Instructor solution manual to accompany physical chemistry 7th ed by peter atkins chap10

10 Equilibrium electrochemistry Solutions to exercises Discussion questions E10.1(b) The DebyeHăuckel 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 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, , of the solution Looking at the details of the theory as outlined in Justification 10.2 we see that 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± E10.2(b) 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 E10.3(b) 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 10.43 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 E10.4(b) NaCl(aq) + AgNO3 (aq) → AgCl(s) + NaNO3 (aq) NaCl, AgNO3 and NaNO3 are strong electrolytes; therefore the net ionic equation is Ag+ (aq) + Cl− (aq) → AgCl(s) rH −− = fH −− (AgCl, s) − fH −− (Ag+ , aq) − fH −− (Cl− , aq) = (−127.07 kJ mol−1 ) − (105.58 kJ mol−1 ) − (−167.16 kJ mol−1 ) = −65.49 kJ mol−1 E10.5(b) Pb2+ (aq) + S2− (aq) PbS(s) aJνJ KS = J Since the solubility is expected to be low, we may (initially) ignore activity coefficients Hence KS = b(Pb2+ ) b(S2− ) × b −− b −− b(Pb2+ ) = b(S2− ) = S INSTRUCTOR’S MANUAL 148 KS = S2 (b −− )2 S = (KS )1/2 b −− − r G−− to obtain KS RT Use ln KS = rG −− = fG −− (S2− , aq) + = (+85.8 kJ mol −1 fG −− (Pb2+ , aq) − ) + (−24.43 kJ mol −1 rG −− (PbS, s) ) − (−98.7 kJ mol−1 ) = 160.07 kJ mol−1 ln KS = −160.07 × 103 J mol−1 (8.314 J K−1 mol−1 ) × (298 K) = −64.61 KS = e−64.61 = 8.7 × 10−29 KS = E10.6(b) S2 b −− S = (KS )1/2 b −− = (8.735 × 10−29 )1/2 = 9.3 × 10−15 mol kg−1 The ratio of hydration Gibbs energies is −− (NO− ) G−− (Cl− ) hyd G hyd We have So E10.7(b) hyd G I = 21 hyd G −− i −− = r(Cl− ) r(NO− 3) = 181 pm = 0.958 189 pm (Cl− ) = −379 kJ mol−1 [Exercise 10.6a] −1 −1 (NO− ) = (0.958) × (−379 kJ mol ) = −363 kJ mol (bi /b −− )zi2 [10.18] and for an Mp Xq salt, b+ /b −− = pb/b −− , b− /b −− = qb/b −− , so 2 + qz− )b/b −− I = 21 (pz+ E10.8(b) E10.9(b) (a) I (MgCl2 ) = 21 (1 × 22 + × 1)b/b −− = 3b/b −− (b) I (Al2 (SO4 )3 ) = 21 (2 × 32 + × 22 )b/b −− = 15b/b −− (c) I (Fe2 (SO4 )3 ) = 21 (2 × 32 + × 22 )b/b −− = 15b/b −− b(K3 [Fe(CN)6 ]) b(KCl) b(NaBr) + + I = I (K3 [Fe(CN)6 ]) + I (KCl) + I (NaBr) = 21 (3 + 32 ) b −− b −− b −− = (6) × (0.040) + (0.030) + (0.050) = 0.320 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? b I = I (KNO3 ) = −− (KNO3 ) = 0.110 b Therefore, the ionic strengths of the added salts must be 0.890 (a) b I (KNO3 ) = −− , so b(KNO3 ) = 0.890 mol kg−1 b and (0.890 mol kg−1 ) × (0.500 kg) = 0.445 mol KNO3 So (0.445 mol) × (101.11 g mol−1 ) = 45.0 g KNO3 must be added EQUILIBRIUM ELECTROCHEMISTRY (b) 149 b b I (Ba(NO3 )2 ) = 21 (22 + × 12 ) −− = −− = 0.890 b b b= 0.890 −− b = 0.2967 mol kg−1 and (0.2967 mol kg−1 ) × (0.500 kg) = 0.1484 mol Ba(NO3 )2 So (0.1484 mol) × (261.32 g mol−1 ) = 38.8 g Ba(NO3 )2 E10.10(b) I (Al2 (SO4 )3 ) = 21 ((2 × 33 ) + (3 × 22 ))b/b −− = 15b/b −− I (Ca(NO3 )2 ) = 21 (22 + 2)b/b −− = 3b/b −− 3(0.500 mol kg−1 ) = 15(b(Al2 (SO4 )3 )) (0.500 mol kg−1 ) = 0.100 mol kg−1 b(Al2 (SO4 )3 ) = 15 E10.11(b) p q γ± = (γ+ γ− )1/s s =p+q For Al2 (SO4 )3 p = 2, q = 3, s = γ± = (γ+2 γ−3 )1/5 E10.12(b) Since the solutions are dilute, use the DebyeHăuckel limiting law log = −|z+ z− |AI 1/2 I = 21 i zi2 (bi /b −− ) = 21 {1 × (0.020) + × (0.020) + × (0.035) + × (0.035)} = 0.125 log γ± = −1 × × 0.509 × (0.125)1/2 = −0.17996 (For NaCl) γ± = 10−0.17996 = 0.661 E10.13(b) I (CaCl2 ) = 21 (4 + 2)b/b −− = 3b/b −− log γ± = −2 × × 0.509 × (0.300)1/2 = −0.5576 γ± = 10−0.5576 = 0.2770 = 0.277 0.524 − 0.277 × 100 per cent = 47.1 per cent 0.524 A|z+ z− |I 1/2 E10.14(b) The extended DebyeHăuckel law is log = + BI 1/2 Solving for B Error = B=− A|z+ z− | + log γ± I 1/2 =− 0.509 + log γ± (b/b −− )1/2 INSTRUCTOR’S MANUAL 150 Draw up the following table b/(mol kg−1 ) γ± B 5.0 × 10−3 0.927 1.32 10.0 × 10−3 0.902 1.36 50.0 × 10−3 0.816 1.29 B = 1.3 E10.15(b) KS = 1.4 × 10−8 PbI2 (aq) PbI2 (s) rG −− = −RT ln KS = −(8.314 J K−1 mol−1 ) × (298.15 K) × ln(1.4 × 10−8 ) = 44.83 kJ mol−1 rG −− = fG −− fG −− (PbI2 , aq) − (PbI2 , aq) = rG −− + fG −− (PbI2 , s) fG −1 −− (PbI2 , s) = 44.83 kJ mol − 173.64 kJ mol−1 = −128.8 kJ mol−1 E10.16(b) The Nernst equation may be applied to half-cell potentials as well as to overall cell potentials E(H+ /H2 ) = RT a(H+ ) ln F (fH2 /p −− )1/2 E = E2 − E1 = RT RT a2 (H+ ) γ± b2 [fH2 is constant] = ln ln + F a1 (H ) F γ± b = (25.7 mV) × ln (0.830) × (5.0 × 10−2 ) (0.929) × (5.0 × 10−3 ) = +56.3 mV E10.17(b) Identify electrodes using species with the desired oxidation states L: Cd(s) + 2OH− (aq) → Cd(OH)2 (s) + 2e− R: Ni(OH)3 (s) + e− → Ni(OH)2 (s) + OH− (aq) Cd(s)|Cd(OH)2 (s)|OH− (aq)|Ni(OH)2 (s)|Ni(OH)3 (s)|Pt E10.18(b) 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: Ag2 CrO4 (s) + 2e− → 2Ag(s) + CrO2− (aq) L: Cl2 (g) + 2e− → 2Cl− (aq) Overall (R − L): Ag2 CrO4 (s) + 2Cl− (aq) → 2Ag(s) + CrO2− (aq) + Cl2 (g) +0.45 V +1.36 V −0.91 V (b) R: Sn4+ (aq) + 2e− → Sn2+ (aq) L: 2Fe3+ (aq) + 2e− → 2Fe2+ (aq) Overall (R − L): Sn4+ (aq) + 2Fe2+ (aq) → Sn2+ (aq) + 2Fe3+ (aq) +0.15 V +0.77 V −0.62 V (c) R: MnO2 (s) + 4H+ (aq) + 2e− → Mn2+ (aq) + 2H2 O(l) L: Cu2+ (aq) + 2e− → Cu(s) Overall (R − L): Cu(s) + MnO2 (s) + 4H+ (aq) → Cu2+ (aq) + Mn2+ (aq) + 2H2 O(l) +1.23 V +0.34 V +0.89 V EQUILIBRIUM ELECTROCHEMISTRY 151 Comment Those cells for which E −− > may operate as spontaneous galvanic cells under standard conditions Those for which E −− < may operate as nonspontaneous electrolytic cells Recall that E −− informs us of the spontaneity of a cell under standard conditions only For other conditions we require E E10.19(b) The conditions (concentrations, etc.) under which these reactions occur are not given For 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: 2H2 O(l) + 2e− → 2OH− (aq) + H2 (g) −0.83 V L: 2Na+ (aq) + 2e− → 2Na(s) −2.71 V and the cell is Na(s)|Na+ (aq), OH− (aq)|H2 (g)|Pt +1.88 V or more simply Na(s)|NaOH(aq)|H2 (g)|Pt (b) R: I2 (s) + 2e− → 2I− (aq) +0.54 V L: 2H+ (aq) + 2e− → H2 (g) and the cell is Pt|H2 (g)|H+ (aq), I− (aq)|I2 (s)|Pt +0.54 V or more simply Pt|H2 (g)|HI(aq)|I2 (s)|Pt (c) R: 2H+ (aq) + 2e− → H2 (g) 0.00 V L: 2H2 O(l) + 2e− → H2 (g) + 2OH− (aq) −0.083 V and the cell is Pt|H2 (g)|H+ (aq), OH− (aq)|H2 (g)|Pt 0.083 V or more simply Pt|H2 (g)|H2 O(l)|H2 (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 E10.20(b) See the solutions for Exercise 10.18(b), where we have used E −− = ER−− − EL−− , with standard electrode potentials from Table 10.7 E10.21(b) See the solutions for Exercise 10.19(b), where we have used E −− = ER−− − EL−− , with standard electrode potentials from Table 10.7 E10.22(b) In each case find E −− = ER−− − EL−− from the data in Table 10.7, then use rG (a) R: L: −− = −νF E −− [10.32] 2− − S2 O2− (aq) + 2e → 2SO4 (aq) +2.05 V + 1.51 V − − I2 (s) + 2e → 2I (aq) +0.54 V rG −− = (−2) × (96.485 kC mol−1 ) × (1.51 V) = −291 kJ mol−1 INSTRUCTOR’S MANUAL 152 (b) Zn2+ (aq) + 2e− → Zn(s) −0.76 V Pb2+ (aq) + 2e− → Pb(s) rG −− −0.13 V E −− = −0.63 V = (−2) × (96.485 kC mol−1 ) × (−0.63 V) = +122 kJ mol−1 E10.23(b) (a) A new half-cell may be obtained by the process (3) = (1) − (2), that is (3) 2H2 O(l) + Ag(s) + e− → H2 (g) + 2OH− (aq) + Ag+ (aq) But, E3−− = E1−− − E2−− , 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 G3 −− r G1 = − −− r G2 −ν3 F E3−− = −ν1 F E1−− − (−ν2 F E2−− ) Solving for E3−− we obtain E3−− = ν1 E1−− − ν2 E2−− (2) × (−0.828 V) − (1) × (0.799 V) = −2.455 V = ν3 (b) The complete cell reactions is obtained in the usual manner We take (2) × (2) − (1) to obtain 2Ag+ (aq) + H2 (g) + 2OH− (aq) → 2Ag(s) + 2H2 O(l) E −− (cell) = ER−− − EL−− = E2−− − E1−− = (0.799 V) − (−0.828 V) = +1.627 V Comment The general relation for E −− of a new half-cell obtained from two others is E3−− = E10.24(b) (a) ν1 E1−− ± ν2 E2−− ν3 E = E −− − Q= J RT ln Q ν = νF 2 aJνJ = aH +a − Cl [all other activities = 1] 2 a− = (γ+ b+ )2 × (γ− b− )2 = a+ = (γ+ γ− )2 × (b+ b− )2 = γ±4 b4 Hence, E = E −− − (b) (c) rG b b ≡ −− here and below b [16, b+ = b, b− = b] 2RT RT ln(γ± b) ln(γ±4 b4 ) = E −− − F 2F = −νF E[10.32] = −(2)×(9.6485×104 C mol−1 )×(0.4658 V) = −89.89 kJ mol−1 log γ± = −|z+ z− |AI 1/2 [19] = −(0.509) × (0.010)1/2 [I = b for HCl(aq)] = −0.0509 γ± = 0.889 E −− = E + 2RT ln(γ± b) = (0.4658 V) + (2) × (25.693 × 10−3 V) × ln(0.889 × 0.010) F = +0.223 V The value compares favourably to that given in Table 10.7 EQUILIBRIUM ELECTROCHEMISTRY E10.25(b) 153 R: Fe2+ (aq) + 2e− → Fe(s) L: 2Ag+ (aq) + 2e− → 2Ag(s) 2Ag(s) + Fe2+ (aq) → 2Ag+ (aq) + Fe(s) R − L: E −− = ER−− − EL−− = (−0.44 V) − (0.80 V) = −1.24 V rG −− = −νF E −− = −2 × (9.65 × 104 C mol−1 ) × (−1.24 V) = +239 kJ mol−1 rH −− = f H −− (Ag+ , aq) − fH −− (Fe2+ , aq) = [(2) × (105.58) − (−89.1)] kJ mol−1 = +300.3 kJ mol−1 ∂ r G−− = − r S −− = ∂T p = Therefore, rG −− − T rH −− [ r G−− = rH − T r S] (239 − 300.3) kJ mol−1 = −0.206 kJ mol−1 K −1 298.15 K νF E −− [10.36] RT Sn(s) + CuSO4 (aq) Cu(s) + SnSO4 (aq) 2+ − R: Cu (aq) + 2e → Cu(s) +0.34 V + 0.48 V L: Sn2+ (aq) + 2e− → Sn(s) −0.14 V ln K = (b) −− (308 K) ≈ (239) + (10 K) × (−0.206 K −1 ) kJ mol−1 ≈ +237 kJ mol−1 E10.26(b) In each case ln K = (a) rG (2) × (0.48 V) = +37.4, 25.693 mV K = 1.7 × 1016 Cu2+ (aq) + Cu(s) 2Cu+ (aq) 2+ − R: Cu (aq) + e → Cu+ (aq) +0.16 V − 0.36 V +0.52 V L: Cu+ (aq) + e− → Cu(s) ln K = −0.36 V = −14.0, 25.693 mV K = 8.2 × 10−7 E10.27(b) We need to obtain E −− for the couple (3) Co3+ (aq) + 3e− → Co(s) from the values of E −− for the couples (1) Co3+ (aq) + e− → Co2+ (aq) E1−− = 1.81 V (2) Co2+ (aq) + 2e− → Co(s) E2−− = −0.28 V We see that (3) = (1) + (2); therefore (see the solution to Exercise 10.23(b)) E3 = ν1 E1−− + ν2 E2−− (1) × (1.81 V) + (2) × (−0.28 V) = = 0.42 V ν3 INSTRUCTOR’S MANUAL 154 Then, R: Co3+ (aq) + 3e− → Co(s) L: 3AgCl(s) + 3e− → 3Ag(s) + 3Cl− (aq) R − L: Co3+ (aq) + 3Cl− (aq) + 3Ag(s) → 3AgCl(s) + Co(s) E −− = ER−− − EL−− = (0.42 V) − (0.22 V) = +0.20 V ER−− = 0.42 V EL−− = 0.22 V E10.28(b) First assume all activity coefficients are and calculate KS◦ , the ideal solubility product constant AgI(s) Ag+ (aq) + I− (aq) S(AgI) = b(Ag+ ) = b(I− ) because all stoichiometric coefficients are b(Ag+ )b(I− ) S2 = = (1.2 × 10−8 )2 = 1.44 × 10−16 Thus KS◦ = b −−2 b −−2 2Bi3+ (aq) + 3S2− (aq) (2) Bi2 S3 (s) (1) b(Bi3+ ) = 2S(Bi2 S3 ) b(S2− ) = 3S(Bi2 S3 ) KS◦ = (b(Bi3+ ))2 × (b(S2− ))3 (2S)2 × (3S)3 S = = 108 b −− b −−5 b −−5 = 1.13 × 10−97 For AgI, KS = γ±2 KS◦ log γ± = −|z+ z− |AI 1/2 −− I = Sb , A = 0.509 |z+ z− | = so log γ± = −(0.509) × (1.2 × 10−8 )1/2 = −5.58 × 10−5 γ± = 0.9999 KS = (0.9999)2 KS◦ = 0.9997KS◦ For Bi2 S3 , I = 15b/b −− = 15Sb −− , |z+ z− | = so log γ± = −(0.509) × (6) × [15(1.6 × 10−20 )]1/2 = −1.496 × 10−9 γ± = 1.0 KS = γ±5 KS◦ = KS◦ Neglect of activity coefficients is not significant for AgI and Bi2 S3 E10.29(b) The Nernst equation applies to half-reactions as well as whole reactions; thus for − 2+ 8H+ + MnO− (aq) + 5e → Mn (aq) + 4H2 O E = E −− − E10.30(b) R: L: a(Mn2+ ) RT ln + 5F a(MnO− )a(H ) 2AgI(s) + 2e− → 2Ag(s) + 2I− (aq) −0.15 V 2H+ (aq) + 2e− → H2 (g) 0V Overall(R − L): 2AgI(s) + H2 (g) → 2Ag(s) + 2H+ (aq) + 2I− (aq) Q = a(H+ )2 a(I− )2 + − ν=2 Assume a(H ) = a(I ), Q = a(H+ )4 EQUILIBRIUM ELECTROCHEMISTRY E = E −− − pH = 155 RT 2RT ln a(H+ )4 = E −− − ln a(H+ ) = E −− + × (2.303) × 2F F F × (2.303RT ) × (E − E −− ) = RT F × pH 1.15 V E + 0.15 V = = 9.72 0.1183 V 0.1183 V E10.31(b) The electrode reactions are L: Ag+ (aq) + e− → Ag(s) R: AgI(s) + e− → Ag(s) + I− (aq) Overall(R − L): AgI(s) → 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 = E10.32(b) R: 2Bi3+ (aq) + 6e− → 2Bi(s) L: Bi2 S3 (s) + 6e− → 2Bi(s) + 3S2− (aq) Overall(R − L): 2Bi3+ (aq) + 3S2− (aq) → Bi2 S3 (s) ν=6 νF E −− RT 6(0.96 V) = (25.693 × 10−3 V) ln K = = 224 K = e224 It is convenient to give the solution for (b) first (b) KS = K −1 = e−224 ≈ 10−98 , since the cell reaction is the reverse of the solubility equilibrium (a) KS ≈ 10−98 = S= 10−98 108 b b 3+ 2− (Bi ) × (S ) = (2S)2 × (3S)3 = 108S b −− b −− 1/5 ≈ 10−20 mol L−1 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 10.4) 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: Hg2 SO4 (s) + 2e− → 2Hg(l) + SO2− +0.62 V (aq) L: PbSO4 (s) + 2e− → Pb(s) + SO2− (aq) −0.36 V R − L: Pb(s) + Hg2 SO4 (s) → PbSO4 (s) + 2Hg(l) +0.98 V INSTRUCTOR’S MANUAL 156 Hence, a suitable cell would be Pb(s)|PbSO4 (s)|H2 SO4 (aq)|Hg2 SO4 (s)|Hg(l) or, alternatively, Pb(s)|PbSO4 (s)|H2 SO4 (aq) H2 SO4 (aq) Hg2 SO4 (s)|Hg(l) For the cell in which the only sources of electrolyte are the slightly soluble salts, PbSO4 and Hg2 SO4 , the cell would be Pb(s)|PbSO4 (s)|PbSO4 (aq) Hg2 SO4 (aq)|Hg2 SO4 (s)|Hg(l) The potential of this cell is given by the Nernst equation [10.34] RT ln Q [10.34]; ν = νF aPb2+ aSO2− KS (PbSO4 ) Q= = aHg2+ aSO2− KS (Hg2 SO4 ) E = E −− − E = (0.98 V) − = (0.98 V) − RT KS (PbSO4 ) ln KS (Hg2 SO4 ) 2F 25.693 × 10−3 V × ln 1.6 × 10−8 6.6 × 10−7 [Table 10.6, 4th Edition, or CRC Handbook] = (0.98 V) + (0.05 V) = +1.03 V P10.6 Pt|H2 (g)|NaOH(aq), NaCl(aq)|AgCl(s)|Ag(s) H2 (s) + 2AgCl(s) → 2Ag(s) + 2Cl− (aq) + 2H+ (aq) ν=2 RT ln Q, Q = a(H+ )2 a(Cl− )2 [f/p −− = 1] 2F RT RT RT Kw a(Cl− ) Kw γ± b(Cl− ) = E −− − = E −− − ln a(H+ )a(Cl− ) = E −− − ln ln − F F F a(OH ) γ± b(OH− ) E = E −− − = E −− − Kw b(Cl− ) b(Cl− ) RT RT RT −− = E ln ln K ln − − w F F F b(OH− ) b(OH− ) = E −− + (2.303) RT RT b(Cl− ) × pKw − ln F F b(OH− ) − b(Cl ) ln b(OH − E − E −− ) Hence, pKw = + 2.303RT /F 2.303 = pKw = − log Kw = − ln Kw 2.303 E − E −− + 0.05114 2.303RT /F E −− = ER−− − EL−− = E −− (AgCl, Ag) − E −− (H+ /H2 ) = +0.22 V − [Table 10.7] We then draw up the following table with the more precise value for E −− = +0.2223 V [Problem 10.8] θ/◦ C E/V 2.303RT F V pKw 20.0 1.04774 25.0 1.04864 30.0 1.04942 0.05819 0.05918 0.06018 14.23 14.01 13.79 EQUILIBRIUM ELECTROCHEMISTRY 157 −− d ln Kw rH [9.26] = dT RT rH Hence, then with rH −− = −(2.303)RT d pKw ≈ dT −− d (pKw ) dT pKw T ≈ −(2.303) × (8.314 J K −1 mol−1 ) × (298.15 K)2 × 13.79 − 14.23 10 K = +74.9 kJ mol−1 −− = −RT ln Kw = 2.303RT × pKw = +80.0 kJ mol−1 −− = rG rS rH −− − T rG −− = −17.1 J K−1 mol−1 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 Ag(s)|AgX(s)|MX(b1 )|Mx Hg(s) Hg (Reduction of M+ and formation of amalgam) R: M+ (b1 ) + e− −→ Mx Hg(s) L: AgX(s) + e− → Ag(s) + X− (b1 ) R − L: Hg Ag(s) + M+ (b1 ) + X− (b1 ) −→ Mx Hg(s) + AgX(s) ν=1 a(Mx Hg) a(M+ )a(X− ) RT E = E −− − ln Q F Q= For a pair of such cells back to back, Ag(s)|AgX(s)|MX(b1 )|Mx Hg(s)|MX(b2 )|AgX(s)|Ag(s) RT RT ER = E −− − EL = E −− − ln QR ln QL F F QL (a(M+ )a(X− ))L −RT RT ln ln E= = F QR F (a(M+ )a(X− ))R (Note that the unknown quantity a(Mx Hg) drops out of the expression for E.) a(M+ )a(X− ) = γ+ b+ b −− γ− b− b −− = γ±2 b b −− (b+ = b− ) With L = (1) and R = (2) we have E= γ± (1) 2RT b1 2RT ln + ln b2 F γ± (2) F b1 Take b2 = 0.09141 mol kg−1 (the reference value), and write b = −− b E= 2RT F ln b γ± + ln 0.09141 γ± (ref) INSTRUCTOR’S MANUAL 158 For b = 0.09141, the extended DebyeHăuckel law gives log (ref) = (−1.461) × (0.09141)1/2 + (0.20) × (0.09141) = −0.2735 (1) + (1.70) × (0.09141)1/2 γ± (ref) = 0.5328 then E = (0.05139 V) × ln ln γ± = b γ± + ln 0.09141 0.5328 E b − ln (0.09141) × (0.05328) 0.05139 V We then draw up the following table b/(mol/kg−1 ) E/V γ 0.0555 −0.0220 0.572 0.09141 0.0000 0.533 0.1652 0.0263 0.492 0.2171 0.0379 0.469 1.040 0.1156 0.444 1.350 0.1336 0.486 A more precise procedure is described in the original references for the temperature dependence of E −− (Ag, AgCl, Cl− ), see Problem 10.10 P10.10 The method of the solution is first to determine H2 (g) + AgCl(s) rG −− , rH −− , and rS −− for the cell reaction → Ag(s) + HCl(aq) 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) rG −− = −νF E −− At 298.15 K(25.00◦ C) E −− /V = (0.23659) − (4.8564 × 10−4 ) × (25.00) − (3.4205 × 10−6 ) × (25.00)2 + (5.869 × 10−9 ) × (25.00)3 = +0.22240 V G−− = −(96.485 kC mol−1 ) × (0.22240 V) = −21.46 kJ mol−1 Therefore, rS −− =− ∂E −− ∂θ p V ∂ r G−− = ∂T p ∂E −− × νF = νF ∂T p ∂E −− ◦ C ∂θ p K [dθ/◦ C = dT /K] = (−4.8564 × 10−4 /◦ C) − (2) × (3.4205 × 10−6 θ/(◦ C)2 ) + (3) × (5.869 × 10−9 θ /(◦ C)3 ) ∂E −− ∂θ p V/◦ C = (−4.8564 × 10−4 ) − (6.8410 × 10−6 (θ/◦ C)) + (1.7607 × 10−8 (θ/◦ C)2 ) Therefore, at 25.00◦ C, ∂E −− = −6.4566 × 10−4 V/◦ C ∂θ p and ∂E −− = (−6.4566 × 10−4 V/◦ C) × (◦ C/K) = −6.4566 × 10−4 V K−1 ∂T p (a) EQUILIBRIUM ELECTROCHEMISTRY 159 Hence, from equation (a) rS −− = (−96.485 kC mol−1 ) × (6.4566 × 10−4 V K−1 ) = −62.30 J K−1 mol−1 rH and −− = rG −− + T r S −− = −(21.46 kJ mol−1 ) + (298.15 K) × (−62.30 J K −1 mol−1 ) = −40.03 kJ mol−1 For the cell reaction H2 (g) + AgCl(s) → Ag(s) + HCl(aq) −− = f G−− (H+ ) + f G−− (Cl− ) − rG = f G−− (Cl− ) − f G−− (AgCl) fG Hence, −− (Cl− ) = rG −− + fG −− fG −− (AgCl) −− [ f G (H+ ) = 0] (AgCl) = [(−21.46) − (109.79)] kJ mol−1 = −131.25 kJ mol−1 fH Similarly, −− (Cl− ) = rH −− + fH −− (AgCl) = (−40.03) − (127.07 kJ mol−1 ) = −167.10 kJ mol−1 For the entropy of Cl− in solution we use −− rS −− with S = S −− (Ag) + S −− (H+ ) + S −− (Cl− ) − 21 S −− (H2 ) − S −− (AgCl) (H+ ) = Then, S −− (Cl− ) = rS −− − S −− (Ag) + 21 S −− (H2 ) + S −− (AgCl) = (−62.30) − (42.55) + 21 × (130.68) + (96.2) = +56.7 J K−1 mol−1 P10.12 ∂G = V [5.10] ∂p T ∂ rG we obtain = rV ∂p T Substituting r G = −νF E [10.32] yields (a) From ∂E rV =− ∂p T ,n νF (b) The plot (Fig 10.1) of E against p appears to fit a straight line very closely A linear regression analysis yields Slope = 2.840 × 10−3 mV atm−1 , Intercept = 8.5583 mV, R = 0.999 997 01 (an extremely good fit) From r V standard deviation = × 10−6 mV atm−1 standard deviation = 2.8 × 10−3 mV ∂E (−2.666 × 10−6 m3 mol−1 ) =− ∂p T ,n × 9.6485 × 104 C mol−1 Since J = V C = Pa m3 , C = m3 V Pa m3 or = V C Pa INSTRUCTOR’S MANUAL 160 13 12 11 10 500 1000 1500 Figure 10.1 Therefore ∂E = ∂p T ,n 2.666 × 10−6 9.6485 × 104 V 1.01325 × 105 Pa × = 2.80 × 10−6 V atm−1 Pa atm = 2.80 × 10−3 mV atm−1 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 yields a = 8.5592 mV, b = 2.835 × 10−3 mV atm−1 , c = 3.02 × 10−9 mV atm−2 , R = 0.999 997 11 standard deviation = 0.0039 mV standard deviation = 0.012 × 10−3 mV atm−1 standard deviation = 7.89 × 10−9 mV atm−1 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 The slope changes from ∂E = b = 2.835 × 10−3 mV atm−1 ∂p ∂E = b + 2c(1500 atm) = 2.836 × 10−3 mV atm−1 ∂p max ∂E are very good We conclude that the linear fit and constancy of ∂p to EQUILIBRIUM ELECTROCHEMISTRY 161 (d) We can obtain an order of magnitude value for the isothermal compressibility from the value of c ∂ rV = 2c ∂p T ∂ rV 2νcF (κT )cell = − = V ∂p T V ∂ 2E =− νF ∂p (κT )cell = cm3 atm 2(1) × (3.02 × 10−12 V atm−2 ) × (9.6485 × 104 C mol−1 ) × 82.058 8.3145 J cm3 0.996 g = 3.2 × 10−7 atm−1 g × 18.016 mol 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 ∼ × 10−6 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= a(H2 O)4 a(V4 O12 −4 ) γ (V4 O12 −4 )b(V4 O12 −4 ) ≈ a(H2 VO4 − )4 γ (H2 VO4 − )4 b(H2 VO4 − )4 Let x be b(H2 VO4 − ); then b(V4 O12 −4 ) = (0.010 − x)/4 Then the equilibrium equation can be expressed as x4 Kγ (H2 VO4 − )4 = (0.010 − x)/4 γ (V4 O12 −4 ) which can be solved numerically once the constants are determined The activity coefficients are log γ (H2 VO4 − ) = − 0.5373 = −0.269 and log γ (V4 O12 −4 ) = − so 0.5373(42 ) = −1.075 γ (H2 VO4 − ) = 0.538 so γ (V4 O12 −4 ) = 0.0842 The equation is x (2.5 × 106 ) = (0.010 − x)/4 Its solution is x = 0.0048 mol kg−1 = b(H2 VO4 − ) and b(V4 O12 −4 ) = 0.010 − (0.010 − 0.0048)/4 = 0.0013 mol kg−1 P10.18 The reduction reaction is Sb2 O3 (s) + 3H2 O(l) + 6e− → 2Sb(s) + 6OH− (aq) Q = a(OH− )6 ν=6 Therefore (a) RT RT 2.303RT ln a(OH− )6 = E −− − ln a(OH− ) = E −− + pOH 6F F F [ln a(OH− ) = 2.303 log a(OH− ) = −2.303pOH] E = E −− − INSTRUCTOR’S MANUAL 162 (b) Since pOH + pH = pKw E = E −− + 2.303RT (pKw − pH) F (c) The change in potential is E= 2.303RT (pOHf − pOHi ) = (59.17 mV) × (pOHf − pOHi ) F pOHf = − log(0.050γ± ) = − log 0.050 − log γ± = − log 0.050 + A (0.050) = 1.415 pOHi = − log(0.010γ± ) = − log 0.010 − log γ± = − log 0.010 + A (0.010) = 2.051 E = (59.17 mV) × (1.415 − 2.051) = −37.6 mV Hence, P10.19 We need to obtain rH −− + H2 (g) + Uup (aq) for the reaction → Uup(s) + H+ (aq) We draw up the thermodynamic cycle shown in Fig 10.2 Data are obtained from Table 13.4, 14.3, 2.6, and 2.6b The conversion factor between eV and kJ mol−1 is eV = 96.485 kJ mol−1 The distance from A to B in the cycle is given by rH −− = x = (3.22 eV) + 21 × (4.5 eV) + (13.6 eV) − (11.3 eV) − (5.52 eV) − (1.5 eV) = 0.75 eV rS −− =S −− (Uup, s) + S −− (H+ , aq) − 21 S −− (H2 , g) − S −− (Uup+ , aq) = (0.69) + (0) − 21 × (1.354) − (1.34) meV K −1 = −1.33 meV K−1 i Figure 10.2 rG −− = rH −− − T r S −− = (0.75 eV) + (298.15 K) × (1.33 meV K −1 ) = +1.15 eV which corresponds to +111 kJ mol−1 The electrode potential is therefore − r G−− , with ν = 1, or −1.15 V νF EQUILIBRIUM ELECTROCHEMISTRY 163 Solutions to theoretical problems P10.21 MX(s) M+ (aq) + X− (aq), Ks ≈ b(M+ )b(X− ) b b ≡ −− b b(X− ) = S + C b(M+ ) = S, Ks = S(S + C), or S + CS − Ks = 4Ks 1/2 which solves to S = 21 (C + 4Ks )1/2 − 21 C or S = 21 C + − 2C C C2, If 4Ks 2Ks S ≈ 21 C + C P10.22 Ks − 21 C (1 + x)1/2 ≈ + 21 x + · · · ≈ C Ks = a(M+ )a(X− ) = b(M+ )b(X− )γ±2 ; log γ± = −AI 1/2 = −AC 1/2 γ± = e−2.303AC Ks = S (S + C) × e−4.606AC We solve S + S C − Ks γ±2 1/2 1/2 =0 4Ks to get S = C2 + 2 γ± 1/2 Ks [as in Problem 10.21] − C≈ Cγ±2 Therefore, since γ±2 = e−4.606AC P10.25 b(X− ) = S + C ln γ± = −2.303AC 1/2 γ±2 = e−4.606AC 1/2 b(M+ ) = S , 1/2 S ≈ Ks e−4.606AC C 1/2 The half-reactions involved are: −− R: cyt ox + e− → cytred Ecyt −− − L: Dox + e → Dred ED The overall cell reaction is: R − L = cyt ox + Dred cytred + Dox −− −− − ED E −−− = Ecyt (a) The Nernst equation for the cell reaction is E=E− RT [cyt red ][Dox ] ln F [cyt ox ][Dred ] at equilibrium, E = 0; therefore ln ln [cytred ]eq [Dox ]eq F = [cyt ox ]eq [Dred ]eq RT [Dox ]eq [Dred ]eq = ln Therefore a plot of ln intercept of F RT −− −− − ED Ecyt [cyt]ox [cyt]red [Dox ]eq [Dred ]eq −− −− − ED Ecyt + F RT −− −− − ED Ecyt against ln [cyt]ox [cyt]red is linear with a slope of one and an INSTRUCTOR’S MANUAL 164 (b) Draw up the following table: [Dox ]eq [Dred ]eq [cytox ]eq ln [cyt red ]eq ln The plot of ln −1.2124 Hence −5.882 −4.776 −3.661 −3.002 −2.593 −1.436 −0.6274 −4.547 −3.772 −2.415 −1.625 −1.094 −0.2120 −0.3293 [Dox ]eq [Dred ]eq [cytox ]eq [cyt red ]eq against ln is shown in Fig 10.3 The intercept is RT × (−1.2124) + 0.237 V F = 0.0257 V × (−1.2124) + 0.237 V −− = Ecyt = +0.206 V ln([Dox]eq / [Dred]eq) –1 y = –1.2124 + 1.0116x R = 0.99427 –2 –3 –4 –5 –6 –5 –4 –3 –2 –1 ln([cytox]eq / [cytred]eq) Figure 10.3 Solutions to application P10.27 (a) molalityH2 SO4 = b(d) = a(d − d25 ) + c(d − d25 )2 where d is density in g cm−3 at 25 ◦ C, a = 14.523 mol kg−1 (g cm−3 )−1 , c = 25.031 mol kg−1 (g cm−3 )−2 , and d25 = 0.99707 g cm−3 For kg solvent (mH2 O = kg): mass %H2 SO4 = mH2 SO4 mH2 SO4 + mH2 O 100 = b × 100 mH✘ 2O b + mH ✘ m✘ SO4✘H2 O 100 × b(d) where mH2 SO4 = 0.09807 kg mol−1 b(d) + mH 1SO an equation for the solution molarity is deduced with a unit analysis mass %H2 SO4 (d) = molarityH2 SO4 (d) = b(d) × − cm mass %H2 SO4 (d) 10✚ ✚ d× 100 L kg 3g 10✚ ✚ EQUILIBRIUM ELECTROCHEMISTRY 165 Sulfuric Acid Solutions Molality / (mol/kg) 10 1.1 1.2 Density/(g / mL) 1.3 1.4 Figure 10.4(a) Sulfuric Acid Solutions Mass Percentage Sulfuric Acid 50 40 30 20 10 1.1 1.2 Density/(g / mL) 1.3 1.4 Figure 10.4(b) Sulfuric Acid Solutions Molarity/(mol / L) (b) 1.1 1.2 Density/(g / mL) 1.3 1.4 Figure 10.4(c) cell: Pb(s) | PbSO4 (s) | H2 SO4 (aq) | PbO2 (s) | PbSO4 (s) | Pb(s) − cathode: PbO2 (s) + 3H+ (aq) + HSO− (aq) + 2e → PbSO4 (s) + 2H2 O(l) −− Ecathode = 1.6913 V anode: PbSO4 (s) + H+ (aq) + 2e− → Pb(s) + HSO− −− Eanode = −0.3588 V net: PbO2 (s) + Pb(s) + 2H+ (aq) + 2HSO− (aq) → 2PbSO4 (s) + 2H2 O(l) −− ◦ E −− = Ecathode − Eanode = 2.0501V (eqn 10.38) rG −− = −νF E −− = −(2)(9.64853 × 104 C mol−1 )(2.0501 V) = −3.956 × 105 C V mol−1 = −3.956 × 105 J mol−1 = −395.6 kJ mol−1 INSTRUCTOR’S MANUAL 166 −− fH −− rH values of Table 2.6 and the CRC Handbook of Chemistry and Physics are used in the calculation rH −− = f H −− (PbSO4 ) + f H −− (H2 O(l)) − fH −− (PbO2 ) − fH −− (Pb) −2 f H −− (H+ ) − f H −− (HSO− 4) = 2(−919.94 kJ mol−1 ) + 2(−285.83 kJ mol−1 ) − (−277.4 kJ mol−1 ) −2(−887.34 kJ mol−1 ) rH rS −− −− = −359.5 kJ mol−1 rH = −− − T rG −− = 121 J K−1 mol−1 E −− (15◦ C) = E −− (25◦ C) + = 2.0501 V + = −359.5 kJ mol−1 − (−395.6 kJ mol−1 ) 298.15 K (eqn 4.39) E −− = E −− (25◦ C) + ✘ ✟✘ ✘−1 K−1 mol (121 J ✟ ) (10   K) ✘ −1 ✘ ✘ 2(96485 C mol ) rS −− νF T (eqn 10.45) = 2.0501V + 0.006V = 2.0507V The temperature difference makes a negligibly small difference in the cell potential When Q = 6.0 × 10−5 , RT ln Q (eqn 10.34) νF ✘ ✟✘ ✘−1 K −1 mol (8.31451 J ✟ )(298.15   K) = 2.0501 V − ln(6.0 × 10−5 ) ✘ ✘−1 mol 2(96485 C ✘ ) E = E −− − = 2.1750 V (c) The general form of the reduction half-reaction is: ox + νe− + νH H+ + aA → red + xX using eqn 10.34, E = E −− − x ared aX RT RT ln Q = E −− − ln νH a νF νF aox aH + aA RT ln νH νF aH+ (all species other than acids are at unit activity in a Pourboix diagram) = E −− − = E −− + νH RT νH RT ln(10) ln aH+ = E −− + log aH+ νF νF = E −− − νH ν RT ln(10) F E = E −− − (0.05916V) pH νH pH ν (eqn 9.29) EQUILIBRIUM ELECTROCHEMISTRY For the PbO2 | PbSO4 couple, − PbO2 (s) + 4H+ + SO2− (aq) + 2e → PbSO4 (s) + 2H2 O(l) E −− = 1.6913 V, νH = 4, ν = E = 1.6913 V − (0.11832 V)pH For pH = 5, E = 1.0997 V For pH = 8, E = 0.7447 V For the PbSO4 /Pb couple, PbSO4 (s) + 2e− → Pb(s) + SO2− (aq) Since νH = O, E = E −− = −0.3588 V at all pH values in the Pourboix diagram 167 ... 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? b I = I (KNO3 ) = −− (KNO3 ) =... that (3) = (1) + (2); therefore (see the solution to Exercise 10.23(b)) E3 = ν1 E1−− + ν2 E2−− (1) × (1.81 V) + (2) × (−0.28 V) = = 0.42 V ν3 INSTRUCTOR S MANUAL 154 Then, R: Co3+ (aq) + 3e− → Co(s)... 1/2 E10.14(b) The extended DebyeHăuckel law is log γ± = − + BI 1/2 Solving for B Error = B=− A|z+ z− | + log γ± I 1/2 =− 0.509 + log γ± (b/b −− )1/2 INSTRUCTOR S MANUAL 150 Draw up the following