A Simple Method to Establish the Relationship Between the Equilibrium Polarization Resistance and the Elementary Kinetic Parameters of an Electrocatalysed Reaction

Abstract

A simple and rigorous methodology to establish the relationship between the faradaic equilibrium polarization resistance of an electrocatalytic reaction with the elementary kinetic parameters involved in the reaction mechanism is proposed. It was derived through an alternative method, which avoided the differentiation of the corresponding current-overpotential dependence. This formalism includes the cases where both, reactants and products, exhibit diffusion contributions. It is demonstrated that the equilibrium polarization resistance is the sum of both, faradaic and diffusion, contributions. Each diffusion term has a linear variation with the inverse of the limiting diffusion current density of the species involved. This behavior was exemplified with two different experimental data sets for the hydrogen electrode reaction, obtained on a rotating disc and microelectrodes, respectively.

Appendix

Reaction (3) can be decomposed into two mass transfer processes (towards and from the electrode surface) and the faradaic process that takes place on the electrode surface,

$$ {\displaystyle \begin{array}{c}{A}_{2\left(\mathrm{sol}\right)}\to {A}_{2\left(\mathrm{es}\right)}\kern0.5em \left(\mathrm{diffusion}\kern0.3em \mathrm{to}\kern0.3em \mathrm{electrode}\kern0.3em \mathrm{surface}\right)\kern2.5em \\ {}{A}_{2\left(\mathrm{es}\right)}\rightleftharpoons \kern0.5em 2\kern.2em {P}_{\left(\mathrm{es}\right)}+2\kern.2em {H}^{+}+2\kern.2em {e}^{-}\kern0.5em \left(\mathrm{faradaic}\kern0.34em \mathrm{reaction}\right)\kern1em \\ {}\begin{array}{cc}{P}_{\left(\mathrm{es}\right)}\to {P}_{\left(\mathrm{sol}\right)}& \left(\mathrm{diffusion}\kern0.3em \mathrm{from}\kern0.3em \mathrm{electrode}\kern0.3em \mathrm{surface}\right)\kern1.5em \end{array}\kern0.5em \end{array}} $$

The variation of Gibbs free electrochemical energy of reaction ($\Delta {\overline{g}}_r$) is:

$$ \Delta {\overline{g}}_r=2\ {\overline{\mu}}_{H^{+}}+2{\overline{\mu}}_{e^{-}}+2\ {\mu}_{P\left(\mathrm{sol}\right)}-{\mu}_{A_2\left(\mathrm{sol}\right)} $$

(30)

where ${\overline{\mu}}_i$ and μ _i are the electrochemical and chemical potential respectively of the species i.

Adding and subtracting μ _i(es) (i: A ₂, P), Eq. (30) can be rewritten as:

$$ \Delta {\overline{g}}_r=\left(2{\overline{\mu}}_{H^{+}}+2{\overline{\mu}}_{e^{-}}+2{\mu}_{P\left(\mathrm{es}\right)}-{\mu}_{A_2\left(\mathrm{es}\right)}\right)+2\left({\mu}_{P\left(\mathrm{sol}\right)}-{\mu}_{P\left(\mathrm{es}\right)}\right)+\left({\mu}_{A_2\left(\mathrm{es}\right)}-{\mu}_{A_2\left(\mathrm{sol}\right)}\right) $$

(31)

where the three parentheses define $\Delta {\overline{g}}_r^{\mathrm{es}}$, Δg _P , and $\Delta {g}_{A_2}$, respectively,

$$ \Delta {\overline{g}}_r=\Delta {\overline{g}}_r^{\mathrm{es}}+2\Delta {g}_P+\Delta {g}_{A_2}=-2 F\eta $$

(32)

Derivation of Eqs. (19), (20), and (21)

As the reaction rate increases, the surface concentration of the reactant decreases and that of the product increases. Consequently, a shift in the surface reaction equilibrium potential occurs. This change is given by the potential difference of the reactant (or product) concentration cell,

$$ Pt(L)/{A}_2\left({c}_{\left(\mathrm{sol}\right)}\right)/{P}_{\left(\mathrm{sol}\right)}/{A}_2\left({c}_{\left(\mathrm{es}\right)}\right)/ Pt(R) $$

where Pt(L) and Pt(R) are the left and right electrodes of the cell, respectively. The following reactions are verified:

$$ {A}_{2\left(\mathrm{sol}\right)}\left({c}_{\mathrm{sol}}\right)\rightleftharpoons {P}_{\left(\mathrm{sol}\right)}+2{H}_{\left(\mathrm{sol}\right)}^{+}+2\ {e}_L^{-} $$

(33a)

$$ {P}_{\left(\mathrm{sol}\right)}+2{H}_{\left(\mathrm{sol}\right)}^{+}+2\ {e}_R^{-}\rightleftharpoons {A}_{2\left(\mathrm{es}\right)}\left({c}_{\mathrm{es}}\right) $$

(33b)

Being the overall reaction of the cell,

$$ {A}_{2\left(\mathrm{sol}\right)}\left({c}_{\mathrm{sol}}\right)+2{e}_R^{-}\rightleftharpoons {A}_{2\left(\mathrm{es}\right)}\left({c}_{\mathrm{es}}\right)+2\ {e}_L^{-} $$

(34)

Applying the equilibrium condition $\sum {\upsilon}_i{\overline{\mu}}_i=0$, it is rapidly arrived to the following expression, which defines the concentration overpotential due to the reactant A ₂ $\left({\eta}_{A_2}^d\right)$,

$$ {\mu}_{A_{2\left(\mathrm{es}\right)}}-{\mu}_{A_{2\left(\mathrm{sol}\right)}}=\Delta {g}_{A_2}=-2F{\eta}_{A_2}^d= RTln\frac{c_{A_{2\left(\mathrm{es}\right)}}}{c_{A_{2\left(\mathrm{sol}\right)}}} $$

(35)

Thus,

$$ \frac{c_{A_{2\left(\mathrm{es}\right)}}}{c_{A_{2\left(\mathrm{sol}\right)}}}={e}^{-\Delta {g}_{A_2}/ RT} $$

(36)

Applying the Nernst model of the diffusion layer [1],

$$ \frac{f_{A_2}^d}{f_{A_{2L}}^d}=\left(1-\frac{c_{A_{2\left(\mathrm{es}\right)}}}{c_{A_{2\left(\mathrm{sol}\right)}}}\right) $$

(37)

Substituting Eq. (36) into Eq. (37) and applying MacLaurin series development, the following linear dependence is obtained:

$$ {f}_{A_2}^d=-\frac{f_{A_2L}^d}{RT}\Delta {g}_{A_2}\left({\eta}_{A_2}^d\right) $$

(38)

Equation (38) corresponds to the application of Eq. (13) to species A ₂.

Furthermore, for the reaction product P, the concentration cell is:

$$ Pt(L)/{A}_{2\left(\mathrm{sol}\right)}/P\left({c}_{\left(\mathrm{sol}\right)}\right)/P\left({c}_{\left(\mathrm{es}\right)}\right)/{A}_{2\left(\mathrm{sol}\right)}/ Pt(R) $$

Operating in a similar way as in the previous case:

$$ {\mu}_{P_{\left(\mathrm{sol}\right)}}-{\mu}_{P_{\left(\mathrm{es}\right)}}=\Delta {g}_P=-F{\eta}_P^d= RTln\ \frac{c_{P_{\left(\mathrm{sol}\right)}}}{c_{P_{\left(\mathrm{es}\right)}}} $$

(39)

Finally, replacing Eqs. (35) and (39) into Eq. (32) and taking into account that $\Delta {\overline{g}}_r^{es}=-2F{\eta}^{\mathrm{es}}$ [14], Eq. (32) can be rewritten as:

$$ {\eta}_{A_2}^d+{\eta}_P^d+{\eta}^{\mathrm{es}}=\eta $$

(40)