Rate-determining electron transfer, multistep

The electron waiting-line problem is hence clear. In a particular multistep electron-transfer reaction, the step with the lowest servicing rate or conductivity produces the largest queue and, indeed, the total queue is virtually a simple multiple of the queue at the rds. In other words, in the steady state, all n steps proceed at the rate of the rate-determining step ir, [cf. Eq. (9.4)], and the total net current is... 

In comparing the general and the simple equations, it is seen that the transfer coefficients play the same role in a multistep, n-electron-transfer reaction as the symmetry factor does in one-step, one-electron transfer reaction, i.e., thea s determine how the input electrical energy (Ft)) affects the reaction rate. Table 15 shows the tabulation of values for y, r, v, y, and n, from which a and a have been evaluated. 

The rate expression for the multistep consecutive electron-transfer reaction of Scheme 1 [i.e., Eq. (31)] is able to relate complex consecutive electron-transfer reaction mechanisms to experimental potential vs. logarithmic current-density relations. When p is assumed to be 1/2, the Tafel slopes (1/a/) predicted by this relation can only have values less than or equal to 118 mV dec i (at 25 °C) for electron-transfer limited reactions, since electrons transferred in non-rds steps will add integers (to P) in the expected a values and therefore decrease the Tafel slope below 118 mV dec 1. For instance, the usual cathodic Tafel slope of 118 mV dec-i for a one- electron transfer over a synunetric harrier is decreased to 39 mV dec for one preceding quasi-equilibrium electron transfer and to 24 mV dec for two, etc., and the anodic Tafel slopes are similarly decreased for one and two following (where the reaction steps are still written as reductions, as in Scheme 1) electron transfers, respectively. It should be noted that the Tafel slopes that are determined hy a values involving y-i- P differ substantially and discontinuously from the value for a = P = 1/2, and therefore should be easily distinguishable. 

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