Standard voltages

Rotating electrical machines lEC standard voltages, rating and performance A122im2l 325/1996 BSEN 60034-/1995 -... 

The standard voltage for a given cell is that measured when the current flow is essentially zero, all ions and molecules in solution are at a concentration of 1M, and all gases are at a pressure of 1 atm. To illustrate, consider the Zn-H+ cell. Let us suppose that the half-cells are set up in such a way that the concentrations of Zn2+ and H+ are both 1 M and the pressure of H2(g) is 1 atm. Under these conditions, the cell voltage at very low current flow is +0.762 V. This quantity is referred to as the standard voltage and is given die symbol fi°. 

Any redox reaction can be split into two half-reactions, an oxidation and a reduction. It is possible to associate standard voltages x (standard oxidation voltage) and (standard reduction voltage) with the oxidation and reduction half-reactions. The standard voltage for the overall reaction, °, is the sum of these two quantities... 

To illustrate, consider the reaction between Zn and H+ ions, for which the standard voltage is +0.762 V. 

There is no way to measure the standard voltage for a half-reaction only fi° can be measured directly. To obtain values for x and fi ed> the value zero is arbitrarily assigned to the standard voltage for reduction of H+ ions to H2 gas ... 

Using this convention, it follows that the standard voltage for the oxidation of zinc must be +0.762 V that is,... 

As soon as one half-reaction voltage is established, others can be calculated. For example, the standard voltage for the Zn-Cu2+ cell shown in Figure 18.2 is found to be +1.101 V. Knowing that oX for zinc is +0.762 V, it follows that... 

Standard half-cell voltages are ordinarily obtained from a list of standard potentials such as those in Table 18.1 (page 487). The potentials listed are the standard voltages for reduction half-reactions, that is,... 

To obtain the standard voltage for an oxidation half-reaction, all you have to do is change the sign of the standard potential listed in Table 18.1. For example, knowing that... 

Another way to express this principle is to say that standard voltages for forward and reverse half-reactions are equal in magnitude but opposite in sign. 

As pointed out earlier, the standard voltage for a redox reaction is the sum of the standard voltages of the two half-reactions, reduction and oxidation that is,... 

This simple relation makes it possible, using Table 18.1, to calculate the standard voltages for more than 3000 different redox reactions. 

Redox reactions, like all reactions, eventually reach a state of equilibrium. It is possible to calculate the equilibrium constant for a redox reaction from the standard voltage. To do that, we start with the relation obtained in Chapter 17 ... 

Notice that if the standard voltage is positive, In K is also positive, and K is greater than 1. Conversely, if the standard voltage is negative, In K is also negative, and K is less than 1. 

Only if the standard voltage falls within a rather narrow range, say+0.10 to—0.10 V, will the value of K (and that of AG°) be such that the reaction will produce an equilibrium mixture containing appreciable amounts of both reactants and products. 

To this point, we have dealt only with standard voltages, that is, voltages when all gases are at 1 atm pressure and all species in aqueous solution are at a concentration of 1M. When the concentration of a reactant or product changes, the voltage changes as well. Qualitatively, the... 

In this equation, E is the cell voltage, E° is the standard voltage, n is the number of moles of electrons exchanged in the reaction, and Q is the reaction quotient. Notice that—... 

Nemst equation An equation relating cell voltage E to the standard voltage ° and the concentrations of reactants and products E = E° — (0.0257/ )(lnQ), 493,508-509q application, 494-495 ion concentrations, 494-495 Nernst, Walther, 493... 

The positive value of the standard voltage obtained in the example indicates that the cell reaction shown is spontaneous. Thus, the standard potentials in Table 6.11 can be used to predict whether a particular reaction will occur, or not. The advantage of Table 6.11 is that it provides quantitative as well as qualitative information. It not only conveys that nickel is a stronger oxidizing agent than silver (because nickel is positioned below silver in the electrochemical series), but it also conveys how much stronger, in terms of the cell emf of+1.05 V. 

Although the reactions themselves can be added to obtain the desired equation, the E° for this equation is not the sum of the E° values for the above three reactions. The E° for the desired equation is actually the weighted average of the E° values for reactions (1) to (3). It can be obtained by summing up the free energy changes for the three reactions. (For reactions of the same type, standard voltages are not additive AG° values are additive, however). When (1) and (2) (each multiplied by two) are added to (3) we obtain ... 

Recall that AG° = -nFE° and that AG° values, not standard voltages are additive for reactions in which the number of electrons do not cancel out. 

We begin by calculating the standard voltages for the two steps in the decomposition... 

The standard voltage for this half-reaction is given in Appendix D ... 

To finish up this problem, we just need to calculate the standard voltage ( ) for the half-reaction (ii) ... 

S° = standard entropy H° = standard enthalpy G° = standard free energy E° = standard voltage T = absolute temperature... 

Standard voltages for oxidation halfreactions are obtained by changing the sign of the standard reduction potentials. When the reaction is reversed, the sign of the ° is changed to the opposite sign. 

In this mode of operation, the standard voltage ramp is replaced by a series of brief pulses (50 to 100 ms) of increasing potential. The pulses are delivered at the... 

Calculate the standard voltage (E°) needed to decompose water into its elements by electrolysis. What does the word standard mean in this question ... 

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