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In both Volta potential and Nernst potential equations, besides some constants, temperature and valence parameters, the potential is a function of the logarithm of the ratio of two densities (or concentrations).

I have a problem understanding why a logarithm. In the ideal gas law, pressure difference is linearly proportional to the number of moles, so a differential (e.g. transmembrane) pressure is linearly proportional to the substraction of the two compartments number of moles, not a log of their ratio. So this must have to do, I guess, with the fundamental difference between a voltage and a pressure. A voltage is joule per coulomb, a pressure is (among other SI definitions) joule per cubic meter. Since coulomb force adds another pressure of some kind on top of the motion of particles and it has a longer range than collisions, this could be why.

I could understand a log as the integration of a series of divisions, and that would fit with a $1/r$ interaction potential decaying with distance. Yet, the log in said potentials equations is applied to a ratio.

Any hint?

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  • $\begingroup$ The Wikipedia articles have derivations. $\endgroup$ Sep 29, 2019 at 17:07
  • $\begingroup$ @Stevens: you are absolutely right, I missed that, sorry. For those interested, it is in the Nernst equation article (not Volta potential). $\endgroup$
    – Winston
    Sep 29, 2019 at 17:17

2 Answers 2

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The Boltzman distribution from statistical mechanics tells us that the charge $e$ particle density varies as $$ n\propto \exp\{-eV/kT\}. $$ and the gas constant $R$ and Boltzmann's constant $k$ are related by $R=N_{\rm Avagadro} k$. Similarly the Fraday unit is $F= eN_{\rm Avagadro}$ so we can write this as $$ n\propto \exp\{-FV/RT\}. $$ So, taking a difference to get rid of the unspecified factor of proprtionality, we get $$ \ln(n_A/n_B)=\ln n_A-\ln n_B= -\frac{F(V_A-V_B)}{RT}, $$
which is easily rearranged to get the formula in Wikipedia.

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  • $\begingroup$ Excellent, many thanks. To make sure I understand it completely though, "eV" here is coulombs x volts = coulomb x joules / coulombs = joules, and so you feed this energy of the state you're looking for into the Boltzmann distribution formula? I just read en.m.wikipedia.org/wiki/Boltzmann_distribution $\endgroup$
    – Winston
    Sep 29, 2019 at 16:57
  • $\begingroup$ Yes $eV$ is the potential energy of a charge $e$ Coulomb particle (needs minus sign if an electron) in potential of $V$ volts. $\endgroup$
    – mike stone
    Sep 29, 2019 at 17:10
  • $\begingroup$ Ok great, thanks again. $\endgroup$
    – Winston
    Sep 29, 2019 at 17:18
  • $\begingroup$ I should have added that I do not see what the Wikipedia formula has to do with the contact potential between two dissimilar metals. That voltage is determined by the work function, and that depends on band theory and chemisty. The equation seem more applicable to electrolytes. $\endgroup$
    – mike stone
    Sep 29, 2019 at 18:23
  • $\begingroup$ Do you suggest another formula? $\endgroup$
    – Winston
    Sep 30, 2019 at 6:10
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The log comes from the way gibbs free energy is related to the potential and the reaction quotient $$ \Delta G = \Delta G^{\ominus} + RT\ln Q \\ \Delta G = -zFE $$ And we know $$ {d}G = \sum_{i=1}^k \mu_i \,\mathrm{d}N_i $$ Where the chemical potential is defined and related to the activity/concentration as: $$ a_i = e^{\frac{\mu_i - \mu^{\ominus}_i}{RT}} \\ \mu_i = \mu_i^{\ominus} + RT\ln{a_i} $$

So you have a logarithm

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  • $\begingroup$ Please improve your answer. Make $\ln Q$ plausible. $\endgroup$
    – Rainald62
    Dec 30, 2019 at 0:45
  • $\begingroup$ I have you see the chemical potential is related to the activity(effective concentration) through the ln operator and the gibbs free energy is the addition of the chemical potential multiplied by dN and since logarithms added or subtracted are just logs of the terms multiplied and divided it is Q the reaction quotient $\endgroup$
    – ChemEng
    Dec 30, 2019 at 1:22
  • $\begingroup$ Please, make $a_i = e^{\frac{\mu_i - \mu^{\ominus}_i}{RT}}$ plausible. Trivial math doesn't help. $\endgroup$
    – Rainald62
    Dec 30, 2019 at 1:39
  • $\begingroup$ It is the definition of activity see here en.wikipedia.org/wiki/Thermodynamic_activity $\endgroup$
    – ChemEng
    Dec 30, 2019 at 18:03
  • $\begingroup$ Please, make plausible why this term is useful as 'a measure of the "effective concentration" '. $\endgroup$
    – Rainald62
    Dec 31, 2019 at 18:13

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