10 Equilibrium electrochemistry

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1 10 Equilibrium electrochemistry E10.1b E10.2b E10.3b E10.4b Solutions to exercises Discussion questions The Debye Hückel theory is a theory of the activity coefficients of ions in solution. It is the coulombic electrostatic interaction of the ions in solution with each other and also the interaction of the ions with the solvent that is responsible for the deviation of their activity coefficients from the ideal value of 1. The electrostatic ion ion interaction is the stronger of the two and is fundamentally responsible for the deviation. Because of this interaction there is a build up of charge of opposite sign around any given ion in the overall electrically neutral solution. The energy, and hence, the chemical potential of any given ion is lowered as a result of the existence of this ionic atmosphere. The lowering of the chemical potential below its ideal value is identified with a non-zero value of RT ln γ ±. This non-zero value implies that γ ± will have a value different from unity which is its ideal value. The role of the solvent is more indirect. The solvent determines the dielectric constant, ɛ, of the solution. Looking at the details of the theory as outlined in Justification 10.2 we see that ɛ enters into a number of the basic equations, in particular, Coulomb s law, Poisson s equation, and the equation for the Debye length. The larger the dielectric constant, the smaller in magnitude is ln γ ±. The potential difference between the electrodes in a working electrochemical cell is called the cell potential. The cell potential is not a constant and changes with time as the cell reaction proceeds. Thus the cell potential is a potential difference measured under non-equilibrium conditions as electric current is drawn from the cell. Electromotive force is the zero-current cell potential and corresponds to the potential difference of the cell when the cell not the cell reaction is at equilibrium. The ph of an aqueous solution can in principle be measured with any electrode having an emf that is sensitive to H + aq concentration activity. In principle, the hydrogen gas electrode is the simplest and most fundamental. A cell is constructed with the hydrogen electrode being the right-hand electrode and any reference electrode with known potential as the left-hand electrode. A common choice is the saturated calomel electrode. The ph can then be obtained from eqn by measuring the emf zero-current potential difference, E, of the cell. The hydrogen gas electrode is not convenient to use, so in practice glass electrodes are used because of ease of handling. Numerical exercises NaClaq + AgNO 3 aq AgCls + NaNO 3 aq NaCl, AgNO 3 and NaNO 3 are strong electrolytes; therefore the net ionic equation is Ag + aq + Cl aq AgCls r H = f H AgCl, s f H Ag +, aq f H Cl, aq = kj mol kj mol kj mol 1 = kj mol 1 E10.5b PbSs Pb 2+ aq + S 2 aq K S = a ν J J J Since the solubility is expected to be low, we may initially ignore activity coefficients. Hence K S = bpb2+ b bs2 b bpb 2+ = bs 2 = S

2 148 INSTRUCTOR S MANUAL K S = S2 b 2 S = K S 1/2 b Use ln K S = rg RT to obtain K S r G = f G S 2, aq + f G Pb 2+, aq r G PbS, s = kj mol kj mol kj mol 1 = kj mol J mol 1 ln K S = J K 1 mol K = K S = e = K S = S2 b 2 S = K S 1/2 b = /2 = mol kg 1 E10.6b The ratio of hydration Gibbs energies is hyd G NO 3 hyd G Cl = rcl 181 pm rno = pm = We have hyd G Cl = 379 kj mol 1 [Exercise 10.6a] So hyd G NO 3 = kj mol 1 = 363 kj mol 1 E10.7b I = 2 1 b i /b zi 2 [10.18] i and for an M p X q salt, b + /b = pb/b, b /b = qb/b,so a b c I = 2 1 pz2 + + qz2 b/b IMgCl 2 = b/b = 3b/b IAl 2 SO 4 3 = b/b = 15b/b Ie 2 SO 4 3 = b/b = 15b/b E10.8b I = IK 3 [ecn 6 ] + IKCl + INaBr = bk 3[eCN 6 ] b = = bkcl b + bnabr b Question. Can you establish that the statement in the comment following the solution to Exercise 10.8a in the Student s Solutions Manual holds for the solution of this exercise? E10.9b I = IKNO 3 = b b KNO 3 = Therefore, the ionic strengths of the added salts must be a IKNO 3 = b b, so bkno 3 = mol kg 1 and mol kg kg = mol KNO 3 So mol g mol 1 = 45.0 g KNO 3 must be added.

3 EQUILIBRIUM ELECTROCHEMISTRY 149 b IBaNO 3 2 = b b b = b = mol kg 1 3 = 3 b = b and mol kg kg = mol BaNO 3 2 So mol g mol 1 = 38.8gBaNO 3 2 E10.10b IAl 2 SO 4 3 = b/b = 15b/b ICaNO 3 2 = b/b = 3b/b mol kg 1 = 15bAl 2 SO 4 3 bal 2 SO 4 3 = mol kg 1 = mol kg 1 E10.11b γ ± = γ p + γ q 1/s s = p + q or Al 2 SO 4 3 p = 2, q = 3, s = 5 γ ± = γ 2 + γ 3 1/5 E10.12b Since the solutions are dilute, use the Debye Hückel limiting law log γ ± = z + z AI 1/2 I = 2 1 zi 2 b i/b = 2 1 { } i = log γ ± = /2 = or NaClγ ± = E10.13b = ICaCl 2 = b/b = 3b/b log γ ± = /2 = γ ± = = = Error = 100 per cent = 47.1 per cent E10.14b The extended Debye Hückel law is log γ ± = A z +z I 1/2 1 + BI 1/2 Solving for B 1 B = I 1/2 + A z +z = log γ ± b/b + 1/2 log γ ±

4 150 INSTRUCTOR S MANUAL Draw up the following table b/mol kg γ ± B B = 1.3 E10.15b PbI 2 s PbI 2 aq K S = r G = RT ln K S = J K 1 mol K ln = kj mol 1 r G = f G PbI 2, aq f G PbI 2, s f G PbI 2, aq = r G + f G PbI 2, s = kj mol kj mol 1 = kj mol 1 E10.16b The Nernst equation may be applied to half-cell potentials as well as to overall cell potentials. EH + /H 2 = RT ln ah + f H2 /p 1/2 E = E 2 E 1 = RT ln a 2H + a 1 H + [f H 2 is constant] = RT ln γ ±b 2 γ ± b 1 [ ] = 25.7mV ln = +56.3mV E10.17b Identify electrodes using species with the desired oxidation states. L: Cds + 2OH aq CdOH 2 s + 2e R: NiOH 3 s + e NiOH 2 s + OH aq Cds CdOH 2 s OH aq NiOH 2 s NiOH 3 s Pt E10.18b The cell notation specifies the right and left electrodes. Note that for proper cancellation we must equalize the number of electrons in half-reactions being combined. a R: Ag 2 CrO 4 s + 2e 2Ags + CrO 2 4 aq V L: Cl 2 g + 2e 2Cl aq V Overall R L: Ag 2 CrO 4 s + 2Cl aq 2Ags + CrO 2 4 aq + Cl 2g 0.91 V b R: Sn 4+ aq + 2e Sn 2+ aq V L: 2e 3+ aq + 2e 2e 2+ aq V Overall R L: Sn 4+ aq + 2e 2+ aq Sn 2+ aq + 2e 3+ aq 0.62 V c R: MnO 2 s + 4H + aq + 2e Mn 2+ aq + 2H 2 Ol V L: Cu 2+ aq + 2e Cus V Overall R L: Cus + MnO 2 s + 4H + aq Cu 2+ aq + Mn 2+ aq + 2H 2 Ol V

5 EQUILIBRIUM ELECTROCHEMISTRY 151 Comment. Those cells for which E > 0 may operate as spontaneous galvanic cells under standard conditions. Those for which E < 0 may operate as nonspontaneous electrolytic cells. Recall that E informs us of the spontaneity of a cell under standard conditions only. or other conditions we require E. E10.19b The conditions concentrations, etc. under which these reactions occur are not given. or the purposes of this exercise we assume standard conditions. The specification of the right and left electrodes is determined by the direction of the reaction as written. As always, in combining half-reactions to form an overall cell reaction we must write half-reactions with equal number of electrons to ensure proper cancellation. We first identify the half-reactions, and then set up the corresponding cell. a R: 2H 2 Ol + 2e 2OH aq + H 2 g 0.83 V L: 2Na + aq + 2e 2Nas 2.71 V and the cell is Nas Na + aq, OH aq H 2 g Pt V or more simply Nas NaOHaq H 2 g Pt b R: I 2 s + 2e 2I aq V L: 2H + aq + 2e H 2 g 0 and the cell is Pt H 2 g H + aq, I aq I 2 s Pt V or more simply Pt H 2 g HIaq I 2 s Pt c R: 2H + aq + 2e H 2 g 0.00 V L: 2H 2 Ol + 2e H 2 g + 2OH aq V and the cell is Pt H 2 g H + aq, OH aq H 2 g Pt V or more simply Pt H 2 g H 2 Ol H 2 g Pt Comment. All of these cells have E > 0, corresponding to a spontaneous cell reaction under standard conditions. If E had turned out to be negative, the spontaneous reaction would have been the reverse of the one given, with the right and left electrodes of the cell also reversed. E10.20b See the solutions for Exercise 10.18b, where we have used E = ER E L, with standard electrode potentials from Table E10.21b See the solutions for Exercise 10.19b, where we have used E = ER E L, with standard electrode potentials from Table E10.22b In each case find E = ER E L from the data in Table 10.7, then use r G = νe [10.32] a R: S 2 O 2 8 aq + 2e 2SO 2 } 4 aq V L: I 2 s + 2e 2I V aq V r G = kc mol V = 291 kj mol 1

6 152 INSTRUCTOR S MANUAL b Zn 2+ aq + 2e } Zns 0.76 V Pb 2+ aq + 2e E = 0.63 V Pbs 0.13 V r G = kc mol V = +122 kj mol 1 E10.23b a A new half-cell may be obtained by the process 3 = 1 2, that is 3 2H 2 Ol + Ags + e H 2 g + 2OH aq + Ag + aq But, E3 E1 E 2, for the reason that the reduction potentials are intensive, as opposed to extensive, quantities. Only extensive quantities are additive. However, the r G values of the half-reactions are extensive properties, and thus r G 3 = r G 1 rg 2 ν 3 E 3 = ν 1 E 1 ν 2 E 2 Solving for E 3 we obtain E3 = ν 1E1 ν 2 E V V = = V ν 3 1 b The complete cell reactions is obtained in the usual manner. We take to obtain 2Ag + aq + H 2 g + 2OH aq 2Ags + 2H 2 Ol E cell = E R E L = E 2 E 1 = V V = V Comment. The general relation for E of a new half-cell obtained from two others is E 3 = ν 1E 1 ± ν 2 E 2 ν 3 E10.24b a E = E RT ν ln Q ν = 2 b c Q = J a ν J J = a 2 H + a 2 Cl [all other activities = 1] [ = a+ 2 a2 = γ +b + 2 γ b 2 b b b ] here and below = γ + γ 2 b + b 2 = γ 4 ± b4 [16, b + = b, b = b] Hence, E = E RT 2 lnγ ± 4 b4 = E 2RT lnγ ±b r G = νe[10.32] = C mol V = kj mol 1 log γ ± = z + z AI 1/2 [19] = /2 [I = b for HClaq] = γ ± = E = E + 2RT lnγ ±b = V V ln = V The value compares favourably to that given in Table 10.7.

7 EQUILIBRIUM ELECTROCHEMISTRY 153 E10.25b R: e 2+ aq + 2e es L: 2Ag + aq + 2e 2Ags R L: 2Ags + e 2+ aq 2Ag + aq + es E = ER E L = 0.44 V 0.80 V = 1.24 V r G = νe = C mol V = +239 kj mol 1 r H = 2 f H Ag +, aq f H e 2+, aq = [ ] kj mol 1 = kj mol 1 r G = r S = rg r H T p T = kj mol K [ r G = r H T r S] = kj mol 1 K 1 Therefore, r G 308 K K K 1 kj mol kj mol 1 νe E10.26b In each case ln K = [10.36] RT a Sns + CuSO 4 aq Cus + SnSO 4 aq R: Cu 2+ aq + 2e } Cus V L: Sn 2+ aq + 2e V Sns 0.14 V ln K = V mv =+37.4, K = b Cu 2+ aq + Cus 2Cu + aq R: Cu 2+ aq + e Cu + } aq V L: Cu + aq + e 0.36 V Cus V ln K = E10.27b We need to obtain E for the couple 0.36 V = 14.0, K = mv 3 Co 3+ aq + 3e Cos from the values of E for the couples 1 Co 3+ aq + e Co 2+ aq E 1 = 1.81 V 2 Co 2+ aq + 2e Cos E 2 = 0.28 V We see that 3 = 1 + 2; therefore see the solution to Exercise 10.23b E 3 = ν 1E1 + ν 2 E V V = = 0.42 V ν 3 3

8 154 INSTRUCTOR S MANUAL Then, R: Co 3+ aq + 3e Cos ER = 0.42 V L: 3AgCls + 3e 3Ags + 3Cl aq EL = 0.22 V R L: Co 3+ aq + 3Cl aq + 3Ags 3AgCls + Cos E = ER E L = 0.42 V 0.22 V = V E10.28b irst assume all activity coefficients are 1 and calculate KS, the ideal solubility product constant. 1 AgIs Ag + aq + I aq SAgI = bag + = bi because all stoichiometric coefficients are 1. Thus KS = bag+ bi b 2 = S2 b 2 = = Bi 2 S 3 s 2Bi 3+ aq + 3S 2 aq bbi 3+ = 2SBi 2 S 3 bs 2 = 3SBi 2 S 3 K S = bbi3+ 2 bs 2 3 b 5 = 2S2 3S 3 b 5 = 108 = or AgI, K S = γ 2 ± K S log γ ± = z + z AI 1/2 A = I = Sb, z + z =1 so log γ ± = /2 = γ ± = K S = KS = K S S 5 b or Bi 2 S 3, I = 15b/b = 15Sb, z + z =6 so log γ ± = [ ] 1/2 = γ ± = 1.0 K S = γ± 5 K S = K S Neglect of activity coefficients is not significant for AgI and Bi 2 S 3. E10.29b The Nernst equation applies to half-reactions as well as whole reactions; thus for 8H + + MnO 4 aq + 5e Mn 2+ aq + 4H 2 O E = E RT 5 ln amn 2+ amno 4 ah+ 8 E10.30b R: 2AgIs + 2e 2Ags + 2I aq 0.15 V L: 2H + aq + 2e H 2 g 0V OverallR L: 2AgIs + H 2 g 2Ags + 2H + aq + 2I aq Q = ah + 2 ai 2 ν = 2 Assume ah + = ai, Q = ah + 4

9 EQUILIBRIUM ELECTROCHEMISTRY 155 E = E RT 2 ln ah+ 4 = E 2RT ph = RT E10.31b The electrode reactions are L: Ag + aq + e Ags R: AgIs + e Ags + I aq ln ah+ = E RT E E = E V V = 1.15 V V = 9.72 ph OverallR L: AgIs Ag + aq + I aq Since the cell reaction is a solubility equilibrium, for a saturated solution there is no further tendency to dissolve and so E = 0 E10.32b R: 2Bi 3+ aq + 6e 2Bis L: Bi 2 S 3 s + 6e 2Bis + 3S 2 aq OverallR L: 2Bi 3+ aq + 3S 2 aq Bi 2 S 3 s ν = 6 νe ln K = RT V = V = 224 K = e 224 It is convenient to give the solution for b first. b K S = K 1 = e , since the cell reaction is the reverse of the solubility equilibrium. [ ] a K S b 2 [ ] b 3 = b Bi3+ b S2 = 2S 2 3S 3 = 108S /5 S = mol L Solutions to problems Solutions to numerical problems P10.1 We require two half-cell reactions, which upon subtracting one left from the other right, yields the given overall reaction Section The half-reaction at the right electrode corresponds to reduction, that at the left electrode to oxidation, though all half-reactions are listed in Table 10.7 as reduction reactions. E R: Hg 2 SO 4 s + 2e 2Hgl + SO 2 4 aq V L: PbSO 4 s + 2e Pbs + SO 2 4 aq 0.36 V R L: Pbs + Hg 2 SO 4 s PbSO 4 s + 2Hgl V

10 156 INSTRUCTOR S MANUAL Hence, a suitable cell would be Pbs PbSO 4 s H 2 SO 4 aq Hg 2 SO 4 s Hgl or, alternatively, Pbs PbSO 4 s H 2 SO 4 aq H 2 SO 4 aq Hg 2 SO 4 s Hgl or the cell in which the only sources of electrolyte are the slightly soluble salts, PbSO 4 and Hg 2 SO 4, the cell would be Pbs PbSO 4 s PbSO 4 aq Hg 2 SO 4 aq Hg 2 SO 4 s Hgl The potential of this cell is given by the Nernst equation [10.34]. E = E RT ln Q [10.34]; ν = 2 ν a Pb 2+a SO 2 4 Q = a Hg 2+ a 2 SO 2 4 = K SPbSO 4 K S Hg 2 SO 4 E = 0.98 V RT 2 ln K SPbSO 4 K S Hg 2 SO V = 0.98 V 2 ln [Table 10.6, 4th Edition, or CRC Handbook] = 0.98 V V = V P10.6 Pt H 2 g NaOHaq, NaClaq AgCls Ags H 2 s + 2AgCls 2Ags + 2Cl aq + 2H + aq ν = 2 E = E RT 2 ln Q, Q = ah+ 2 acl 2 [f/p = 1] = E RT ln ah+ acl = E RT ln K wacl aoh = E RT ln K wγ ± bcl γ ± boh = E RT ln K wbcl boh = E RT = E RT Hence, pk w = E E ln 2.303RT/ + ln K w RT ln bcl boh pk w RT ln bcl boh pk w = log K w = ln K w bcl boh = E E RT/ E = E R E L = E AgCl, Ag E H + /H 2 =+0.22 V 0 [Table 10.7] We then draw up the following table with the more precise value for E = V [Problem 10.8] θ/ C E/V RT V pk w

11 EQUILIBRIUM ELECTROCHEMISTRY 157 dlnk w dt = rh RT 2 [9.26] Hence, r H = 2.303RT 2 d dt pk w then with dpk w dt pk w T r H J K 1 mol K 2 = kj mol 1 r G = RT ln K w = 2.303RT pk w = kj mol 1 r S = rh r G T = 17.1JK 1 mol K See the original reference for a careful analysis of the precise data. P10.7 The cells described in the problem are back-to-back pairs of cells each of the type Ags AgXs MXb 1 M x Hgs R: M + b 1 + e Hg M x Hgs Reduction of M + and formation of amalgam L: AgXs + e Ags + X b 1 R L: Ags + M + b 1 + X b 1 Hg M x Hgs + AgXs ν = 1 Q = am xhg am + ax E = E RT ln Q or a pair of such cells back to back, Ags AgXs MXb 1 M x Hgs MXb 2 AgXs Ags E R = E RT ln Q R E L = E RT ln Q L E = RT ln Q L = RT Q R ln am+ ax L am + ax R Note that the unknown quantity am x Hg drops out of the expression for E. am + ax γ+ b + γ b = b b = γ± 2 b 2 b b + = b With L = 1 and R = 2 we have E = 2RT ln b 1 + 2RT b 2 ln γ ±1 γ ± 2 Take b 2 = mol kg 1 the reference value, and write b = b 1 b ln E = 2RT b ln γ ± γ ± ref

12 158 INSTRUCTOR S MANUAL or b = , the extended Debye Hückel law gives log γ ± ref = γ ± ref = then E = V / = /2 b ln ln γ ± E ln γ ± = V ln b We then draw up the following table b/mol/kg E/V γ A more precise procedure is described in the original references for the temperature dependence of E Ag, AgCl, Cl, see Problem P10.10 The method of the solution is first to determine r G, r H, and r S for the cell reaction 1 2 H 2g + AgCls Ags + HClaq and then, from the values of these quantities and the known values of f G, f H, and S for all the species other than Cl aq, to calculate f G, f H, and S for Cl aq. r G = νe At K25.00 C E /V = = V Therefore, G = kc mol V = kj mol 1 r S r G E E C = = ν = ν T p T p θ p K [dθ/ C = dt/k] a E θ p = / C θ/ C 2 V θ 2 / C 3 E θ p V/ C = θ/ C θ/ C 2 Therefore, at C, E = V/ C θ p and E T = V/ C C/K = VK 1 p

13 EQUILIBRIUM ELECTROCHEMISTRY 159 Hence, from equation a r S = kc mol VK 1 = J K 1 mol 1 and r H = r G + T r S or the cell reaction = kj mol K J K 1 mol 1 = kj mol H 2g + AgCls Ags + HClaq r G = f G H + + f G Cl f G AgCl = f G Cl f G AgCl [ f G H + = 0] Hence, f G Cl = r G + f G AgCl = [ ] kj mol 1 = kj mol 1 Similarly, f H Cl = r H + f H AgCl = kj mol 1 = kj mol 1 or the entropy of Cl in solution we use r S = S Ag + S H + + S Cl 2 1 S H 2 S AgCl with S H + = 0. Then, S Cl = r S S Ag S H 2 + S AgCl = = +56.7JK 1 mol 1 G P10.12 a rom = V [5.10] p T r G we obtain = r V p T Substituting r G = νe [10.32] yields E = rv p T,n ν b The plot ig of E against p appears to fit a straight line very closely. A linear regression analysis yields Slope = mv atm 1, standard deviation = mv atm 1 Intercept = mv, standard deviation = mv R = an extremely good fit rom r V E p T,n = m 3 mol C mol 1 Since J = VC= Pa m 3,C= Pa m3 V or m3 C = V Pa

14 160 INSTRUCTOR S MANUAL igure 10.1 Therefore E V = p T,n Pa Pa atm This compares closely to the result from the potential measurements. c A fit to a second-order polynomial of the form E = a + bp + cp 2 = V atm 1 = mv atm 1 yields a = mv, standard deviation = mv b = mv atm 1, standard deviation = mv atm 1 c = mv atm 2, standard deviation = mv atm 1 R = This regression coefficient is only marginally better than that for the linear fit, but the uncertainty in the quadratic term is > 200 per cent. E = b + 2cp p T E The slope changes from = b = mv atm 1 p min E to = b + 2c1500 atm = mv atm 1 p max E We conclude that the linear fit and constancy of are very good. p

15 EQUILIBRIUM ELECTROCHEMISTRY 161 d We can obtain an order of magnitude value for the isothermal compressibility from the value of c. 2 E p 2 = 1 r V = 2c ν p T κ T cell = 1 r V = 2νc V p T V V atm C mol cm 3 atm J κ T cell = 1cm g g 1 mol = atm 1 standard deviation 200 per cent where we have assumed the density of the cell to be approximately that of water at 30 C. Comment. It is evident from these calculations that the effect of pressure on the potentials of cells involving only liquids and solids is not important; for this reaction the change is only V atm 1. The effective isothermal compressibility of the cell is of the order of magnitude typical of solids rather than liquids; other than that, little significance can be attached to the calculated numerical value. P10.15 The equilibrium is K = ah 2O 4 av 4 O 4 12 ah 2 VO 4 4 γv 4O 4 12 bv 4 O 4 12 γh 2 VO 4 4 bh 2 VO 4 4 Let x be bh 2 VO 4 ; then bv 4 O 4 12 = x/4. Then the equilibrium equation can be expressed as x 4 KγH 2 VO 4 4 γv 4 O 4 = x/4 12 which can be solved numerically once the constants are determined. The activity coefficients are log γh 2 VO 4 = = so γh 2 VO 4 = and log γv 4 O 4 12 = = so γv 4 O 4 12 = The equation is x = x/4 Its solution is x = mol kg 1 = bh 2 VO 4 and bv 4 O 12 4 = /4 = mol kg 1 P10.18 The reduction reaction is Sb 2 O 3 s + 3H 2 Ol + 6e 2Sbs + 6OH aq Q = aoh 6 ν = 6 Therefore a E = E RT 6 ln aoh 6 = E RT [ln aoh = log aoh = 2.303pOH] ln aoh = E RT poh

16 162 INSTRUCTOR S MANUAL b Since poh + ph = pk w E = E RT pk w ph c The change in potential is E = 2.303RT poh f poh i = mv poh f poh i poh f = log0.050γ ± = log log γ ± = log A = poh i = log0.010γ ± = log log γ ± = log A = Hence, E = mv = 37.6mV P10.19 We need to obtain r H for the reaction 1 2 H 2g + Uup + aq Uups + H + aq We draw up the thermodynamic cycle shown in ig Data are obtained from Table 13.4, 14.3, 2.6, and 2.6b. The conversion factor between ev and kj mol 1 is 1eV= kj mol 1 The distance from A to B in the cycle is given by r H = x = 3.22 ev eV eV 11.3eV 5.52 ev 1.5eV = 0.75 ev r S = S Uup, s + S H +, aq 2 1 S H 2, g S Uup +, aq = mev K 1 = 1.33 mev K 1 i igure 10.2 r G = r H T r S = 0.75 ev K 1.33 mev K 1 =+1.15eV which corresponds to +111 kj mol 1 The electrode potential is therefore rg, with ν = 1, or 1.15V ν

17 EQUILIBRIUM ELECTROCHEMISTRY 163 Solutions to theoretical problems [ P10.21 MXs M + aq + X aq, K s bm + bx b b ] b bm + = S, bx = S + C K s = SS + C, or S 2 + CS K s = 0 which solves to S = 2 1 C2 + 4K s 1/2 2 1 C or S = 2 1 C 1 + 4K 1/2 s C C If 4K s C 2, S 2 1 C 1 + 2K s C 2 [1 12 C + x 1/ x + ] K s C P10.22 K s = am + ax = bm + bx γ± 2 ; bm+ = S, bx = S + C log γ ± = AI 1/2 = AC 1/2 ln γ ± = 2.303AC 1/2 γ ± = e 2.303AC1/2 γ 2 ± = e 4.606AC1/2 K s = S S + C e 4.606AC1/2 We solve S 2 + S C K s γ± 2 = 0 1/2 to get S = 1 C 2 + 4K s 2 γ± C K s Cγ± 2 [as in Problem 10.21] Therefore, since γ 2 ± = e 4.606AC1/2 P10.25 The half-reactions involved are: S K se 4.606AC1/2 C R: cyt ox + e cyt red Ecyt L: D ox + e D red ED The overall cell reaction is: R L = cyt ox + D red cyt red + D ox a The Nernst equation for the cell reaction is E = E RT ln [cyt red ][D ox] [cyt ox ][D red ] at equilibrium, E = 0; therefore ln [cyt red ] eq[d ox ] eq = Ecyt [cyt ox ] eq [D red ] eq RT E D [Dox ] eq [cyt]ox ln = ln + [D red ] eq [cyt] red RT [Dox ] eq Therefore a plot of ln [D red ] eq intercept of RT E cyt E D E = E cyt E D Ecyt E D [cyt]ox against ln is linear with a slope of one and an [cyt] red

18 164 INSTRUCTOR S MANUAL b Draw up the following table: [Dox ] eq ln [D red ] eq [cytox ] eq ln [cyt red ] eq [Dox ] eq [cytox ] eq The plot of ln against ln is shown in ig The intercept is [D red ] eq [cyt red ] eq Hence Ecyt = RT V = V V = V 0 1 y = x R = ln[d ox ] eq /[D red ] eq ln[cyt ox ] eq /[cyt red ] eq igure 10.3 Solutions to application P10.27 a molality H2 SO 4 = bd = ad d 25 + cd d 25 2 where d is density in g cm 3 at 25 C, a = mol kg 1 gcm 3 1, c = mol kg 1 gcm 3 2, and d 25 = g cm 3. or 1 kg solvent m H2 O = 1kg: m H2 SO mass % H2 SO 4 = 4 b = m H2 SO 4 + m H2 O b + m H2 O m H2 SO 4 m H2 O 100 bd mass % H2 SO 4 d = bd + m 1 H2 SO 4 an equation for the solution molarity is deduced with a unit analysis. molarity H2 SO 4 d = bd where m H2 SO 4 = kg mol 1 1 mass % H 2 SO 4 d 100 d 10 3 cm 3 L kg 10 3 g

19 EQUILIBRIUM ELECTROCHEMISTRY Sulfuric Acid Solutions Molality / mol/kg Density/g /ml igure 10.4a 50 Sulfuric Acid Solutions Mass Percentage Sulfuric Acid Density/g / ml igure 10.4b Molarity/mol / L Sulfuric Acid Solutions Density/g / ml igure 10.4c b cell: Pbs PbSO 4 s H 2 SO 4 aq PbO 2 s PbSO 4 s Pbs cathode: PbO 2 s + 3H + aq + HSO 4 aq + 2e PbSO 4 s + 2H 2 Ol E cathode = V anode: PbSO 4 s + H + aq + 2e Pbs + HSO 4 E anode = V net: PbO 2 s + Pbs + 2H + aq + 2HSO 4 aq 2PbSO 4s + 2H 2 Ol E = E cathode E anode = V eqn r G = νe = C mol V = C V mol 1 = J mol 1 = kj mol 1

20 166 INSTRUCTOR S MANUAL f H values of Table 2.6 and the CRC Handbook of Chemistry and Physics are used in the r H calculation. r H r H r S = 2 f H PbSO f H H 2 Ol f H PbO 2 f H Pb 2 f H H + 2 f H HSO 4 = kj mol kj mol kj mol kj mol 1 = kj mol 1 = rh r G T = 121 J K 1 mol 1 eqn 4.39 = kj mol kj mol K E 15 C = E 25 C + E = E 25 C + rs T eqn ν 121 J K 1 mol 1 = V C mol 1 10 K = V V = V The temperature difference makes a negligibly small difference in the cell potential. When Q = , E = E RT ν = V = V ln Q eqn J K 1 mol K C mol 1 ln c The general form of the reduction half-reaction is: ox + νe + ν H H + + aa red + xx using eqn 10.34, E = E RT ν ln Q = E RT ared ν ln ax x a ox a ν H H +aa a = E RT ν ln 1 a ν H H + all species other than acids are at unit activity in a Pourboix diagram = E + ν HRT ν ln a H + = E + ν HRT ln 10 log a ν H + = E ν H RT ln 10 ph eqn 9.29 ν E = E νh V ν ph

21 EQUILIBRIUM ELECTROCHEMISTRY 167 or the PbO 2 PbSO 4 couple, PbO 2 s + 4H + + SO 2 4 aq + 2e PbSO 4 s + 2H 2 Ol E = V,ν H = 4,ν = 2 E = V VpH or ph = 5, E = V or ph = 8, E = V or the PbSO 4 /Pb couple, PbSO 4 s + 2e Pbs + SO 2 4 aq Since ν H = O, E = E = V at all ph values in the Pourboix diagram.

8. ELECTROCHEMICAL CELLS. n Electrode Reactions and Electrode Potentials a. H 2 2H + + 2e. Cl 2 + 2e 2Cl. H 2 + Cl 2 2H + + 2Cl ; z = 2

8. ELECTROCHEMICAL CELLS. n Electrode Reactions and Electrode Potentials a. H 2 2H + + 2e. Cl 2 + 2e 2Cl. H 2 + Cl 2 2H + + 2Cl ; z = 2 8. ELECTROCHEMICAL CELLS n Electrode Reactions and Electrode Potentials 8.1. a. H H + + e Cl + e Cl H + Cl H + + Cl ; z = E = E RT F ln ( a H +a Cl ) b. Hg(l)+ Cl Hg Cl + e H + + e H Hg + H + + Cl Hg Cl