Charging the Double Layer

Even in the absence of Faradaic current, ie, in the case of an ideally polarizable electrode, changing the potential of the electrode causes a transient current to flow, charging the double layer. The metal may have an excess charge near its surface to balance the charge of the specifically adsorbed ions. These two planes of charge separated by a small distance are analogous to a capacitor. Thus the electrode is analogous to a double-layer capacitance in parallel with a kinetic resistance. 

If the reaction is too fast for this procedure, a double-pulse method can be used The current pulse is preceded by a short but high pulse which is designed to charge the double layer. The height of the pulse is adjusted in such a way that the transient 77(f) is horizontal at the... 

FIG. 22. Atomic and mesoscopic structure of the reconstructed and uiuecon-structed Au(lOO) surface. The sharp peak in the voltammogram demarks the phase transition and is physically associated with charging the double layer due to the difference in the pzc of the respective phases. (From Ref. 216.)... 

Clearly, charging a capacitor requires charge, so how does charging the double-layer around the electrode affect a coulometric measurement ... 

To reiterate, the amount of charge consumed during charging the double-layer is a function of potential, so as soon as coulometry commences, the charge held within the double-layer will change as ions adsorb and/or desorb in response to the change in potential. For convenience, we will discuss these processes in terms of a reduction reaction, with electrons constantly leaving the electrode. 

The proportion of the overall charge needed to charge the double layer is readily computed as charge per area x area , so Qdoubie-iayer charging = 0.196 C. 

Clay is just one example of a material used to modify the electrochemical properties of electrodes to form a chemically modified electrode (CME) (as described belovt/). A porous-clay CME has an area of 5 cm, and charging the double-layer requires a charge of 1.43 C per square centimetre. Repeat the calculations shown above in Worked Example 5.3 to determine the respective faradaic efficiencies. 

Fig. 7.42. A potentiostatic transient. The current (A-B) ascends almost vertically after being switched on, because all of it goes to charge the double layer. In B-C, the current is increasingly used in the form of electrons crossing the double layer. After C the current should decline slowly as diffusion control sets in. In reality, at solid polycrystalline electrodes, in reactions involving adsorbed intermediates, there is often some further variation of /, owing to, e.g., surface crystalline rearrangements and the effect of impurities from the solution.

Fig. 8.2. An early transient. Current density is constant. Potential builds up first through charging of the double layer, but at a higher potential, electrons pass across the interface, i.e., current flows and the double layer behaves as a leaky capacitor. The very early sections of the transient (double-layer condenser not leaking) can be used to obtain the capacity of the double layer because, there, there is a negligible Faradaic current through the interfacial region and the current goes overwhelmingly to charging the double layer. C = (dq/dV) = (idt/dV).

The lower time limit is usually the time needed to charge the double layer to the chosen potential for the measurement. 

Use two pulses first a large one charges the double layer second a smaller one runs the reaction... 

Residual Current A small current that flows in the solution free of electroactive species (see curve 1 in Fig. 5.10). The residual current in DC polarography is mainly the charging current, which is for charging the double layer on the surface of the DME (Fig. 5.13).6)... 

A capacitive current /DL, which is used to charge the double layer present at the solid/liquid interface [12], It corresponds to the charge of a capacitor (electrons in the electrode side, ions from the electrolytes on the electrolyte side) with a capacitance of about 15 pF/cm2. 

Colloidal dispersions can be stabilized by attaching polymer chains to their surface [1-9]. When neutral polymer chains grafted on two parallel plates interpenetrate, a steric repulsion is generated. If the polymer chains grafted to the plates are charged, the double layer interaction between the two plates is also affected by the presence of the chains. 

A current step applied to an electrode provokes a change in its potential. The flux of electrons is used first to charge the double layer, and then for the faradaic reactions. The study of the variation of the potential with... 

A better alternative is to modify the experiment. The use of a double current step10, as demonstrated in Fig. 10.7, can reduce the problem. The first step has sufficient length to charge the double layer the current is then reduced to a lower value at which there is not capacitive component. 

Fig. 10.7. The double current step. The first pulse is to charge the double layer.

We shall treat the simplest case, in which the pulse duration is very short compared to the time constant for charging the double-layer capacitance (r /x 1), and diffusion limitation can be ignored. 

The main purpose of using fast transients is to deal with very fast reactions, for which diffusion limitation is significant, even at short times. We have discussed the time required to charge the double layer, in terms of the relaxation time for charge transfer x, given by... 

The derivation of the Frumkin isotherm is rather involved and is not given here. We note only that it is based on ealculating the difference in electrostatic energy of charging the double-layer capacitor with and without the adsorbed species. The final result is as follows ... 

It would be nice to use very short pulses, in the range of nano-seconds to study this reaction. This is unfortunately not possible because of the sluggishness of the interphase, which is due to the need to charge the double-layer capacitor. If we wish to change the potential by, say, 10 mV, we need to add a charge of... 

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