Driving potential difference solution

In Eq 4.11, (E"x - E"M) is positive since E"x at the cathodic site is greater than E"u at the anodic site. Thus, the potential in the solution at the anodic site, ())s a, is greater than the potential in the solution at the cathodic site, < )s c, which is consistent with the overall electrochemical corrosion circuit. It follows that the driving potential difference for conventional current flow (Icorr) in the solution is ... 

The fourth fully developed membrane process is electrodialysis, in which charged membranes are used to separate ions from aqueous solutions under the driving force of an electrical potential difference. The process utilizes an electrodialysis stack, built on the plate-and-frame principle, containing several hundred individual cells formed by a pair of anion- and cation-exchange membranes. The principal current appHcation of electrodialysis is the desalting of brackish groundwater. However, industrial use of the process in the food industry, for example to deionize cheese whey, is growing, as is its use in poUution-control appHcations. 

A reverse osmosis membrane acts as the semipermeable barrier to flow ia the RO process, aHowiag selective passage of a particular species, usually water, while partially or completely retaining other species, ie, solutes such as salts. Chemical potential gradients across the membrane provide the driving forces for solute and solvent transport across the membrane. The solute chemical potential gradient, —is usually expressed ia terms of concentration the water (solvent) chemical potential gradient, —Afi, is usually expressed ia terms of pressure difference across the membrane. 

For the solute flux, it is assumed that chemical potential difference owing to pressure is negligible. Thus the driving force is almost entirely a result of concentration differences. The solute flux, J), is defined as in equation 6 ... 

Galvanic Corrosion. Galvanic corrosion occurs when two dissimilar metals are in contact in a solution. The contact must be good enough to conduct electricity, and both metals must be exposed to the solution. The driving force for galvanic corrosion is the electric potential difference that develops between two metals. This difference increases as the distance between the metals in the galvanic series increases. 

This is the difference in potential that one must use from an external source in order to balance the driving force of the forward cell reaction. If one uses a potential difference just in excess of this value, one can bring a reversal of the cell reaction and commencement of the electrolysis of the HC1 solution, 2 HC1 —> H2 + Cl2. 

Case I Pure Liquids and Inert Electrolytes. In the absence of significant impurity currents, no faradaic current will flow if the applied bias between the tip and substrate, AEt, is less than the total potential difference, AEp rev, required to drive faradaic reactions at the STM tip and at the substrate. This condition can be easily calculated from the electrochemical potential data for the solvent/electrolyte system under study. This situation is most likely to exist in pure liquids or in solutions of nonelectroactive electrolytes where the faradaic reactions at both electrodes are... 

Initially, both electrodes are at equilibrium. Since the anode has accumulated electrons and the cathode has depleted electrons, electrons begin to flow from electrode from the anode to the cathode. The thermodynamic driving force for the electron flow is the electrode potential difference, which for the fuel cell reaction is 1.23 Y at standard conditions. In addition to electron flow, H + ions produced at the anode diffuse through the bulk solution and react at the cathode. The reaction is able to continue as long as H2 is fed at the anode and 02 at the cathode. Hence, the cell is not at equilibrium. The shift in electrode potential from equilibrium is called the overpotential (>/). 

In the expressions of the driving force above, E is, strictly speaking, the potential difference between the electrode and the reaction site. It is usually not exactly the same as the potential difference between the electrode and the solution as illustrated by the potential profile across the double layer represented in Figure 1.6. In other words, E = M — (j)rs rather than E = (f>M, thus resulting in a double-layer effect on the electron transfer kinetics10 that ought to be taken into account. The reaction site is... 

Equilibrium is reached when the driving force for the diffusion (the concentration gradient) is compensated for by the electric field (the potential gradient). Under these equilibrium conditions, there is an equilibrium net charge on each side of the junction and an equilibrium potential difference d< >e. This process is analogous to the way charge transfer across a nonpolarizable electrode/solution interface results in the establishment of an equilibrium potential difference across the interface. 

The photoinduced difference between the quasi-Fermi level for electrons in the Ti02 and the quasi-Fermi level for holes in solution, EFn = EFn, - EFp.solution, sets an upper limit to the photovoltage, Voc, because it is this potential difference and the fact that electrons and holes are confined to separate chemical phases that drives electrons toward the substrate electrode and holes toward the counterelectrode. Although VEFn is mainly comprised of V x in DSSCs, there is, nevertheless, a possible role for q at interfaces where the field cannot be entirely screened by mobile electrolyte. 

LIF is used to filter any large molecules (e.g., proteins) present in a solution by using an appropriate membrane. Although the driving potential in UF is the hydraulic pressure difference, the mass transfer rates will often affect the rate of UF due to a phenomenon known as concentration polarization (this is discussed later in the chapter). 

RO is a membrane-based process that is used to remove solutes of relatively low molecular weight that are in solution. As an example, almost pure water can be obtained from sea water by using RO, which will filter out molecules of NaCl and other salts. The driving potential for water permeation is the difference in hydraulic pressure. A pressure that is higher than the osmotic pressure ofthe solution (which itself could be quite high if the molecular weights of the solutes are small) must be applied to the solution side of the membrane. RO also involves the concentration polarization of solute molecules. 

The driving potential for UF - that is, the filtration of large molecules - is the hydraulic pressure difference. Because of the large molecular weights, and hence the low molar concentrations of solutes, the effect of osmotic pressure is usually minimal in UF this subject is discussed in Section 8.5. 

In addition to simply solving the differential equation, we seek to use the solution to understand and quantify the heat transfer between the fluid and the duct walls. The heat flux q" (W/m2) can be described in terms of a heat-transfer coefficient h (W/m2 K), with the thermal driving potential being the difference between the wall temperature and the mean fluid temperature ... 

The driving force for diffusion of C+ from the membrane to the aqueous solution is the favorable solvation of the ion by water. As C+ diffuses from the membrane into the water, there is a buildup of positive charge in the water immediately adjacent to the membrane. The charge separation creates an electric potential difference ( ou,cr) across the membrane. The free-energy difference for C+ in the two phases is AG = —nFE(Mcr, where F is the Faraday constant and n is the charge of the ion. At equilibrium, the net change in free energy for diffusion of C+ across the membrane boundary must be 0 ... 

This would be accomplished by immersing the chemically modified electrode, a reference electrode, and an auxiliary electrode into an appropriate electrolyte solution (e.g., 0.1 M NaC104 in acetonitrile). The potential difference between the modified electrode (the working electrode) and the reference would then be adjusted to a value appropriate to drive this reaction, using a commercially available potentiostat, and the resulting anodic current would be measured. 

Consider two half-cell reactions, one for an anodic and the other for a cathodic reaction. The exchange current densities for the anodic and the cathodic reactions are lO-6 A/cm2 and 1(T2 A/cm2, respectively, with transfer coefficients of 0.4 and 1, respectively. The equilibrium potential difference between the two reactions is 1.5 V. (a) Calculate the cell potential when the current density of 1CT5 A/cm2 flows through the self-driving cell, neglecting the concentration overpotentials. The solution resistance is 1000 Q cm2, (b) What is the cell potential when the current density is 10-4 A/cm2 (Kim)... 

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