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I know that measuring the open-circuit voltage gives a good indication of the state of charge of a battery, both for rechargeables and single-use. But why is this so? As far as I can see the voltage is defined by the electron energies during the chemical reactions at the electrodes, which are constants. I think I can see why the internal resistance increases, since the molecules have to travel further before reacting, but the sub-microamp drain of a DVM is surely insufficient to cause a voltage drop of hundreds of millivolts.

So what am I missing?

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    $\begingroup$ > "the voltage is defined by the electron energies during the chemical reactions at the electrodes, which are constants. " Why do you think so? As the electric cell discharges, its chemical composition and physical properties on the electrodes change. More of unwanted stuff appears on the electrodes or in the electrolyte. Presence of this stuff can influence energies electrons gain in the electromotive chemical reactions. $\endgroup$
    – Ján Lalinský
    Apr 18, 2020 at 21:49
  • $\begingroup$ This probably is only vaguely similar, but compare it with a capacitor, where capacitance is constant and voltage is directly proportional to the stored charge. When you use some of the charge, the voltage drops proportionally. $\endgroup$
    – Krumuvecis
    Apr 19, 2020 at 3:42
  • $\begingroup$ @Ján Lalinský "Why do I think so?" because I'm not very clever ;-) I had heard of electronegativity and assumed it would be dominant. This seems not to be the case (see ChemEng's answer, about the concentrations) $\endgroup$
    – NL_Derek
    Apr 19, 2020 at 21:42

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The theoretical voltage of a cell is given by the Nernst Equation: $$\Delta G=-nFE=-RT\ln K_\mathrm{eq}+RT\ln Q$$ The Q is the reaction quotient and is dependent on the concentration. You want the concentration of reactants high and products low for a high cell potential

When the concentrations are 1M and the temperature is 25 degC and all the other conditions found here: https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page) are satisfied then the lnQ term is zero and the cell potential can be calculated through the standard electrode table simply as the sum of E_oxidation+E_reduction where E_oxidation=-E_reduction for the same reaction which is also the first term with the equilibrium constant divided by -nF

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  • $\begingroup$ This needs more explanation to answer the question. $\endgroup$
    – Andrew
    Apr 19, 2020 at 5:37
  • $\begingroup$ I agree with @Andrew, can you elaborate? I googled "Nernst Equation" and it seems that the concentration of the reagents is important (which actually answers my question), but even the Wikipedia entry is way over my head. $\endgroup$
    – NL_Derek
    Apr 19, 2020 at 21:49

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