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I know there's lots of questions that address similar situations, (Batteries connected in Parallel, Batteries and fields?, Naive Question About Batteries, and the oft-viewed I don't understand what we really mean by voltage drop).

However, I have a question, and after examining just the battery structure, I have been wondering, exactly what structure/process maintains the constant voltage drop within batteries? I mean, certain chemical reactions are occurring in each half-cell, and electrolytes maintain charge conservation, I get that there's some motivation for the electron to move from one cell to the other.

But why is this motivation so constant? I really want to get this.

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Consider for a moment, a cell that is not connected to a circuit, i.e., there is no path for current external to the cell.

The chemical reactions inside the cell remove electrons from the cathode and add electrons to the anode.

Thus, as the chemical reactions proceed, an electric field builds between the anode and cathode due to the differing charge densities.

It turns out that this electric field acts to reduce the rate of the chemical reactions within the cell.

At some point, the electric field is strong enough to effectively stop the chemical reactions within the cell.

The voltage across the terminals of the cell, due to this electric field, is then constant and this is the open-circuit voltage of the cell.

If an external circuit is connected to the cell, electrons flow from the anode through the external circuit and into the cathode, reducing the difference in charge densities which in turn reduces the electric field just enough such that the chemical reactions can once again take place to maintain the electric current through the circuit.

The larger the external current, the greater the required rate of chemical reactions and thus, the lower the voltage across the terminals.

As long as the circuit current is significantly less than the maximum current the chemicals reactions can sustain, the voltage across the battery terminals will be close to the open circuit voltage.

As the external current approaches the maximum current, the voltage across the terminals rapidly falls and when the voltage is zero, the cell is supplying maximum current. This current is called the short-circuit current.

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    $\begingroup$ +1, and I'd add that the electric field required to stop the chemical reaction is related to the redox potential. $\endgroup$ – Phil Frost Jul 17 '14 at 18:52
  • $\begingroup$ Thanks for the answer. In the above example, are the two electrodes connected by a salt bridge? Also, I thought that what slowed down the reaction was the electric field between the interior of the electrode and the electrolytic solution, not the electric field between the anode and cathode. $\endgroup$ – Andres Salas Jul 17 '14 at 19:02
  • $\begingroup$ Also, I see that an electron going from point a (on an anode) to point b (on a cathode) will receive a given amount of energy and thus has a voltage, but what about a roundabout journey from a to b following a circuit? Is potential still the same in this case? $\endgroup$ – Andres Salas Jul 17 '14 at 19:28
  • $\begingroup$ @AndresSalas, the example is general. I don't quite understand what you mean by "interior of the electrode and the electrolytic solution". Also, your 2nd comment is unclear. We measure voltage across or between two points in a circuit. I'm not quite sure how to interpret "an electron ... thus has a voltage". Try this link: av8n.com/physics/lead-acid.htm $\endgroup$ – Alfred Centauri Jul 17 '14 at 19:53
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    $\begingroup$ @AndresSalas, your intuition needs work is all. Intuition is not automatic. Intuition comes from studying and working with the subject, correcting erroneous premises and intuition along the way. If you keep studying and asking questions, at some point you'll see where you're misconceptions are and, after correcting those, this will be intuitive. $\endgroup$ – Alfred Centauri Jul 17 '14 at 22:21

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