User:PAR/Work4

Plot of Gibbs energy vs. reaction coordinate for a redox reaction involving n electrons.

O + ne- = R

The reaction will proceed towards the lowest Gibbs energy, at a rate determined by the activation energy G^*, which will only be overcome by molecules with a Maxwell-distributed energy greater than G^*. The activation energy is found at the intersection of the cathode and anode energy curves. Imposing a voltage E on the electrode lowers the cathode curve by nFE, and lowers the activation energy from G^*(0) to G^*(E). If the curves can be considered to be practically linear in the transition region, and their slopes are C and -A for the cathode and anode energy curves respectively, then the transfer coefficient α is defined as A/(A+C).

For an applied potential E, the activation energy (energy barrier) to be overcome on the oxidizing side is seen to be:

$\Delta ^{\ddagger }G_{c}=\Delta ^{\ddagger }G_{oc}+\alpha nFE$

and on the reduction side:

$\Delta ^{\ddagger }G_{a}=\Delta ^{\ddagger }G_{oa}-(1-\alpha )nFE$

The forward and backward reaction rate constants (kc and ka) are determined by the activation energy, which only allows the fraction of molecules with sufficient energy overcome the barrier and to react. Assuming that this fraction, and therefore the reaction rate constants, are well represented by an Arrhenius equation:

$k_{c}=A_{c}\exp(-\Delta ^{\ddagger }G_{c}/RT)$

$k_{a}=A_{a}\exp(-\Delta ^{\ddagger }G_{a}/RT)$

so that:

$k_{c}=A_{c}\exp(-\Delta ^{\ddagger }G_{oc}/RT)\exp(-\alpha E/V_{T})$

$k_{a}=A_{a}\exp(-\Delta ^{\ddagger }G_{oa}/RT)\exp((1-\alpha )E/V_{T})$

Where V_T=RT/F is the thermal voltage. There exists an equilibrium voltage E^o at which the forward and reverse reaction rates are equal, and is given by:

$E^{o}=-(\Delta Go-\Delta Gr>/nF$

In this case, kf and kr are equal and ko is defined as ko=kf=kr.

$k_{o}=A_{c}\exp(-\Delta ^{\ddagger }G_{oc}/RT)\exp(-\alpha E_{o}/V_{T})$

$k_{o}=A_{a}\exp(-\Delta ^{\ddagger }G_{oa}/RT)\exp((1-\alpha )E_{o}/V_{T})$

So that for a general E,

$k_{c}=ko\exp(-\alpha (E-Eo)/V_{T})$

$k_{a}=ko\exp((1-\alpha )(E-Eo)/V_{T})$

The net rate of reaction is then v=k_c C_c - k_a C_a, where C's are concentrations, and by Faradays law, this is related to the current density by:

$j=nFv=nFko(C_{c}\exp(-\alpha (E-Eo)/V_{T})-C_{a}\exp((1-\alpha )(E-Eo)/V_{T}))$

which is just the Butler-Volmer equation.

Problem: Wikipedia says C/C*