1. Electron transfer in heterogeneous systems (on electrodes)

2. Impacts on our society Most of the modern method of generating electricity are inefficient, and the development of fuel cells could revolutionize the production and deployment of energy. The inefficiency can be improved by knowing more about the kinetics of electrochemical processes. Applications in organic and inorganic electrosynthesis. Waste treatments.

3. 25.8 The electrode-solution interface Electrical double layer Helmholtz layer model 

4. Gouy-Chapman Model This model explains why measurements of the dynamics of electrode processes are almost always done using a large excess of supporting electrolyte. ( Why??)

5. The Stern model of the electrode- solution interface The Helmholtz model overemphasizes the rigidity of the local solution. The Gouy-Chapman model underemphasizes the rigidity of local solution. The improved version is the Stern model.

6. The electric potential at the interface 1.Outer potential 2. Inner potential 3. Surface potential The potential difference between the points in the bulk metal (i.e. electrode) and the bulk solution is the Galvani potential difference which is the electrode potential discussed in chapter 7.

7. The origin of the distance- independence of the outer potential

8. The connection between the Galvani potential difference and the electrode potential Electrochemical potential (û) û = u + zFø Discussions through half-reactions

9. 25.9 The rate of charge transfer Expressed through flux of products: the amount of material produced over a region of the electrode surface in an interval of time divided by the area of the region and the duration of time interval. The rate laws Product flux = k [species] The rate of reduction of Ox, v ox = k c [Ox] The rate of oxidation of Red, v Red = k a [Red] Current densities: j c = F k c [Ox] for Ox + e - → Red j a = F k a [Red] for Red →Ox + e - the net current density is: j = j a – j c = F k a [Red] - F k c [Ox]

10. The activation Gibbs energy Write the rate constant in the form suggested by activated complex theory: Notably, the activation energies for the catholic and anodic processes could be different!

11. The Butler-Volmer equation Variation of the Galvani potential difference across the electrode – solution interface

12. The reduction reaction, Ox + e → Red

14. The parameter α is called the transient coefficient and lies in the range 0 to 1. Based on the above new expressions, the net current density can be expressed as:

15. Example 25.1 Calculate the change in cathodic current density at an electrode when the potential difference changes from ΔФ’ to ΔФ Self-test 25.5 calculate the change in anodic current density when the potential difference is increased by 1 V.

16. Overpotential When the cell is balanced against an external source, the Galvani potential difference,, can be identified as the electrode potential. When the cell is producing current, the electrode potential changes from its zero-current value, E, to a new value, E’. The difference between E and E’ is the electrode’s overpotential, η. η = E’ – E The ∆Φ = η + E, Expressing current density in terms of η j a = j 0 e (1-a)fη and j c = j 0 e -afη where j o is called the exchange current density, when j a = j c

17. The butler-Volmer equation: j = j 0 (e (1-a)fη - e -afη ) The lower overpotential limit ( η less than 0.01V) The high overpotential limit (η ≥ 0.12 V)

18. The low overpotential limit The overpotential η is very small, i.e. fη <<1 When x is small, e x = 1 + x + … Therefore j a = j 0 [1 + (1-a) fη] j c = j 0 [1 + (-a fη)] Then j = j a - j c = j 0 [1 + (1-a) fη] - j 0 [1 + (-a fη)] = j 0 fη The above equation illustrates that at low overpotential limit, the current density is proportional to the overpotential. It is important to know how the overpotential determines the property of the current.

19. Calculations under low overpotential conditions Example 25.2: The exchange current density of a Pt(s)|H 2 (g)|H + (aq) electrode at 298K is 0.79 mAcm -2. Calculate the current density when the over potential is +5.0mV. Solution: j 0 = 0.79 mAcm -2 η = 5.0mV f = F/RT = j = j 0 fη Self-test 25.6: What would be the current at pH = 2.0, the other conditions being the same?