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Electrolysis half-cell reaction

We suggest that these results arise from adsorption effects which are the cathodic complements to the anodic phenomena outlined above. The electrolysis half-cell reaction at the photocathode ... [Pg.322]

The nature of the electrodes matters. An inert electrode, such as platinum, provides a surface on which an electrolysis half-cell reaction occurs, but the reactants themselves must come from the electrolyte solution. An active electrode is one that can itself participate in the oxidation or reduction half-reaction. The distinction between inert and active electrodes is explored in Figure 19-23 and Example 19-12. [Pg.902]

A review of photo-assisted electrolysis studies performed with p-type semiconductor photocathode/dark Pt anode systems suggests that a complementary phenomena arising from the presence of OH ions produced during the reduction half-cell reaction,... [Pg.323]

In Section 17.6 we discussed applications of electrolysis in the extraction and purification of metals from their ore sources. Here we examine the electrolysis of water and aqueous solutions. Consider first the electrolysis of water between inert electrodes such as platinum, for which the half-cell reactions are... [Pg.735]

Thus, the half-cell reactions in the electrolysis of aqueous sodium chloride are... [Pg.787]

Fluorine (F2) is obtained by the electrolysis of liquid hydrogen fluoride (HF) containing potassium fluoride (KF). (a) Write the half-cell reactions and the overall reaction for the process, (b) What is the purpose of KF (c) Calculate the volume of F2 (in liters) eollected at 24.0°C and 1.2 atm after electrolyzing the solution for 15 h at a current of 502 A. [Pg.797]

Industrially, copper is purified by electrolysis. The impure copper acts as the anode, and the cathode is made of pure copper. The electrodes are immersed in a CUSO4 solution. During electrolysis, copper at the anode enters the solution as Cu while Cu ions are reduced at the cathode, (a) Write half-cell reactions and the overall reaction for the electrolytic process, (b) Suppose the anode was contaminated with Zn and Ag. Explain what happens to these impurities during electrolysis, (c) How many hours will it take to obtain 1.00 kg of Cu at a current of 18.9 A ... [Pg.797]

Both reaction 2.1 and reaction 2.2 represent a half-cell reaction. If the strips of copper and zinc are connected by a wire so that electrolysis proceeds, the solution in the zinc compartment shows an increase in zinc-ion concentration, while the solution in the copper compartment is depleted in cupric ions. It is the purpose of the porous barrier separating the two compartments to allow the migration of ions between the two compartments so that there is no buildup of charge inhomogeneity, but without the gross mixing of solutions. [Pg.13]

Bulk electrolysis for the purpose of electrosynthesis or for a coulometric determination of the number of electrons (w) associated with a half-cell reaction of the kind A B + neT (A is the compound being oxidized and B is product, with charges being omitted for simplicity, and n is the overall number of electrons transferred per molecule of A oxidized as determined by coulometry and application of Faraday s Law of electrolysis). [Pg.198]

Setup The half-cell reactions for the electrolysis of water are... [Pg.783]

Fluorine ip2) is obtained by the electrolysis of liquid hydrogen fluoride (HF) containing potassium fluoride (KF). (a) Write the half-cell reactions and the overall reaction for the process. [Pg.792]

Describe the half-cell reactions and the overall reactions that occur in electrolysis. [Pg.543]

The reactants very often are in nonstandard states. In the industrial electrolysis of NaCl(aq), [CP] 5.5 M, not the unit activity ([Cl ] 1 M) implied in half-cell reaction (19.31) therefore Eci2/cr = 1-31 V (not 1.36 V). Also, the pH in the anode half-cell is adjusted to 4, not the unit activity ([H3O ] 1M) implied in half-cell reaction (19.32) hence E02/H2O = 0.99 V (not 1.23 V)- The net effect of these nonstandard conditions is to favor the production of O2 at the anode. In practice, however, the Cl2(g) obtained contains less than 1% 02(g), indicating the overpowering effect of the high overpotential of 02(g). Not surprisingly, the proportion of 02(g) increases significantly in the electrolysis of very dilute NaCl(aq). [Pg.902]

On page 901 we described the electrolysis of NaCl(aq) through the reduction half-cell reaction (19.30) and the oxidation half-cell reaction (19.31). [Pg.906]

To predict the products of an electrolysis involving an aqueous solution, you must examine all possible half-reactions and their reduction potentials. Then, you must find the overall reaction that requires the lowest external voltage. That is, you must find the overall cell reaction with a negative cell potential that is closest to zero. The next Sample Problem shows you how to predict the products of the electrolysis of an aqueous solution. [Pg.529]

However, the cell in Fig. 9.2(b) has a disadvantage in that the concentration of the electrogenerated substance decreases with increasing distance from the OTE surface. Because of this, simulation of the reaction is very difficult, except for the first order (or pseudo-first-order) reactions. For more complicated reactions, it is desirable that the concentration of the electrogenerated species is kept uniform in the solution. With a thin-layer cell, a solution of uniform concentration can be obtained by complete electrolysis, but it takes 30 s. Thus, the thin-layer cell is applicable only for slow reactions. For faster reactions, a column-type cell for rapid electrolysis is convenient. Okazaki et al. [7] constructed a stopped-flow optical absorption cell using one or two column-type cells (Fig. 9.4) and used it to study the dimerization of the radical cations (TPA +) of triphenylamine and the reactions of the radical cation (DPA +) of 9,10-diphenylanthracene with water and alcohols. Using the stopped-flow cell, reactions of substances with a half-life of 1 s can be studied in solutions of uniform concentrations. [Pg.274]

Charles Hall discovered that by mixing the mineral cryolite, Na,AlF6, with alumina he got a mixture that melted at a much more economical temperature, 950°C, instead of the 2050°C of pure alumina. The melt is electrolyzed in a cell that uses graphite (or carbonized petroleum) anodes and a carbonized steel-lined vat that serves as the cathode (Fig. 14.28). The electrolysis half-reactions are... [Pg.822]


See other pages where Electrolysis half-cell reaction is mentioned: [Pg.323]    [Pg.323]    [Pg.87]    [Pg.288]    [Pg.328]    [Pg.860]    [Pg.941]    [Pg.704]    [Pg.1395]    [Pg.924]    [Pg.950]    [Pg.422]    [Pg.87]    [Pg.719]   


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