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Engel and Reid's Physical Chemistry [1, p. 308] defines the electrochemical potential in terms of the charge in units of the electron charge $(z),$ Faraday constant $(F),$ and electrical potential $(\phi)$ as follows:

The electrochemical potential is a generalization of the chemical potential to include the effect of an electrical potential on a charged particle. It is the sum of the normal chemical potential $\mu$ and a term that results from the nonzero value of the electrical potential:

$$\tilde{\mu} = \mu + zF\phi \tag{11.3}$$

  1. Is the electrical potential $\phi$ due to electric field generated from charge separation between electrode and solution, or is it an externally applied electric field?

  2. Isn't $zF\phi$ electrical potential energy? How can we simply add it to chemical potential?

Reference

  1. Engel, T.; Reid, P. Thermodynamics, Statistical Thermodynamics, and Kinetics: Physical Chemistry, 4th ed.; Pearson: New York, 2019. ISBN 978-0-13-480458-3.
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  • $\begingroup$ As long as the dimensional analysis is correct, the math or equation will never know the labels of those terms (whatever the humans call the first term or the second term). $\endgroup$
    – AChem
    Jun 28, 2023 at 4:57
  • $\begingroup$ I did check the dimensions of both the terms and they do come out to be equal. But we are adding electrical potential energy to chemical potential. Isn't it like adding oranges and apples? $\endgroup$
    – Natasha J
    Jun 28, 2023 at 5:03

1 Answer 1

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The chemical potential of a substance $i$ in a system is generally defined in relation to the Gibbs energy differential:

$$\mathrm{d}G = \left( \frac{\partial G}{\partial n_i} \right)_{T,p}\mathrm{d}n_i = \mu_i\ \mathrm{d}n_i \tag{1}$$

The related Gibbs energy differential for a charge is:

$$\mathrm{d}G = \left( \frac{\partial G}{\partial q} \right)_{T,p}\mathrm{d}q = \phi \mathrm{d}q = \phi (zF \mathrm{d}n_i) = (\phi zF) \mathrm{d}n_i = \mu_{i, E}\cdot \mathrm{d}n_i \tag{2}$$

Then, it can be summed up:

$$\mathrm{d}G = \tilde{\mu_i}\cdot \mathrm{d}n_i = \left(\mu_i + \mu_{i, E}\right)\cdot \mathrm{d}n_i =\left(\mu_i + \phi zF\right)\cdot \mathrm{d}n_i \tag{3}$$

By other words, the electrochemical potential of a charged molecular entity is its chemical potential dependent on local electrostatic potential:

$$\tilde{\mu_i}=\tilde{\mu_i}(\phi)\tag{4}$$

This allows us to modify chemical potentials of molecular entities at the electrode surface by externally forced potential. This is used during electrolysis, when reactions that would not be otherwise spontaneous become spontaneous.


Commenting the feedback:

I have incorrectly evaluated your level from your question. :-) I think I have not written nothing extraordinary.

The contribution of electrostatic potential can be written as dG/dn the same way as is for the chemical potential. It formally contributes to the summary electrochemical potential.


Regarding the 1st question, it is potential of any origin. It may be forced externally, it may be caused local charge disbalance, it may be caused by the electrode surface dual layer, it may be the bulk solution volume potential, there may be Ohmic potential gradient....

But the potential at electrode/solution boundary is the most important, as that is the place where redox reactions occur.

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  • $\begingroup$ I see... Thank you very much. Can you comment on my first question (1.)? I just want to have a clear understanding about this topic. Thanks again!! $\endgroup$
    – Natasha J
    Jun 28, 2023 at 17:14
  • $\begingroup$ thank you again! $\endgroup$
    – Natasha J
    Jun 29, 2023 at 4:13

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