Presentation on theme: "Chapter 25 Electron transfer in heterogeneous systems (Processes at electrodes)"— Presentation transcript:
Chapter 25 Electron transfer in heterogeneous systems (Processes at electrodes)
25.8 The electrode-solution interface Electrical double layer Helmholtz layer model
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.
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.
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.
The origin of the distance- independence of the outer potential
The connection between the Galvani potential difference and the electrode potential Electrochemical potential (û) û = u + zFø Discussions through half-reactions
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]
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!
The Butler-Volmer equation Variation of the Galvani potential difference across the electrode – solution interface
The reduction reaction, Ox + e → Red
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:
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.
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
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)
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.
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?
The high overpotential limit The overpotential η is large, but could be positive or negative ! When η is large and positive j c = j 0 e -afη = j 0 /e afη becomes very small in comparison to j a Therefore j ≈ j a = j 0 e (1-a)fη ln(j) = ln(j 0 e (1-a)fη ) = ln(j 0 ) + (1-a)fη When η is large but negative j a is much smaller than j c then j ≈ - j c = - j 0 e -afη ln(-j) = ln(j 0 e -afη ) = ln(j 0 ) – afη Tafel plot: the plot of logarithm of the current density against the over potential.
Applications of a Tafel plot The following data are the anodic current through a platinum electrode of area 2.0 cm 2 in contact with an Fe 3+, Fe 2+ aqueous solution at 298K. Calculate the exchange current density and the transfer coefficient for the process. η/mV I/mA Solution: Asked to calculate j 0 and α First, I needs to be converted to J second: choose ln(j) = ln(j 0 e (1-a)fη ) = ln(j 0 ) + (1-a)fη
Self-test 25.7: Repeat the analysis using the following cathodic current data: η/mV I/mA In general exchange currents are large when the redox process involves no bond breaking or if only weak bonds are broken. Exchange currents are generally small when more than one electron needs to be transferred, or multiple or strong bonds are broken.
The general arrangement for electrochemical rate measurement