistor (maybe a 10meg resistor, maybe infinite,) the electrons AREN'T shuttled out. They build up, but just enough to produce the usual output-voltage. When the charge is enough to hit that voltage, the chemical reactions halt. Adding a resistor will turn on the reactions but not eliminate the buildup unless it acts as a short. Don't confuse a shorted-out battery with one that has a 100Meg resistor across it (or 100ohm, etc., whatever doesn't reduce the batt-volts to zero). Also try my amasci.com/elect/battvolt.html [Edited] \$\endgroup\$
    – wbeaty
    Mar 18, 2020 at 5:01
  • \$\begingroup\$ @wbeaty sorry, I didn't think to include that information. For a 1.5 V battery and a 10 ohm resistor. (Arbitrary values). I'm not sure I know what a shorted-out battery is. Additionally, I'm not sure I understand what you mean by "electrons aren't shuttled out". Do electrons from the anode NOT leave the battery and enter the wire of the circuit when the circuit is closed? \$\endgroup\$
    – S.C.
    Mar 18, 2020 at 5:12
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    \$\begingroup\$ "Do electrons from the anode NOT leave the battery and enter the wire of the circuit when the circuit is closed?" - Yes, they do. But not quickly. Imagine if you could watch the electrons as they made that journey, bouncing around as they slowly drift through the wires from one battery terminal to the other at perhaps only a few millimeters per hour! \$\endgroup\$ Mar 18, 2020 at 5:31
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    \$\begingroup\$ @S.Cramer Possible confusion: "electrons leave the battery" implies that ALL electrons leave. They don't. Instead, electrons build up on one terminal (and positives build up on the other) until 1.5V appears, then this voltage will halt the chemical reactions. Now connect a 10ohm resistor. It slightly lowers the 1.5V, chemical reactions turn back on, and 150mA flows. The build-up is still there. But all new electrons leave the battery to pass through the resistor. (To "short out" a battery, connect some 8-gauge wire across your c-cell, then measure zero volts at battery terminals, not 1.5V.) \$\endgroup\$
    – wbeaty
    Mar 19, 2020 at 5:38

1 Answer 1


All the stuff happening inside batteries is a bit complicated, even if we greatly simplify everything. After all, a battery is actually two batteries back to back, each one composed of a water-metal junction. The water, the electrolyte, is a conductor that connects the two. They're in series, so they both have to produce exactly equal current (if a flashlight bulb is connecting the two metal plates,) which means they have to communicate with each other via electrostatic forces.

It's not going to be pretty.

Copper-zinc batteries are simpler than modern manganeze-oxide flashlight batteries, so let's do those. .

If we place one copper and one zinc plate into some salt solution, but don't touch them together, then the chemical reactions will dissolve both metals. This chemical corrosion is pumping some positive ions out of both electrodes, and into the salt water. It leaves excess electrons behind on both electrodes.

Basically, some positive protons are briefly flowing into the water, as carried by the dissolving metal atoms, which leaves excess electrons behind the the metals. At the start, these reactions only occur briefly when the battery is first assembled, and they halt when the growing charge-imbalance between water and metal will finally become large enough to prevent more ions from leaving the metal. (The neg-charged metal attracts positives, preventing any more from leaving, which causes the energy-producing corrosion reactions to halt.)

But the water attacks a zinc surface more violently than it attacks copper, so the reactions at the zinc electrode will only halt after the zinc metal builds up a larger imbalance, with more electrons left behind, than does the copper. (Energetic corrosion steals MORE positive ions from metals like zinc, lithium, sodium etc., creating larger neg voltage than it would with copper, gold, etc.) So, the charges and the voltages on the zinc and copper, wrt the water, are not equal.

From electrochem tables, if the salt water is our common terminal, then the zinc metal is charged to -5.2 volts, and the copper charged to -4.1 volts.[1] That gives 1.1V between them, as measured by a conventional voltmeter touching the metals.

BOTH terminals always have extra electrons as long as the output voltage of 1.1V exists and the battery is not "dead," (as long as the zinc plate has not corroded away.)

But this excess is unequal between the two metal plates, and that's exactly the effect which creates the 1.1V on the terminals of any copper-zinc cell. Note that the differential-excess always appears, whether or not the battery is creating a current in an external circuit. Even a battery sitting on a shelf is actively maintining it's output voltage.

Now, if we connect a light bulb across the battery terminals, the excess electrons on the two electrodes don't go away. The lightbulb does start to discharge the excess charges, and the battery-voltage starts decreasing. The excess electrons only diminish slightly, just enough to let the chemical reactions start up again. The slight decrease in voltage "tells" the battery to start pumping charge through itself, but only enough to maintain the 1.1V output voltage. (Whenever there is slightly lower voltage between water and metal, then the aggressive water can again start tearing pos ions out of the metal.) The re-started reactions keep the battery-voltage from being further discharged by the light-bulb. In this way, the battery-plates are actively maintaining the 1.1V output, regardless of the presence of the external current-draw.

Now the reaction at the zinc metal is dominating, and it's output-voltage is pumping an output current (since in the series circuit, the zinc-water voltage is the higher one,) and this higher voltage forces the copper-electrode's chemical reaction to run backwards! The zinc still dissolves, but the copper now gets electroplated, as the positive ions in the salt-water are forced to re-join the copper.[2]

So in all, we have protons (pos metal ions) flowing in the electrolyte, and electrons flowing in the metal battery plates (and in the external wires.) At the zinc surface, positive charges (the positive zinc ions) are torn out of the metal and pushed through the salt water, leaving extra electrons behind in the zinc. Because the zinc now has extra electrons, it produces an electrostatic force which pushes electrons out through the external circuit. But at the same time, protons (positive metal ions) in the water are forced to merge with the copper surface, using up some of the energy coming from the zinc-corrosion, and charging the whole copper plate less negative than before. This produces electrostatic attraction, which pulls electrons in from the external circuit.

Wall of text! Is your brain numb yet? Yet barely any math! Un-numbering un-numb-brains. (At least it does so in comparison to the wall-of-equations style explain-ations. I hope!)

Since all this stuff is part of a series circuit, each part has to "magically" adjust itself so the exact same series-current flows everywhere. (The "magic" is found in the negative feedback between the rate of chemical reactions and the slight voltage-changes between the salt water and each metal surface.) If the voltage at the zinc plate should ever drop, this lets the reactions crank up higher, which pumps out a big current, which raises the voltage again. But the voltage remains slightly below the unloaded 1.1V. In this way, each plate "feels" the presence of that external load, and adjusts its current-pumping in order to obey Ohm's law. The complicated neg-feedback process will cause the battery to have an "internal resistor," which is simply the ratio between the voltage-decrease and the zinc-plate's rate of current-pumping, delta-V/delta-I.


Is the copper terminal charged positive? Unknown. Unknown, since we never know the absolute net charge of the entire battery! If we started out with uncharged materials, then we end up with voltages on the three sections of -5.2V, 0V, -4.1V (for zinc, water, copper.) But what if we bump the negative battery terminal briefly against a true earth-ground stake? The battery puts out a tiny blip of current, actively maintains its open-circuit voltage, and so the three voltages (zinc, water, copper) become 0V, +5.2V, +1.1V, and the copper plate is now truly positive ...rather than just less negative than the zinc. Normally we never measure this stuff, and just ignore it. We PRETEND that the common-bus of the schematic (and of the circuit) is truly neutral-charged, as if it were actually connected to an earthing-stake hammered into the ground. But in reality, any un-Earthed circuit will have many tens of volts upon its "ground" bus, with voltages which swerve wildly around, influenced by nearby plastic-covered 120V cables, by people rubbing their butts against plastic chairs, by incoming rf from distant transmitters, "radio static" which is actually some rf coming from lightning bolts in Africa, etc.


In carbon-zinc flashlight batteries ...they are not carbon-zinc batteries! The two plates are Manganese dioxide and zinc metal. The carbon is just a conductor. The carbon-rod is not a battery plate; instead it's a thick current-spreader. The MnO2 powder is mixed with carbon powder, the carbon providing a "tree structure" connection to the center rod. "Intercalcated" battery plate, or "interleaved" contact-conductor with chemical layer. Such electrodes have a very large surface area, something like a conductive sponge. The carbon rod and the carbon powder are behaving much like the metal-pattern seen on solar cells, with their heavy tin-strip running down the center, and thin branches spreading across the surface. (Yer brain un-numbed yet?)

[1] These voltages are from the table of electrochemical potentials, as modified by the 4.4V of the "Absolute Hydrogen Potential." Typical tables use the platinum/hydrogen electrode as a reference, pretending that it's water/metal potential difference is zero. It's not. Actually it's roughly 4.4V, with the electrolyte being the positive terminal. This makes a difference in energy calculations, and reveals that the zinc corrosion is the energy source of the battery, while the copper electrode "steals" the majority of energy as it electroplates and thickens itself. ALL batteries do this: "wasting" a large portion of their energy in the positive-plate reactions. The outside world only sees the excess energy which remains after the pos-plate electroplating energy is subtracted from the neg-plate corrosion energy.

[2] Note about these backwards reactions on the positive-plate surface: we don't want our copper electrode to get plated with zinc! That would eliminate the copper/water contact, making both plate-surfaces the same, and wiping out the imbalanced voltages. So, just pre-load the salt water with some copper sulfate, or copper chloride, etc. Then, only copper is plated onto the copper surface. In the telegraph era, the "Crow's Foot" batteries would have two layers of dense salt water: some copper chloride near the copper electrode, and some zinc chloride near the zinc electrode. Eventually the zinc plate dissolves away, and also the copper ions in the salt water all get plated onto the copper electrode. Using a DC dynamo, they could drive the reactions backwards, to un-plate the copper and re-plate the zinc. Easier to just replace the zinc plate and the two fluids (heh then sell the thickened copper electrode to a scrap dealer!)

  • \$\begingroup\$ thank you for the lovely response. Question, though. Several times in your post, you talk about "positive protons". What exactly do you mean? Are you literally referring to hydrogen atoms that do not have an electron? Or do you mean positively charged ions? \$\endgroup\$
    – S.C.
    Mar 20, 2020 at 2:27
  • \$\begingroup\$ Also, you sort of lost me towards the end of the post. Perhaps I am misunderstanding, but you initially state that the voltage difference between the two electrodes is due to the zinc electrode "holding on to" more electrons than the copper electrode...resulting in a voltage of 1.1 V. However, later on you state that the copper plate ADDITIONALLY begins to get coated with protons. Does this mean that the voltage between the copper electrode and zinc electrode is now GREATER than 1.1 V (as a consequence of the excess negative charge on the copper being canceled out by positive ions)? \$\endgroup\$
    – S.C.
    Mar 20, 2020 at 2:34
  • 2
    \$\begingroup\$ @S.Cramer yep, "proton flow" is just the current of positive metal ions: like an un-cancelled proton which is dragging an entire atom along as it moves (I try to always call it "proton flow," even if it's pos metal ions, because too many students are convinced that all currents are electrons-flow only.) \$\endgroup\$
    – wbeaty
    Mar 20, 2020 at 8:46
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    \$\begingroup\$ @S.Cramer zinc plate has larger negative because water dragged off more positive ions. (The water-zinc contact-area is like a self-charged capacitor, with the water as the pos plate. A more energetic corrosion (metal like zinc, lithium, sodium, etc.) will have higher volts between the water and the metal, than the less-corroding metals (copper, gold, etc.) The pos ions may be near a plate, but they're in the conductive salt water. And, the 1.1V is entirely created by these corrosion effects, by water stealing pos ions. The copper has neg volts wrt the water, but pos volt wrt the zinc.. \$\endgroup\$
    – wbeaty
    Mar 20, 2020 at 8:55
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    \$\begingroup\$ @S.Cramer The total charge on each plate is actually unknown! After all, the entire battery will always end up charged wrt ground. For example, briefly bump the zinc end against ground. The battery responds, actively maintaining the 1.1V output. Now the neg terminal has become 0v wrt ground. Our initial three voltages (-5.2v, 0v -4.1v) have now become 0v, +5.2v, +1.1v, and the copper end is now true positive! Since the battery actively maintains the DIFFERENCE between output terminals, normally we ignore this stuff, and PRETEND that the common-bus in any circuit is truly neutral-charged. \$\endgroup\$
    – wbeaty
    Mar 22, 2020 at 18:26

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