Chapter 8 Electrochemical Cells

1. Chapter 8 Electrochemical Cells ~An electrochemical cell, also a voltaic cell or a galvanic cell, is a device in which a chemical reaction occurs with the production of an electric potential difference between two electrodes. ~If the electrodes are connected to an external circuit there results a flow of current, which can lead to the performance of mechanical work so that the electrochemical cell transforms chemical energy into work. ~Besides, being of considerable practical importance, electrochemical cells are valuable laboratory instruments, because they provide some extremely useful scientific data. ~For example, they lead to thermodynamic quantities such as enthalpies and Gibbs energies, and they allow us to determine transport numbers and activity coefficients for ions in solution.

2. 8.1 The Daniell Cell ~The voltage produced by the cell depends on the activities of the Zn2+ and Cu2+ ions in the two solutions. ~If the molalities of the two solutions are 1 mol kg-1 (1 m), the cell is called a standard Daniell cell. ~When a Daniell cell is set up, there is a flow of electrons from the zinc to the copper electrode in the outer circuit. This means that positive current is moving from left to right in the cell itself. ~By convention, a potential difference corresponding to an external flow of electrons from the left-hand electrode to the right-hand electrode is said to be a positive potential difference. ~Within the cell, zinc metal dissolves to form Zn2+ ions, Zn Zn2++2e- copper atom deposits to copper, Cu2++2e-Cu two electrons travel round the outer circuit. Daniell cell consists of a zinc electrode immersed in dilute zinc sulfate solution and a copper electrode immersed in a copper sulfate solution. Figure 8.1

3. Figure 8.2 ~When the electric potential of a cell is exactly balanced, the cell is operating reversibly and its potential is then referred to as the electromotive force (emf) of the cell. ~If the counter-potential in the slide wire is slightly less than the emf of the cell, there is a small flow of electrons from left to right in the external circuit. ~If the counter-potential is adjusted to be slightly greater than the cell emf, the cell is forced to operate in reverse; zinc is deposited at the left-hand electrode, Zn2++2e-  Zn and copper dissolves at the right, CuCu2++2e-

4. Figure 8.1 ~The factor that the electrons normally flow from the zinc to the copper electrode indicates that the tendency for Zn Zn2++2e- to occur is greater than for the reaction CuCu2++2e-, which is forced to occur in the reverse direction. ~The magnitude of the emf developed is a measure of the relative tendencies of the two processes. ~The emf varies with the activities of the Zn2+ and Cu2+ ions in the two solutions. ~The tendency for Zn Zn2++2e- to occur is smaller when the concentration of Zn2+ is large, while the tendency for Cu2++2e-Cu to occur increases when the concentration of Cu2+ is increased.

5. 8.2 Standard Electrode Potentials If would be very convenient if we could measure the potential of a single electrode, such as the right-hand electrode in Figure 8.1, which we will write as The potential of such an electrode would be a measure of the tendency of the process to occur. However, there is no way to measure the emf of a single electrode, since in order to obtain an emf there must be two electrodes, with an emf associated with each one. The procedure used is to choose one electrode as a standard and to measure emf values of other electrodes with reference to that standard. Figure 8.3 ~The standard hydrogen electrode consists of a platinum electrode immersed in a 1 m solution of hydrogen ions maintained at 25 C and 1 bar pressure. ~Hydrogen gas is bubbled over the electrode and passes into solution, forming hydrogen ions and electrons: ~The emf corresponding to this electrode is arbitrarily assigned the value of zero, and this electrode is used as a standard for other electrodes.

6. ~If the hydrogen electrode is placed on the left-hand side, the emf of the cells is the standard electrode potentials or standard reduction potentials (given the symbol Eo) of the other electrode. ~The standard hydrogen electrode may be placed on the right-hand side; the potential so obtained is known as the standard oxidation potential. ~The voltaic cell shown in Figure 8.4 can be represented as follows: where the double vertical dashed lines represent the salt bridge. The observed emf is 0.34V; the sign is positive by convention since electrons flow from left to right in the outer circuit. ~There is a greater tendency for the process to occur than for to occur; the latter process is forced to go in the reverse direction. Figure 8.4 not allowing bulk mixing of the two solutions

7. The individual reactions can be written as The overall process is This process is accompanied by the passage of two electrons around the outer circuit.

8. Other Standard Electrodes Standard silver-silver chloride electrode, in which a silver electrode is in contact with solid silver chloride, which is highly insoluble salt. The whole is immersed in potassium chloride solution in which the chloride-ion concentration is 1 m. This electrode can be represented as We can set up a cell involving this electrode and the hydrogen electrode, with a salt bridge connecting the two solutions. The emf at 25 C is found to be 0.22233V. The individual reactions are and the overall process is

9. calomel electrode In calomel electrode, mercury is in contact with mercurous chloride (calomel, Hg2Cl2) immersed either in a 0.1 m solution of potassium chloride or in a saturated solution of potassium chloride. If the cell Figure 8.5 is set up, the individual reactions are and the overall process is The emf at 25 C is 0.3337V, which is the standard electrode potential Eo. If a saturated solution of KCl is used with the calomel electrode, the standard electrode potential is 0.2412V.

10. glass electrode ~The glass electrode consists of a tube terminating in a thin-walled glass bulb; the glass is reasonably permeable to hydrogen ions. ~The glass bulb contains a 0.1 m hydrochloric acid solution and a tiny silver-silver chloride electrode. ~The theory of the glass electrode is somewhat complicated, but when the bulb is inserted into an acid solution, it behaves like a hydrogen electrode. ~The glass electrode is particularly convenient for making pH determination. Figure 8.5

11. Ion-Selective Electrodes Figure 8.6 ~The sample solution is separated from an internal solution by an ion- selective membrane, and an internal reference electrode is placed within the internal solution. ~An external reference electrode, such a silver-silver chloride electrode, is also immersed in the sample solution, and a measurement is made of the reversible emf of the assembly.

12. Problem 8.5 Design electrochemical cells in which each of the following reactions occurs a. Ce4+(aq)+Fe2+(aq)  Ce3+(aq)+Fe3+(aq) b. Ag+(aq)+Cl-(aq)  AgCl(s) c. HgO(s)+H2(g) Hg(l)+H2O(l) In each case, write the representation of the cell and the reactions at the two electrodes. Solution (a) (b) (c)

13. 8.3 Thermodynamic of Electrochemical Cells For a standard cell, Every time 1 mol of H2 reacts with 1 mol of Cu2+, 2 mol of electron pass through the outer circuit. According to Faraday’s laws, this means the transfer of 296485 C of electricity. The emf developed is +0.3419 V, and the passage of 296485 C across this potential drop means that 2964850.3419 C V=6.598104 J of work has been done by the system. Thus, for this cell process, Go=-6.598104 J non-PV work is equal to the decrease in Gibbs energy In general, for any standard-cell reaction associated with the passage of z electrons and an emf of Eo, the change in Gibbs energy is relating to a cell in which the molalities are unity standard Gibbs energy

14. The same argument applies to any cell; if the emf is E, the Gibbs energy change is the change in Gibbs energy when the reaction occurs with the concentrations having the values employed in the cell. If E is positive, G is negative; a positive E means that the cell is operating spontaneously with the reactions occurring in the forward direction (e.g., H2+Cu2+2H++Cu) For any reaction at 25 C These equations provide a very important method for calculating Gibbs energy changes and equilibrium constants.

15. Example 8.1 Calculate the equilibrium constant at 25 C for the reaction occurring in the Daniell cell, if the standard emf is 1.10 V. Solution The reaction is Example 8.2 Using the data in Table 8.1, calculate the equilibrium constant for the reaction Solution From Table 8.1, the standard electrode potential are

16. Note: emf is an intensive property If the problem had been to calculate the equilibrium constant Ko’ for the process

17. Example 8.3 Calculate Eo for the process making use of the following Eo values: Solution z=1

18. Consider the following Eo values involving both a one-electron and a two-electron process: z=2

19. Problem 8.3 Calculate the standard electrode potential for the reaction Cr2++2e-Cr at 298 K. The necessary Eo values are Solution z=2

20. Nernst Equation Consider cells in which the concentrations are other than unity. For example, the cell A hydrogen electrode has been combined with a copper electrode immersed in a Cu2+ solution, the concentration of which is other than unity. The overall cell reaction is The Gibbs energy is

21. In general, we may consider any cell for which the overall reaction has the general form Nernst Equation Suppose that we apply the Nernst equation to the cell for which the overall reaction is Eo for the overall process is -0.257-(-0.762)=0.505 V at 25 C Note: Increasing the ratio [Zn2+]/[Ni2+] decreases the cell emf; this is understandable in view of the fact that a positive cell emf means that the cell is producing Zn2+ and that Ni+2 ions are being removed.

22. Example 8.4 Calculate the emf of the cell if the concentrations are a.[Ni2+]=1.0 m and [Co2+]=0.1 m b.[Ni2+]=0..010 m and [Co2+]=1.0 m Solution The cell reaction is The standard emf is -0.257-(-0.280)=0.023 V and z=2. (a) (b)

23. Problem 8.14 b. Calculate the emf at 25 C of the cell Solution (b) The electrode reactions are The cell reaction is From the Nernst equation

24. Problem 8.20 Calculate the emf of the cell Solution The electrode reactions are Overall reaction: Cell emf:

25. Problem 8.21 Suppose that the cell in Problem 8.20 is setup but that the two solutions are separated by a membrane that is permeable to H+ ion but impermeable to Cl- ions. What will be the emf of the cell at 25 C? Solution The electrode reactions are Every H+ ion produced in the left-hand solution will have to pass through the membrane to preserve electrical neutrality. The net reaction is Cell emf is

26. Problem 8.28 Suppose that the cell is set up and that the membrane separating the two solutions is permeable only to H+ ions. What is the emf of the cell at 25 C? Solution The electrode reactions are The electrical neutrality is maintained by the passage of H+ ions from right to left: The net reaction is Cell emf is

27. Nernst Potentials Figure 8.7 Suppose, for example, that solutions of potassium chloride were separated by membrane, as shown in Figure 8.7 ~If both ions could cross the membrane, the concentrations would eventually become equal and there would be no potential difference (Figure 8.7a). ~If only the potassium ions can cross, and the membrane is not permeable to solvent, there will very little change in the concentration on the two sides of the membrane. Some K+ ions will cross from the more concentrated side (the left-hand side in Figure 8.7b), and as a result the left-hand side will have a negative potential with respect to the right-hand side. This effect will be to prevent more K+ ions crossing. An equilibrium will therefore to be established at which the electric potential will exactly balance the tendency of the concentration to become equal. This potential is referred to the Nernst potential.

28. The Gibbs energy Ge arising from the potential difference  is z is the charge on the permeable ions In this particular example (Figure 8.7b), the potential is higher on the right-hand side (>0), because a few K+ ions have crossed, and there will thus be higher Gibbs energy, as far as K+ ions are concerned, on the right-hand side (Ge >0). The Gibbs energy Gc arising from the concentration difference is At equilibrium there is no net G across the membrane and therefore The Nernst potential is due to the transfer of only an exceedingly small fraction of the diffusible ions, so that there is no detectable concentration change.

29. Example 8.5 Mammalian muscle cells are freely permeable to K+ ions but much less permeable to Na+ and Cl- ions. Typical concentration of K+ ions are Inside the cell: [K+]=155 mM Outside the cell: [K+]=4 mM Calculate the Nernst potential at 310 K (37 C) on the assumption that the membrane is impermeable to Na+ and Cl-. Solution The potential is negative inside the cell and positive outside. In reality, such potential are much like -85 mV, because there is a certain amount of diffusion of Na+ and Cl-.

30. Suppose, for example, that two solutions 1 dm3 in volume are separated by a membrane 1 dm2 in area (A) and 1 cm in thickness (l), and suppose that 0.1 M and 1.0 M solutions of KCl are present on the two sides (Figure 8.7c), the membrane again being permeable only to the K+ ions. The capacitance C of the membrane is Figure 8.7 , the dielectric constant of the membrane, will be taken to have a value of 3 (this is typical of biological and other organic membrane). The capacitance in this example is therefore F is the farad.

31. The Nernst potential at 25 C is With a capacitance of 2.6610-11F, the net charge on each side of the wall, required to maintain this potential, is The number of K+ ions required to produce this charge is However, 1 dm3 of the 0.1 M solution contains 0.16.0221023=6.022 1022 ions. The fraction involved in establishing the potential difference is therefore exceedingly small (<10-14) and no concentration change could be detected. The fraction is larger if the system is smaller, but even for a biological system it is no more than 10-6.

32. Problem 8.22 A typical biological cell has a volume of 10-9 cm3, a surface area of 10-6 cm2, and a membrane thickness of 10-6 cm; the dielectric constant of the membrane may be taken as 3. Suppose that the concentration of K+ ions inside the cell is 0.115 M and that the Nernst potential across the cell wall is 0.085V. a. Calculate the net charge on either side of the wall, and b. Calculate the fraction of the K+ ions in the cell that are required to produce this charge. Solution The capacitance of the cell membrane is (a) The net charge on either side of the wall is (b) The number of K+ ions required to produce this charge is The fraction of K+ ions required to produce this charge is

33. If more than one ion can pass through the membrane, the situation is more complex. Considering Example 8.5, the Cl- ion are also somewhat permeable, and typical concentrations are Inside the cell: [Cl-]=4.3 mM Outside the cell: [Cl-]=104 mM The Nernst potential corresponding to this distribution is, since z=-1, close to the observed potential Typical concentrations for Na+ ions are Inside the cell: [Na+]=12 mM Outside the cell: [Na+]=145 mM If these ions were permeable, their Nernst potential would be This is of the opposite sign to the true potential, and this distribution of Na+ ions only arises because of their inability to cross the membrane.

34. Example 8.6 A 0.10 M solution of sodium palmitate is separated from an equal volume of a 0.20 M solution of sodium chloride by a membrane that is permeable to Na+ and Cl- but not to palmitate ions. Calculate the final concentrations and the Nernst potential at 25 C, assuming ideal behavior. Solution Suppose that x M of Na+ and Cl- ions move to the palmitate side; the final concentrations are Palmitate side: [Na+]=(0.1+x) M; [Cl-]=x M Other side: [Na+]=(0.2-x) M; [Cl-]=(0.2-x) M At equilibrium, (0.2-x)2=x(0.1+x) x=0.08 M The final concentration are therefore Palmitate side: [Na+]=0.18 M; [Cl-]=0.08 M Other side: [Na+]=[Cl-]=0.12 M The Nernst potential arising from the distribution of Na+ ions is, at 25 C, The palmitate side, having the higher concentration of Na+, is the negative side, since Na+ ions will trend to cross from that side.

35. Temperature Coefficients of Cell emfs The basic relationship is For an overall reaction The enthalpy change is The measure of emf values at various temperatures provides a very convenient method of obtaining thermodynamic values for chemical reactions and has frequently been employed.

36. Example 8.7 The emf of the cell is 0.2002 V at 25 C; E/T is -8.66510-5 V K-1. Write the cell reaction and calculate G, S, and H at 25 C. Solution The electrode reactions are The cell reaction is The Gibbs energy change is The entropy change is The enthalpy change is

37. Problem 8.33 The Weston standard cell is a. Write the cell reaction b. At 25 C, the emf is 1.01832 V and Eo/T=-5.0010-5 VK-1. Calculate Go, Ho, and So. Solution (a) The electrode reactions are The cell reaction is (b)

38. Problem 8.34 Salstrom and Hidebrand reported the following data for the cell Find the temperature coefficient for this cell assuming a linear dependence of the cell potential with temperature. What is the entropy change for the cell reaction? Solution

39. 8.4 Types of Electrochemical Cells Chemical cells There is a net chemical change (a chemical reaction takes place between electrodes and solute ions). Concentration cells ~The driving force is a dilution process. ~The changes in concentration can occur either in the electrolyte or at the electrodes. ~Example of concentration changes at electrodes are found with electrodes made of amalgams or consisting of alloys and with gas electrodes (e.g., the Pt, H2 electrode) where there are different gas pressures at the two electrodes. Figure 8.8 Cell without transference ~There is not a boundary between two solutions. Cell with transference ~There is a boundary between two solutions.

40. Concentration Cells A simple example of a concentration cell is obtained by connecting two hydrogen electrode by means of a salt bridge: The reactionat the left-hand electrodeis The reactionat the right-hand electrodeis The net process is therefore simply the transfer of hydrogen ions from solution of molality m2 to one of molality m1. If m2 is greater than m1, the process will actually occur in this direction, and a positive emf is produced; if m2 is less than m1, the emf is negative and electrons flow from the right-hand to the left-hand electrode. The emf produced is E>0 when m2>m1

41. Example 8.8 Calculate the emf at 25 C of a concentration cell of this type in which the molalities are 0.2 m and 3.0 m. Solution

42. Redox Cells Both the oxidized and reduced species are in solution, their interconversion is effected by an inert electrode such as one of platinum. Consider, for example, the cell ~The left-hand electrode is the standard hydrogen electrode. ~The right-hand electrode consists simply of a platinum electrode immersed in a solution containing both Fe2+ and Fe+3 ions. ~The platinum electrode is able to catalyze the interconversion of these ions. The electrode reactions are The overall reaction is The emf of the cell is Eo relates to PH2=1bar and [H+]=[Fe2+]=[Fe3+]=1 m

43. The interconversion of oxidized and reduced forms frequently involves also the participation of hydrogen ions. Thus, the half-reaction for the reduction of fumarate ions to succinate ions is succinate fumarate If we wised to study this system, we could set up the following cell: F2- and S2- represent fumarate and succinate, respectively. overall cell reaction The emf of the cell is The Eo for this system is related to a standard Gibbs energy change Go by the usual equation

44. This standard Gibbs energy is related to the equilibrium constant: at some specified hydrogen-ion concentration Often this standard concentration is taken to be 10-7 M, corresponding to a pH of 7. In that case In many cases the K’ value corresponds to a fairly well-balanced equilibrium at pH 7, whereas K will be larger by the factor 1014; the K’ value and the corresponding Go’ at pH 7 therefore give a clearer indication of the situation at that pH. Eo’ is a modified standard potential

45. Example 8.9 The enzyme glycollate oxidase is a catalyst for the reduction of cytochrome c in its oxidized form-denoted as cytochrome c (Fe+3)-by glycollate ions. The relevant standard electrode potential Eo’, relating to 25 C and pH 7, are as follows: a. Calculate Go’ for the reduction, at pH 7 and 25 C. b. Then calculate the equilibrium ratio at pH 7 and 25 C c. What is the equilibrium ratio at pH 7.5 and 25 C?

46. Solution a. The balanced reaction is and this equation corresponds to z=2. b. The equilibrium ratio at pH 7 is c. The equilibrium ratio at pH 7.5 is conveniently calculated in terms of the true (pH- independent) equilibrium constant Kture. If K” is the equilibrium ratio at pH 7.5,

47. 8.5 Applications of emf Measurements pH Determinations Figure 8.4 ~Since the emf of a cell such as that shown in Figure 8.4 depends on the hydrogen-ion concentration, pH values can be determined by dipping hydrogen electrode into solutions and measuring the emf with reference to another electrode. ~In commercial pH meters the electrode immersed in the unknown solution is often a glass electrode, and the other electrode may be the silver-silver chloride or the calomel electrode. ~In accordance with the Nernst equation the emf varies logarithmically with the hydrogen-ion concentration and therefore varies linearly with the pH. ~Commercial instruments are calibrated so as to give a direct reading of the pH.

48. Problem 8.39 The emf of a cell was found to be 0.517 V at 25 C. Calculate the pH of the HCl solution. Solution

49. Activity Coefficients ~Up to now we have expressed the Gibbs energy changes and emf values of cells in terms of molalities. This is an approximation and the errors become serious as concentrations are increased. ~For a correct formulation, activities must be employed, and the emf measurements over a range of concentrations lead to values for the activity coefficients. Consider the cell The overall process is and the Gibbs energy change is a+ and a- are the activities of the H+ and Cl- ions, and Go is the standard Gibbs energy, when the activities are unity. The emf is, since z=1,

50. This equation can be written as Figure 8.9 ~If E is measured over a range of molalities of HCl, the quantity on the left-hand side can be calculated at various molalities. ~If this quantity is plotted against m, as shown schematically in Figure 8.9, the value extrapolated to zero m gives Eo, since at zero m the activity coefficient  is unity so that the final term vanishes. ~At any molality the ordinate minus Eo then yields from which the activity coefficient can be calculated.