Electrochemical Reduction of Aqueous Imidazolium on Pt(111) by Proton Coupled Electron Transfer

Abstract

Recent electrochemical studies have reported aqueous CO₂ reduction to formic acid, formaldehyde and methanol at potentials of ca. −600 mV versus SCE, when using a Pt working electrode in acidic pyridine solutions. In those experiments, pyridinium is thought to function as a one-electron shuttle for the underlying multielectron reduction of CO₂. DFT studies proposed that the critical step of the underlying reaction mechanism is the one-electron reduction of pyridinium at the Pt surface through proton coupled electron transfer. Such reaction forms a H adsorbate that is subsequently transferred to CO₂ as a hydride, through a proton coupled hydride transfer mechanism where pyridinium functions as a Brønsted acid. Here, we find that imidazolium exhibits an electrochemical behavior analogous to pyridinium, as characterized by the experimental and theoretical analysis of the initial reduction on Pt. A cathodic wave, with a cyclic voltammetric half wave potential of ca. −680 mV versus SCE, is consistent with the theoretical prediction based on the recently proposed reaction mechanism suggesting that positively charged Brønsted acids could serve as electrocatalytic one-electron shuttle species for multielectron CO₂ reduction.

Appendix

The reduction potential of the Brønsted acid (HA = PyrH⁺, ImH⁺) relative to the SHE, is obtained as the cell potential for the equilibrium:

$${\text{AH}} \to {\text{A}}^{ - } + {\text{ H}}^{ + } ,$$

(1)

which can be expressed as two half reactions:

$${\text{AH }} + {\text{ e}} - \to {\text{A}}^{ - } + \, \raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} {\text{ H}}_{ 2}$$

(2)

and

$$\raise.5ex\hbox{$\scriptstyle 1$}\kern-.1em/ \kern-.15em\lower.25ex\hbox{$\scriptstyle 2$} {\text{ H}}_{ 2} \to {\text{H}}^{ + } + {\text{ e}} -$$

(3)

The Gibbs free energy change ΔG for the overall reaction of Eq. (1) is given by

$$\Delta {\text{G }} = \Delta {\text{G}}^{\text{o}} + {\text{ RT ln }}\left[ {{\text{A}}^{ - } } \right]\left[ {{\text{H}}^{ + } } \right]/\left[ {\text{AH}} \right]$$

(4)

Which is essentially the Nernst equation upon substitution of ΔG^o = −F E^o and ΔG = −F E:

$${\text{E }} = {\text{ E}}^{\text{o}} {-} \, \left( {{\text{RT}}/{\text{F}}} \right){ \ln }\left[ {{\text{A}}^{ - } } \right]\left[ {{\text{H}}^{ + } } \right]/\left[ {\text{AH}} \right]$$

(5)

At equilibrium, ΔG = 0 and [A⁻][H⁺]/[AH] = K_a. Therefore,

$${\text{E}}^{0} = \, \left( {{\text{RT}}/{\text{F}}} \right){\text{ ln K}}_{{{\text{a}}({\text{AH}})}} = \, - \left( {{\text{RT}}/{\text{ F log e}}} \right){\text{ pK}}_{{{\text{a}}({\text{AH}})}} = \, - 5 9 {\text{ mV pK}}_{{{\text{a}}({\text{AH}})}}$$

(6)

When T = 298.15 K. Considering that E⁰ = E ⁰_{PyrH +}/_H(Pt)Pyr−E ⁰_{H +}/_H(Pt), we obtain:

$${\text{E}}^{\text{o}}_{\text{AH}} /_{{{\text{H}}\left( {\text{Pt}} \right){\text{A}} - }} = {\text{ E}}^{\text{o}}_{{{\text{H}} + }} /_{{{\text{H}}({\text{Pt}})}} - {\text{ 59 mV pKa}}_{( {{\text{AH}}})}$$

(7)