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How does the Butler-Volmer equation collapse into the Nernst equation when applying the equilibrium potential current? I don't know where to start, and this source I used (https://www.sciencedirect.com/topics/engineering/butler-volmer-equation#:~:text=It%20agrees%20with%20the%20Nernst,true%20under%20high%2Dresistance%20conditions.) uses Bard and Faulkner as given, but I would like to know how to derive it properly. Thanks in advance

Edit: I have to do the first task:

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Maybe you are confused with the concept of equilibrium potential current, which seems strange. BV equation collapses to Nernst equation when the current density is zero, i.e., is thermodynamically consistent since it is an extension of transition theory to electrochemical reactions. An analogy would be recovering the law of mass action ($K = \prod_j a_j^{\nu_j}$) when setting the rate $r = 0$ for a chemically reversible reaction in a homogeneous phase.

When this is the case, the potential of the electrode with respect to the bulk solution is named the equilibrium potential. The math is immediate from the BV expression and setting $j = 0$

\begin{align} j &= j_0\left\{\exp\left(-\frac{\alpha F \eta}{RT}\right) - \exp\left[\frac{(1 - \alpha) F \eta}{RT}\right] \right\} \\ 0 &= j_0\left\{\exp\left(-\frac{\alpha F \eta}{RT}\right) - \exp\left[\frac{(1 - \alpha) F \eta}{RT}\right] \right\} \\ 0 &= \exp\left(-\frac{\alpha F \eta}{RT}\right) - \exp\left[\frac{(1 - \alpha) F \eta}{RT}\right] \\ \exp\left[\frac{(1 - \alpha) F \eta}{RT}\right] &= \exp\left(-\frac{\alpha F \eta}{RT}\right) \\ \exp\left(\frac{F \eta}{RT}\right) &= 1 \\ \frac{F \eta}{RT} &= 0 \\ \eta &= 0 \\ E - E^\text{eq} &= 0 \rightarrow \boxed{E = E^\text{eq} = E^\circ - \frac{RT}{nF} \ln \left(\prod_j a_j^{\nu_{j}}\right)} \tag{1} \\ \end{align}

Although we cannot directly measure the potential of a conducting solid phase, nor the potential of a liquid phase, the meaning of Eq. (1) is as follows. In an electrochemical cell in which no current is circulating between the electrodes, the electrode potential $E = \phi_\text{electrode} - \phi_\text{solution}$ in any of the electrodes, is only a function of the activities of the species participating in that electrochemical reaction in the bulk electrolyte solution.

I suggest reading chapter 2 from the book you cited, but if you need more precisions, feel free to comment and I will add more information.


Edit

Let us study the academic electrochemical reaction $$ \ce{\nu_0O + ne^- <=> \nu_RR} \tag{2} $$

We employ the current-potential equation, equivalent to the BV equation, but before making manipulations in order to put in in terms of the overpotential and the exchange current density \begin{equation} j = Fk^\circ\left\{ a_O^{\nu_O}(0,t)\exp\left[-\frac{\alpha nF(E - E^\circ)}{RT}\right] - a_R^{\nu_R}(0,t)\exp\left[\frac{(1 - \alpha) nF(E - E^\circ)}{RT}\right] \right\} \tag{3} \end{equation}

When the electrode potential is the equilibrium potential, i.e. $E = E^\text{eq}$ and $j = 0$, from Eq. (3) we obtain \begin{gather} 0 = Fk^\circ\left\{ a_O^{\nu_O}(0,t)\exp\left[-\frac{\alpha nF(E^\text{eq} - E^\circ)} {RT}\right] - a_R^{\nu_R}(0,t)\exp\left[\frac{(1 - \alpha)nF(E^\text{eq} - E^\circ)} {RT}\right] \right\} \\ 0 = \color{blue}{Fk^\circ a_O^{\nu_O}(0,t)\exp\left[-\frac{\alpha nF(E^\text{eq} - E^\circ)} {RT}\right]} - \color{red}{Fk^\circ a_R^{\nu_R}(0,t)\exp\left[\frac{(1 - \alpha)nF(E^\text{eq} - E^\circ)} {RT}\right]} \\ 0 = a_O^{\nu_O}(0,t)\exp\left[-\frac{\alpha nF(E^\text{eq} - E^\circ)} {RT}\right] - a_R^{\nu_R}(0,t)\exp\left[\frac{(1 - \alpha)nF(E^\text{eq} - E^\circ)} {RT}\right] \\ a_O^{\nu_O}(0,t)\exp\left[-\frac{\alpha nF(E^\text{eq} - E^\circ)} {RT}\right] = a_R^{\nu_R}(0,t)\exp\left[\frac{(1 - \alpha)nF(E^\text{eq} - E^\circ)} {RT}\right] \\ \dfrac{a_O^{\nu_O}(0,t)}{a_R^{\nu_R}(0,t)} = \dfrac{\exp\left[\dfrac{(1 - \alpha)nF(E^\text{eq} - E^\circ)} {RT}\right]} {\exp\left[-\dfrac{\alpha nF(E^\text{eq} - E^\circ)}{RT}\right]} \\ \dfrac{a_O^{\nu_O}(0,t)}{a_R^{\nu_R}(0,t)} = \exp\left[\dfrac{nF(E^\text{eq} - E^\circ)}{RT}\right] \\ \ln\left[\dfrac{a_O^{\nu_O}(0,t)}{a_R^{\nu_R}(0,t)}\right] = \frac{nF(E^\text{eq} - E^\circ)}{RT} \\ \frac{nF(E^\text{eq} - E^\circ)}{RT} = \ln\left[\dfrac{a_O^{\nu_O}(0,t)}{a_R^{\nu_R}(0,t)}\right] \\ E^\text{eq} = E^\circ\ + \frac{RT}{nF} \ln\left[\dfrac{a_O^{\nu_O}(0,t)}{a_R^{\nu_R}(0,t)}\right] \rightarrow \boxed{E^\text{eq} = E^\circ\ - \frac{RT}{nF} \ln\left[\dfrac{a_R^{\nu_R}(0,t)}{a_O^{\nu_O}(0,t)}\right]} \tag{4} \end{gather}

  • Eq. (3) comes by studying a free energy diagram vs the reaction coordinate. You can relate the changes in free energy with changes in the potential $E$. You may want to see chapter 3 from Bard and Faulkner.
  • In said book you will see that instead the standard reduction potential $E^\circ$, Bard writes the formal potential $E^{0'}$. This is because, almost in all cases, we don't know the activity coefficients. So the $E^{0'}$ incorporates its dependence (somewhat loosely), so we can write Eq. (3) with concentrations instead of activities. You can write it with activity coefficients as I did, to have a more general form.
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  • $\begingroup$ Nice explanation. Is there a way of deriving the Nernst equation from Butler-Volmer directly instead of knowing that E=Nernst equation? For example, setting E=E(eq) and going from here? That's what my sheet asks me to do $\endgroup$
    – Mäßige
    Nov 16, 2023 at 15:46
  • $\begingroup$ @Mäßige Hi!I think this is the way of deriving it. I imagine you know that the definition of the overpotential is $\eta \equiv E - E^\text{eq}$. So if the overpotential is zero, then the electrode potential reduces to the electrode potential dictated by the Nernst equation. However, maybe I am mistaking your question. If you can upload a photo of the exercise in the comments, I can see what might be going on. $\endgroup$ Nov 16, 2023 at 16:07
  • $\begingroup$ Wait I think you might be right again. Wait let me upload it as an edit $\endgroup$
    – Mäßige
    Nov 16, 2023 at 16:44
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    $\begingroup$ Woww that's a very nice derivation thanks a lot :D $\endgroup$
    – Mäßige
    Nov 17, 2023 at 8:57
  • $\begingroup$ One more question; why is ΔE=E(eq)-E(0) necessarily? Why the standard reduction potential, I thought the Butler-Volmer equation applied to any electrode, i.e. the redox pair not necessarily at standard conditions, just how the potential differs from equilibrium potential of the redox pair (which does not need to be at standard conditions)? $\endgroup$
    – Mäßige
    Nov 17, 2023 at 11:23

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