Cell Potential, Free Energy, and the Equilibrium Constant

5 Cell Potential, Free Energy, and the Equilibrium Constant 

A solution contains both Nal and NaBr. Which oxidizing agent could you add to the solution to selectively oxidize T aq) but not Br (ag)  

In Chapter 15, we learned that acids dissolve metals. Most acids dissolve metals by the reduction of ions to hydrogen gas and the corresponding oxidation of the metal to its ion. 

For example, if solid Zn is dropped into hydrochloric acid, the following reaction occurs  

We observe the reaction as the dissolving of the zinc and the bubbling of hydrogen gas. The zinc is oxidized and the ions are reduced. Notice that this reaction involves the pairing of a reduction half-reaction (the reduction of H ) with the reverse of a halfreaction that falls below it in Table 18.1. Therefore, this reaction is spontaneous. What happens, however, if we pair the reduction of with the oxidation of Cu The reaction is not spontaneous, because it involves pairing the reduction of with the reverse of a half-reaction that is listed above it in the table. Consequently, copper does not react with H+ and does not dissolve in acids such as HCl. In general, metals whose reduction halfreactions are listed below the reduction ofH to H2 in Table 18.1 dissolve in acids, while metals listed above it do not. 

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C19-0039. Write a paragraph explaining the linkages among cell potential, free energy, and the equilibrium constant. 

Before we discuss redox titration curves based on reduction-oxidation potentials, we need to learn how to calculate equilibrium constants for redox reactions from the half-reaction potentials. The reaction equilibrium constant is used in calculating equilibrium concentrations at the equivalence point, in order to calculate the equivalence point potential. Recall from Chapter 12 that since a cell voltage is zero at reaction equilibrium, the difference between the two half-reaction potentials is zero (or the two potentials are equal), and the Nemst equations for the halfreactions can be equated. When the equations are combined, the log term is that of the equilibrium constant expression for the reaction (see Equation 12.20), and a numerical value can be calculated for the equilibrium constant. This is a consequence of the relationship between the free energy and the equilibrium constant of a reaction. Recall from Equation 6.10 that AG° = —RT In K. Since AG° = —nFE° for the reaction, then... 

In Chapter 20, we discussed the relationship of useful work, free energy, and the equilibrium constant. In this section, we examine this relationship in the context of electrochemical cells and see the effect of concentration on cell potential. 

We leam to determine the standard reduction potentials based on the standard hydrogen electrode reference and use them to calculate the emf of a cell and hence the spontaneity of a cell reaction. A relationship exists between a cell s emf, the change in the standard Gibbs free energy, and the equilibrium constant for the ceU reaction. (19.3 and 19.4)... 

Figure 21.10 summarizes the interconnections among the standard free energy change, the equilibrium constant, and the standard cell potential. The procedures... 

Eigure 21.10 summarizes the interconnections among the standard free energy change, the equilibrium constant, and the standard cell potential. In Chapter 20, we determined K from AG°, which we found either from A//° and A5° values or from AGf values. Now, for redox reactions, we have a direct experimental method for determining K and AG° measure E eii-... 

Relate the standard cell potential (E n) to the standard Gibbs free energy change (AG ) and the equilibrium constant (K)... 

The above important relationship now allows evaluation of the thermodynamic driving force of a redox reaction in terms of a measurable cell emf. Moreover, it is possible to utilize the relationship between the standard state potential and the standard state free energy to arrive at an expression for the equilibrium constant of a redox reaction in terms of the emf. Thus... 

Provided the reaction is, in some sense, reversible, so that equilibrium can be attained, and provided the reactants and products arc all gas-phase, solution or solid-state species with well-defined free energies, it is possible to define the free energies for all such reactions under any defined reaction conditions with respect to a standard process this is conventionally chosen to be the hydrogen evolution/oxidation process shown in (1.11). The relationship between the relative free energy of a process and the emf of a hypothetical cell with the reaction (1.11) as the cathode process is given by the expression AC = — nFE, or, for the free energy and potential under standard conditions, AG° = — nFEl where n is the number of electrons involved in the process, F is Faraday s constant and E is the emf. 

The quantity AG° is the free energy of the half-reaction when the activities of the reactant and product have values of unity and is directly proportional to the standard half-cell potential for the reaction as written. It also is a measure of the equilibrium constant for the half-reaction assuming the activity of electrons is unity ... 

If the reactants and products are in their standard states, the resulting cell potential is called the standard cell potential. This latter quantity is related to the standard free-energy change for the reaction and thus to the equilibrium constant by... 

Here we have divided both sides of the equation by nF to isolate the cell potential in the equation. This equation also resembles the Nernst equation (Equation 13.4), and it is easy to see how it arises. At equilibrium, the free energy change is zero and the reaction quotient, Q, is equal to the equilibrium constant, K. 

The free energies in (18) are illustrated in Fig. 10. It can be seen that GA is that part of AG ° available for driving the actual reaction. The importance of this relation is that it allows AGXX Y to be calculated from the properties of the X and Y systems. In thermodynamics, from a list of n standard electrode potentials for half cells, one can calculate j (m — 1) different equilibrium constants. Equation (18) allows one to do the same for the %n(n— 1) rate constants for the cross reactions, providing that the thermodynamics and the free energies of activation for the symmetrical reactions are known. Using the... 

Electrochemical equilibrium is established at each interface of the cell when the -> electrochemical potentials of the common components of the two phases (a and f) forming the interface are equal, that is pf = ji , and the electrochemical free energy change (AG) for the process occurring at the interface is zero. For the net cell reaction given above, such considerations lead to an expression for the electrochemical equilibrium constant Ka given by [i]... 

The potential difference between two of these half-cells (or of their half-reactions) is AE° (or simply, E°), and it represents the driving force that exists for the electron transfer to happen. This difference is related to the free energy change of the system and to its equilibrium constant by the well-known expression... 

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