Exchange current density reduction

Table 21.16 Exchange current densities for several noble metals and a platinum-rhodium alloy in the reduction of oxygen from perchloric acid solution ...

It is clear from the calculated limiting-current curves in Fig. 3a that the plateau of the copper deposition reaction at a moderate limiting-current level like 50 mA cm 2 is narrowed drastically by the surface overpotential. On the other hand, the surface overpotential is small for reduction of ferri-cyanide ion at a nickel or platinum electrode (Fig. 3b). At noble-metal electrodes in well-supported solutions, the exchange current density appears to be well above 0.5 A/cm2 (Tla, S20b, D6b, A3e). At various types of carbon, the exchange current density is appreciably smaller (Tla, S17a, S17b). 

Fig. 2. Current-potential curves in Evans diagram [29] format for reduction of Cu2+ ions and oxidation of H2CO. and are the equilibrium, or open circuit, potentials for the Cu2+ reduction and H2CO oxidation reactions, respectively. Assuming negligible interfering reactions, the vertical dashed lines indicate the exchange current densities for the two half reactions, and the deposition current for the complete electroless solution. Adapted from ref. 23.

The improvement in cell performance at higher pressure and high current density can be attributed to a lower diffusion polarization at the cathode and an increase in the reversible cell potential. In addition, pressurization decreases activation polarization at the cathode because of the increased oxygen and water partial pressures. If the partial pressure of water is allowed to increase, a lower acid concentration will result. This will increase ionic conductivity and bring about a higher exchange current density. The net outcome is a reduction in ohmic losses. It was reported (33) that an increase in cell pressure (100% H3PO4, 169°C (336°F)) from 1 to 4.4 atm (14.7 to 64.7 psia) produces a reduction in acid concentration to 97%, and a decrease of about 0.001 ohm in the resistance of a small six cell stack (350 cm electrode area). 

The most important electrokinetic data pertinent to fuel cell models are the specific interfacial area in the catalyst layer, a, the exchange current density of the oxygen reduction reaction (ORR), io, and Tafel slope of ORR. The specific interfacial area is proportional to the catalyst loading and inversely proportional to the catalyst layer thickness. It is also a strong function of the catalyst layer fabrication methods and procedures. The exchange current density and Tafel slope of ORR have been well documented in refs 28—31. 

Figure 5. Measurement and analysis of steady-state i— V characteristics, (a) Following subtraction of ohmic losses (determined from impedance or current-interrupt measurements), the electrode overpotential rj is plotted vs ln(i). For systems governed by classic electrochemical kinetics, the slope at high overpotential yields anodic and cathodic transfer coefficients (Ua and aj while the intercept yields the exchange current density (i o). These parameters can be used in an empirical rate expression for the kinetics (Butler—Volmer equation) or related to more specific parameters associated with individual reaction steps.(b) Example of Mn(IV) reduction to Mn(III) at a Pt electrode in 7.5 M H2SO4 solution at 25 Below limiting current the system obeys Tafel kinetics with Ua 1/4. Data are from ref 363. (Reprinted with permission from ref 362. Copyright 2001 John Wiley Sons.)...

Extrapolation of the linear portion of the curves to the standard oxygen reduction potential gives the corresponding exchange current density, a measure of the equilibrium rate of the reaction [35]. The results are given in Table I. Within... 

Exchange Current Density. Let us now return to our electrochemical cell shown in Figure 3.8. This cell is a combination of two half-cells, with the oxidation reaction occurring at the anode and the reduction reaction occurring at the cathode resulting in a net flow of electrons from the anode to the cathode. Equilibrium conditions dictate that the rate of oxidation and reduction, roxid and rred, be equal, where both rates can be obtained from Faraday s Law ... 

As in Eq. (3.22), F is the Faraday constant, n is the number of electrons taking part in the reaction, but iq is a new quantity called the exchange current density. These rates have units of mol/cm s, so the exchange current density has units of A/cm. Typical values of io for some common oxidation and reduction reactions of various metals are shown in Table 3.4. Like reversible potentials, exchange current densities are influenced by temperature, surface roughness, and such factors as the ratio of oxidized and reduced species present in the system. Therefore, they must be determined experimentally. 

The slowest step, or rate-determining step, can be either (a) electron transfer at the electrode-solution interface or (b) formation of atoms at the electrode surface. The activation polarization component of the overpotential, r)a, is related to the actual rate of oxidation or reduction, i, and the exchange current density ... 

Given that the rates of oxidation and reduction of the half-reactions are controlled by activation polarization only, that = 4-0.07 and = —0.08, and that the exchange current densities for both the oxidation of Fe and reduction of hydrogen in acidic solution are identical, use the data in Tables 3.3 and 3.4 to determine the following quantities. Recall that the potential for each half-cell is the sum of the equilibrium potential and the corresponding overpotential, in this case, r]a-... 

To do this, it may be helpful to plot the derivative dE/di as a function of 1/i, deducing from it the coefficients a and b. One can then calculate the value of E0 from the initial curves, (b) Establish theoretically the law E =JQn i). Find from the experimental data the resistance of the electrolyte / p, the product OCj n2, and (he exchange current density for the reduction of oxygen, (c) Calculate the maximum current and maximum power of the cell, (d) What electrode surface would be necessary to obtain twice the normal power of the cell You are given ... 

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