768 Lecture #11 of 18

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Bernard Lester Hall
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1 Lecture #11 of

2 769 Q: What s in this set of lectures? A: B&F Chapter 2 main concepts: Section 2.1 : Section 2.3: Salt; Activity; Underpotential deposition Transference numbers; Liquid junction potentials Sections 2.2 & 2.4: Donnan potential; Membrane potentials; ph meter; Ion-selective electrodes

Question: What if the redox species were negatively charged like [Fe III (CN) 6 ] 3 /[Fe II (CN) 6 ] 4, and we increase the concentration of supporting electrolyte to 0.1 M?

3 Question: What if the redox species were negatively charged like [Fe III (CN) 6 ] 3 /[Fe II (CN) 6 ] 4, and we increase the concentration of supporting electrolyte to 0.1 M? [Fe III ] 3 + 1e [Fe II ] E = E 0 RT nf ln γ R γ O RT nf ln C R C O Fe(CN) 6 3- /Fe(CN) 6 4- with added salt No added salt (note i pc and ir) γ R γ O < 1. 0 ln γ R γ O < 0 ANSWER: E 0' shifts to more positive potentials

4 Practical activity even of solids: Underpotential deposition (UPD) 771 E eq Co 0 Co e Co e Co 0

5 Practical activity even of solids: Underpotential deposition (UPD) 772 E eq Co 0 Co e Co e Co 0 but what are these cathodic current bumps?

6 Practical activity even of solids: Underpotential deposition (UPD) 773 E eq Co 0 Co e a M < 1 activity of a solid is proportional to its surface coverage! B&F, pg. 420 Co e Co 0 but what are these cathodic current bumps?

7 Practical activity even of solids: Underpotential deposition (UPD) 774 E eq Co 0 Co e a M < 1 activity of a solid is proportional to its surface coverage! B&F, pg. 420 Co δ+ cobalt gold Co e Co 0 gold UPD of cobalt OPD of cobalt Mendoza-Huizar, Robles, & Palomar-Pardavé, J. Electroanal. Chem., 2003, 545, 39

8 775 Q: What s in this set of lectures? A: B&F Chapter 2 main concepts: Section 2.1 : Section 2.3: Salt; Activity; Underpotential deposition Transference numbers; Liquid junction potentials Sections 2.2 & 2.4: Donnan potential; Membrane potentials; ph meter; Ion-selective electrodes

9 two general liquid junctions that we care about (the most) 776 an SCE an ISE for nitrate ion

10 liquid junctions: when two ionic solutions are separated across an interface that prevents bulk mixing of the ions, but has ionic permeability, a potential (drop) develops called the liquid junction potential. 777 same salt; different conc. one ion in common; same conc. everything else. Bard & Faulkner, 2 nd Ed., Wiley, 2001, Figure 2.3.2

11 example: look at the high concentration side ion of higher mobility will impart its charge to the opposite side of the junction 778 same salt; different conc. Bard & Faulkner, 2 nd Ed., Wiley, 2001, Figure 2.3.2

12 example: look at the high concentration side ion of higher mobility will impart its charge to the opposite side of the junction 779 Conceptually, think about a condition in the limit where t H+ = 1 as H + diffuse down their concentration gradient, an electrostatic force is exerted on Cl to pull it along and at the same time this slows transport of H + this happens until t i effective = 0.5, and at which time the system is at steady-state mass transport and has generated its maximum liquid-junction potential. same salt; different conc. Bard & Faulkner, 2 nd Ed., Wiley, 2001, Figure 2.3.2

13 example: look at the high concentration side ion of higher mobility will impart its charge to the opposite side of the junction 780 Conceptually, think about a condition in the limit where t H+ = 1 as H + diffuse down their concentration gradient, an electrostatic force is exerted on Cl to pull it along and at the same time this slows transport of H + this happens until t i effective = 0.5, and at which time the system is at steady-state mass transport and has generated its maximum liquid-junction potential. same salt; different conc. FYI, this process in a semiconductor results in a Dember potential and the transport process is called ambipolar diffusion Bard & Faulkner, 2 nd Ed., Wiley, 2001, Figure 2.3.2

14 equivalent ionic conductivity 781

15 equivalent ionic conductivity 782 / / / / / / / / / / / / / / / / / = F

16 example: compare dissimilar ion (cations or anions) ion of higher mobility will impart its charge to the opposite side of the junction 783 one ion in common; same conc. Bard & Faulkner, 2 nd Ed., Wiley, 2001, Figure 2.3.2

17 example: compare dissimilar ion (cations or anions) ion of higher mobility will impart its charge to the opposite side of the junction 784 and the sign of the liquid-junction potential is obvious for Types 1 and 2 (but not Type 3) based on the mobilities of the individual ions and so when in doubt, think logically about the sign of the potential to verify answers. one ion in common; same conc. Bard & Faulkner, 2 nd Ed., Wiley, 2001, Figure 2.3.2

18 785

19 recall, a transference number, ti, (or transport number) is based on 786 units: cm 2 /(s V) (Stokes' law) and the ionic conductivity, κ or σ, is defined as Siemens units: S/cm or 1/(Ω cm) so ti is the fraction of the solution conductivity attributable to ion "i"

$so ti is the fraction of the solution conductivity attributable to ion "i" The Kohlrausch law (empirical) and Debye Hückel Onsager equation (theoretical) predicts that the equivalent molar$

20 recall, a transference number, ti, (or transport number) is based on 787 units: cm 2 /(s V) (Stokes' law) and the ionic conductivity, κ or σ, is defined as Siemens Λ = κ/c eq units: S/cm or 1/(Ω cm) so ti is the fraction of the solution conductivity attributable to ion "i" The Kohlrausch law (empirical) and Debye Hückel Onsager equation (theoretical) predicts that the equivalent molar conductivity is proportional to the square root of the salt concentration Λ = a(c) 1/2 Friedrich Wilhelm Georg Kohlrausch ( ) from Wiki Physicist

$fraction of the solution conductivity attributable to ion "i" The Kohlrausch law (empirical) and P-Chemist & Physicist Debye Hückel Onsager equation (theoretical) predicts that the equivalent molar$

21 recall, a transference number, ti, (or transport number) is based on 788 units: cm 2 /(s V) and the ionic conductivity, κ or σ, is defined as Siemens Λ = κ/c eq units: S/cm or 1/(Ω cm) so ti is the fraction of the solution conductivity attributable to ion "i" The Kohlrausch law (empirical) and P-Chemist & Physicist Debye Hückel Onsager equation (theoretical) predicts that the equivalent molar conductivity is proportional to the square root of the salt concentration Λ = a(c) 1/2 from Wiki Lars Onsager ( )

22 we use ti values to determine/define the liquid-junction potential (for derivations, see B&F, pp ): 789 Type 1 Type 2 Type 3 same salt; different concentrations same cation or anion; different counter ion; same concentration. no common ions, and/or one common ion; different concs.

23 we use ti values to determine/define the liquid-junction potential (for derivations, see B&F, pp ): 790 Type 1 (α) (β) Type 2 Type 3

24 we use ti values to determine/define the liquid-junction potential (for derivations, see B&F, pp ): 791 Type 1 Type 2 (α) (β) just the ratio of the conductivity attributable to the dissimilar ions Type 3

25 we use ti values to determine/define the liquid-junction potential (for derivations, see B&F, pp ): 792 Type 1 Type 2 (α) (β) just the ratio of the conductivity attributable to the dissimilar ions (with a few assumptions, pg. 72) Type 3

26 we use ti values to determine/define the liquid-junction potential (for derivations, see B&F, pp ): 793 Type 1 (α) (β) Type 2 just the ratio of the conductivity attributable to the dissimilar ions (with a few assumptions, pg. 72) sign depends on the charge of the dissimilar ion: (+) when cations are dissimilar, and ( ) when anions are dissimilar Type 3

27 we use ti values to determine/define the liquid-junction potential (for derivations, see B&F, pp ): 794 Type 1 (α) (β) Type 2 just the ratio of the conductivity attributable to the dissimilar ions (with a few assumptions, pg. 72) sign depends on the charge of the dissimilar ion: (+) when cations are dissimilar, and ( ) when anions are dissimilar Type 3 the Henderson Eq. (with a few assumptions, pg. 72)

28 we use ti values to determine/define the liquid-junction potential (for derivations, see B&F, pp ): 795 Type 1 (α) (β) Type 2 just the ratio of the conductivity attributable to the dissimilar ions (with a few assumptions, pg. 72) sign depends on the charge of the dissimilar ion: (+) when cations are dissimilar, and ( ) when anions are dissimilar Type 3 the Henderson Eq. (with a few assumptions, pg. 72) as written, these equations calculate E j at β vs α

29 example: B&F Problem 2.14d 796 Calculate E j for NaNO3 (0.10 M) / NaOH (0.10M) 1. What Type?

30 example: B&F Problem 2.14d 797 Calculate E j for NaNO3 (0.10 M) / NaOH (0.10M) 1. What Type? Type 2 2. Polarity?

31 798

32 example: B&F Problem 2.14d 799 Calculate E j for NaNO3 (0.10 M) / NaOH (0.10M) 1. What Type? Type 2 2. Polarity? Polarity should be clear compare mobilities; OH is higher so NaNO 3 side will be ( ). 3. Calculate it: - ( ) due to anions moving

33 example: B&F Problem 2.14d 800 Calculate E j for NaNO3 (0.10 M) / NaOH (0.10M) 1. What Type? Type 2 2. Polarity? Polarity should be clear compare mobilities; OH is higher so NaNO 3 side will be ( ). 3. Calculate it: - ( ) due to anions moving E j = log µ NO 3 µ OH E j = log x x 10 4 = V = mv a very large E j!

34 example: B&F Problem 2.14d 801 Calculate E j for NaNO3 (0.10 M) / NaOH (0.10M) 1. What Type? Type 2 2. Polarity? Polarity should be clear compare mobilities; OH is higher so NaNO 3 side will be ( ). 3. Calculate it: - ( ) due to anions moving E j = log E j = log µ NO 3 µ OH as predicted, a (+) LJ potential correlates with the compartment in the denominator, β, vs α x x 10 4 = V = mv a very large E j!

35 802 so, why is saturated KCl (or KNO 3 ) the preferred salt used to fill reference electrodes? similar mobilities and thus, similar t i s and thus, vanishingly small LJ potentials!

36 question: How do B&F obtain from ? 803 Type 3: Type 2 :

37 question: How do B&F obtain from ? 804 Type 3: Type 2 :

38 805 well, Λ (the equivalent conductivity) is defined as follows: where Ceq is the concentration of positive or negative charges associated with a particular salt in solution, and so working backward from and because conductivity is defined as one gets

39 where Ceq is the concentration of positive or negative charges associated with a particular salt in solution, and so working backward from and because conductivity is defined as one gets Type 2 :

40 where Ceq is the concentration of positive or negative charges associated with a particular salt in solution, and so working backward from and because conductivity is defined as one gets Type 2 : Note: I switched α and β

41 off by a factor of (-1). Let s look at that pre-factor for a specific example: 0.1M HCl (α) 0.1M KCl (β) 808

42 809 off by a factor of (-1). Let s look at that pre-factor for a specific example: 0.1M HCl (α) 0.1M KCl (β) so this really means the Lewis Sargent relation should have a ( ) in front of it when net cations diffuse to switch the sign, as.

43 810 since we know that the β side will be (+) in the previous case, this really means the Lewis Sargent Eq. should have a ( ) sign in front of it when net cations diffuse. if we re sticking to our convention that the potential is the β (product/red. phase) versus the α (reactant/ox. phase)

44 ANYWAY 811

45 812 Q: What s in this set of lectures? A: B&F Chapter 2 main concepts: Section 2.1 : Section 2.3: Salt; Activity; Underpotential deposition Transference numbers; Liquid junction potentials Sections 2.2 & 2.4: Donnan potential; Membrane potentials; ph meter; Ion-selective electrodes

an ionomer film a cell Nafion semipermeable membrane membrane impermeable to

46 Donnan potential: A special liquid-junction potential due to fixed charges here are two systems in which Donnan potentials play a prominent role: 813 an ionomer film a cell Nafion semipermeable membrane membrane impermeable to charged macromolecules

47 consider this model which applies to both scenarios 814 a film of poly(styrene sulfonate) CNaCl R Na + m m s s cna+ ccl- cna+ ccl-

48 Because differences in free energy drive net mass transport, mobile Na + and Cl partition between the membrane and the solution in compliance with their electrochemical potentials: 815 μ i o,m + RT ln γ i m + RT ln c i m + z i Fφ m = μ i o,s + RT ln γ i s + RT ln c i s + z i Fφ s m s

49 Because differences in free energy drive net mass transport, mobile Na + and Cl partition between the membrane and the solution in compliance with their electrochemical potentials: 816 μ i o,m + RT ln γ i m + RT ln c i m + z i Fφ m = μ i o,s + RT ln γ i s + RT ln c i s + z i Fφ s (for ion "i" electrochemical potential in membrane is the same as in solution this is the definition of something that has equilibrated!)

50 Because differences in free energy drive net mass transport, mobile Na + and Cl partition between the membrane and the solution in compliance with their electrochemical potentials: 817 μ i o,m + RT ln γ i m + RT ln c i m + z i Fφ m = μ i o,s + RT ln γ i s + RT ln c i s + z i Fφ s If the standard states are the same inside and outside the membrane, we can solve for the "Galvani" / inner potential difference, ϕ m ϕ s which is exactly what we did for the liquid-junction potentials! φ m φ s = RT z i F ln γ i s s c i γ m m i c i = E Donnan

51 Because differences in free energy drive net mass transport, mobile Na + and Cl partition between the membrane and the solution in compliance with their electrochemical potentials: 818 μ i o,m + RT ln γ i m + RT ln c i m + z i Fφ m = μ i o,s + RT ln γ i s + RT ln c i s + z i Fφ s If the standard states are the same inside and outside the membrane, we can solve for the "Galvani" / inner potential difference, ϕ m ϕ s which is exactly what we did for the liquid-junction potentials! φ m φ s = RT z i F ln γ i s s c i γ m m i c i = E Donnan so we can express E Donnan, an equilibrium (electric) potential, in terms of any ion that has access to both the membrane and the solution: E Donnan = RT 1 F ln a s Na + m = RT 1 F ln a s Cl a m Cl a Na +

52 E Donnan = RT F ln a Na + m s a Na + s = RT F ln a Cl a m Cl 819 Aside #1: Recall Type 1 case of LJ potential but now with t = 0 (α) (β) E j = RT F ln a 1 α a 1 β = E Donnan Na + with β being the membrane

53 E Donnan = RT F ln a Na + m s a Na + s = RT F ln a Cl a m Cl 820 Aside #1: Recall Type 1 case of LJ potential but now with t = 0 (α) (β) E j = RT F ln a 1 α a 1 β = E Donnan Na + with β being the membrane Aside #2: This is what B&F write for this (Donnan) potential Check! Eqn. (2.4.2)

54 E Donnan = RT F ln a Na + m s a Na + s = RT F ln a Cl a m Cl 821 Anyway now divide both sides by RT/F and invert the argument of the ln() on the right to eliminate the negative sign, and we have...

55 E Donnan = RT F ln a Na + m s a Na + s = RT F ln a Cl a m Cl 822 Anyway now divide both sides by RT/F and invert the argument of the ln() on the right to eliminate the negative sign, and we have... or s s a Na + a Cl = a m m Na + a Cl

56 recall the scenario we are analyzing (with R representing the fixed charges) 823 a film of poly(styrene sulfonate) CNaCl R Na + m m s s cna+ ccl- cna+ ccl-

57 s s a Na + a Cl = a m m Na + a Cl 824 if these are dilute electrolytes, we can neglect activity coefficients s s c Na + c Cl = c m m Na + c Cl now, there is an additional constraint: the solution and the membrane must be electrically neutral: c m Na + = c m Cl + c m s R c Na + s = c Cl an equation quadratic in c m Cl- is obtained as follows

58 an equation quadratic in c m Cl- is obtained as follows 825 s s c Na + c Cl = c m m Na + c Cl s c Na + s = c Cl because in the solution, c Na+ = c Clfor goodness sakes! s c Cl 2 = m ccl 2 + c m Cl c m R

59 an equation quadratic in c m Cl- is obtained as follows 826 s s c Na + c Cl = c m m Na + c Cl s c Na + s = c Cl c m Na + = c m Cl + c m R s c Cl 2 = m ccl 2 + c m m R c Cl

60 an equation quadratic in c m Cl- is obtained as follows 827 s s c Na + c Cl = c m m Na + c Cl s c Na + s = c Cl c m Na + = c m Cl + c m R s c Cl 2 = m ccl 2 + c m m R c Cl m 0 = c Cl 2 + c m R c m Cl s 2 c Cl use the quadratic formula to solve for c m Cl- and you get c m Cl = c m R + c m R 2 2 s c Cl = c R m c s Cl R c Cl 2 1

61 828 c m Cl = c m R + c m R 2 2 s c Cl = c R m c s Cl c m R 2 1 if s c Cl c m R (which is the normal case of interest), then

62 829 c m Cl = c m R + c m R 2 2 s c Cl = c R m c s Cl c m R 2 1 if s c Cl c m R (which is the normal case of interest), then c s Cl c m R c s Cl c m R 2 (Taylor/Maclaurin series expansion to the first 3 (or 4) terms)

63 830 c m Cl = c m R + c m R 2 2 s c Cl = c R m c s Cl c m R 2 1 if s c Cl c m R (which is the normal case of interest), then c s Cl c m R c s Cl c m R 2 c m Cl = c R m c s Cl c m R 2 1 = Fixed charge sites are responsible for the electrostatic exclusion of mobile like charges (co-ions) from a membrane, cell, etc. This is Donnan Exclusion.

c m Cl = c R m 2 1 + 2 c s Cl c m R 2 1 = 831 Fixed charge sites are responsible for the electrostatic exclusion of

64 c m Cl = c R m c s Cl c m R 2 1 = 831 Fixed charge sites are responsible for the electrostatic exclusion of mobile like charges (co-ions) from a membrane, cell, etc. This is Donnan Exclusion. the higher C R-m, the lower C Cl-m.

65 so how excluded IS excluded? 832 is s c Cl c m R a reasonable assumption? What is C R-m? well, for Nafion 117, the sulfonate concentration is 1.13 M for CR61 AZL from Ionics, the sulfonate concentration is 1.6 M so, as an example, if C Cl-s = 0.1 M = = 0.01 M an order of magnitude lower than C Cl-s rather excluded! Source: Torben Smith Sørensen, Surface Chemistry and Electrochemistry of Membranes, CRC Press, 1999 ISBN ,

66 so how excluded IS excluded? 833 is s c Cl c m R a reasonable assumption? What is C R-m? well, for Nafion 117, the sulfonate concentration is 1.13 M for CR61 AZL from Ionics, the sulfonate concentration is 1.6 M so, as an example, if C Cl-s = 0.1 M = = 0.01 M an order of magnitude lower than C Cl-s rather excluded! but what if C Cl-s is also large (e.g. 1 M)? No more Donnan exclusion!

67 but Donnan exclusion is amplified in Nafion and some other polymers how? 834 Nafion phase separates into a hydrophobic phase, concentrated in -(CF 2 )- backbone, and hydrophilic clusters of SO 3 solvated by water Mauritz & Moore, Chem. Rev., 2004, 104, 4535

68 but Donnan exclusion is amplified in Nafion and some other polymers how? 835 Nafion phase separates into a hydrophobic phase, concentrated in -(CF 2 )- backbone, and hydrophilic clusters of SO 3 solvated by water and, FYI, SO 3 clusters are interconnected by channels that percolate through the membrane, imparting a percolation network for ionic conduction

69 but Donnan exclusion is amplified in Nafion and some other polymers how? tlplib/fuel-cells/figures/dow.png

70 837 So in Nafion, "amplifying" effects operate in parallel 1. The aqueous volume accessible to ions of EITHER charge is a small fraction of the polymer s overall volume, and 2. The local concentration of SO 3 (C R-m ) is much higher than calculated based on the polymer's density and equivalent weight (i.e. the molecular weight per sulfonic acid moiety).

690 Lecture #10 of 18

690 Lecture #10 of 18 Lecture #10 of 18 690 691 Q: What s in this set of lectures? A: B&F Chapters 4 & 5 main concepts: Section 4.4.2: Section 5.1: Section 5.2: Section 5.3 & 5.9: Fick s Second Law of Diffusion Overview of