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Cathodic Current, i c Anodic Current, i a

Kinetics of Electron Transfer at the Solid Electrode-Solution Interface. Cathodic Current, i c Anodic Current, i a. The rate of consumption/production of a species at an electrode of area A (cm 2 ) is:. Both in units of mol cm -2 s -1 and so k in units of cm s -1.

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Cathodic Current, i c Anodic Current, i a

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  1. Kinetics of Electron Transfer at the Solid Electrode-Solution Interface Cathodic Current, ic Anodic Current, ia The rate of consumption/production of a species at an electrode of area A (cm2) is: Both in units of mol cm-2 s-1 and so k in units of cm s-1 There must be a net current at any given potential, E: Note that kf, kb = F(E), as required by any kinetic theory.

  2. D G DGcis the free energy of activation for cathodic reaction DGais the free energy of activation for anodic reaction D D G G c a State 1 State 2 O +ne R Reaction Coordinate This means that kf = kbandDGc = DGa Kinetics of Electron Transfer at the Solid Electrode-Solution Interface So, how do we find F, for k = F(E)?? Recall: Potential Energy Curves for Reactions Also recall Arrhenius Equation (or energy needed to take 1 mole O/R to activated state) We have chosen E =Eo’ so that CO(0,t) = CR(0,t)= 1, and inet = 0, and then ic = - ia. This is known as "standard state" conditions, and the resulting current, namely the standard exchange current, i0, flows. More about this later.

  3. o’ E for initial Electrode E D D G = - nF E D G o’ D E = (E - E ) D D G G 0, a 0,c D a) D G = (1 - nF E D G a More +E D G c D a D G = - nF E Thus, DG a = DG0,a - (1-a)nFDE The barrier for reduction is now more by the fraction anFDE, so DG c = DG0,c + anFDE Kinetics of Electron Transfer at the Solid Electrode-Solution Interface What if we vary the electrode potential? E for final Electrode E, O +ne R Reaction Coordinate The barrier for oxidation is now less than DG0,a by a fraction (1-a) of the total energy change, where 0≤a≤1 (vide infra). Assume Arrhenius applies:

  4. Kinetics of Electron Transfer at the Solid Electrode-Solution Interface Substituting and letting f = F/RT: and At DE= 0, and When CO* = CR*, E = E0', and thus kf = kb = k0 This is the Standard Rate Constant, which is either kf or kb at E0' Finally, and Now, what is the net current at any givenE??? This is known as the Eyring Equation or Current-Potential Characteristic

  5. o’ E for initial Electrode E D G More +E Kinetics of Electron Transfer at the Solid Electrode-Solution Interface What is the term alpha? E for final Electrode E j q q O +ne R Reaction Coordinate So, a = 0.5 when the two energy wells are symmetric. Thus, a is a measure of the symmetry of the energy barrier. We will now examine the Eyring Equation under various conditions. Case I: Under equilibrium conditions By definition, E = Eeq which means that inet =0 At equilibrium: and So,

  6. Kinetics of Electron Transfer at the Solid Electrode-Solution Interface Taking the natural log: The Nernst Equation, as expected! We are in the standard state, so inet = 0, but ia = ic≠ 0 The equilibrium current is called the standard exchange current, i0 See B&F for derivation Example: O + e ↔ R a = 0.4 What is the current density?

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