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1、4. Corrosion Kinetics,At the electrode/electrolyte interface, a charge separation between the metal surface and the electrolyte occurs. The spatial region corresponding to the charge separation is called the electrical double layer. It is usually separated into two parts, the Helmholtz layer or comp
2、act double layer and the Gouy-Chapman layer or diffuse layer. The charge at the interface establish an electric field. Within the compact layer, the electric field reaches the order of 108 to 109 V/m, and hence has an influence on the charge transfer reaction. Since corrosion is an electrochemical p
3、rocess involving the charge transfer reaction, its rate is significantly influenced by the electrode potential or the electric field across the double layer. .,Helmholtz Model,Stern Model,4.1The Electrical analogue of double layer The electrical double layer is characterized by two layers of opposit
4、e charge facing each other, as in a capacitor. The electrical current can, however, pass across the metal-solution interface, although there is some resistance to it. The electrode can then be represented by an electrical analogue composed of a capacitor parallel to resistance RF called Faradaic res
5、istance. The RF is called also the polarization resistance or charge-transfer resistance.,When an electrical current is impressed on the electrode, the RF must be overcome. This generates additional voltage and causes a shift in the electrode potential. At rest(open circuit), The electrode has a cha
6、rged layer; in the absence of an electric current, the capacitor Cdl is charged. The current impressed on the electrode, it, is divided into two parts. it = iF + ich,M,S,When iF : Faradaic current ich : current of charge accumulated in the capacitor Usually iF ich . The electrode potential is propor
7、tional to the charge Q of the double layer. Thus, the electrode potential changes under an electric current across the double layer ; E = Eeq + (i) (i)= the additional voltage due to the current flow.,4.2 Charge transfer overpotential or Activation potential For charge transfer reaction in metal/sol
8、ution interface: M(in lattice) M+z (hydrated in sol.) + ze-,Actually, M, lattice M, adsorbed M+z, sol.,The metal atoms on the electrode surface are in energy wells associated with the lattice structure, and in order to pass into the solution they have to overcome the activation energy.,The chemical
9、free energy change due to the dissolution or deposition of the metal is balanced by an equivalent quantity of electrical work done by the ions in crossing the electric field imposed by the equilibrium electrode potential. M M+z + ze- . at equilibrium or reversible potential. Reaction rate = k x Conc
10、entration = amount produced per unit area per unit time = moles/ cm2sec,The rate of metal dissolution or reduction may be expressed as a current density (A/cm2) according to the Faradys law; i=Q/At = zFm/At , where z = number of electrons, M = number of moles F = Faradys constant (=96500 C/mole) A =
11、 Surface area on which reaction take places. At the equilibrium electrode potential, the flux of charge through the double layer is the same in both directions and we call this the exchange current density, io.,At equil., ia = ic =io ,and anodic reation rate = cathodic reaction rate = io/zF Anodic r
12、x. rate = ka x CM = f CM exp (-G*/RT) where , ka = rate constant = f exp (-G*/RT) CM = concentration of surface active atom f = frequency of activated complex 1012/sec exp (-G*/RT) = Probability of reaction k = f (pool reactant species) (probability of reaction),Lets calculate CM approximately. If t
13、he lattice constant is 3, the number of surface atoms : Ns = (10162/cm2)/(92/atom) 1015 atoms/cm2 Thus, the moles of atoms/cm2=Ns/NA (NA =Avogadro No.) However, surface atoms adjacent to the crystal defects have a higher probability of reaction than others or more active; Fraction of surface atoms w
14、hich are likely to be surface active, = 10-3 to 10-4 depending on crystal structure, defect density and grain size.,Therefore, CM =(Ns/NA) = moles of surface active atoms/cm2,ia = io = zF x (anodic reaction rate) = zFf CM exp (-G*/RT) Cathodic rx. rate at equil. = ic/zF = kc CM+zVL =fCM+zVLexp (-G*/
15、RT) where, CM+z= conc. of metal ion at OHP VL = volume of the double layer/cm2 10-8 cm3 Thus, ic = io = zFfCM+zVLexp (-Gc*/RT),G* can be changed in electrochemical reactions by externally applied potential (Eapp). The change in electrode potential from the equil. value to acquire a net current ( ie,
16、 externally measurable value) is called polarization. ie = f(Eapp-Eeq) = f(),If anodic polarization is applied to the metal electrode, what happens to the energy well curve? The energy of M(metal) increases by zaF and the metal ions become more unstable (high energy state).,For anodic reaction, G*a= G* +(1-)zFa - zFa = G* - zFa For cathodic reaction, G*c = G* + (1-)zFa ia = zFfCM exp (- G*a/RT) = zFfCM exp - (G*- zFa)
