623 Lecture #9 of 18

Transcription

1 Lecture #9 of

2 624 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 step experiments Potential step under diffusion controlled Ultramicroelectrodes Sections : Chronoamperometry/Chronocoulometry

3 OK that s Step # Solve Fick s Second Law to get C O (x, t), and in the process of doing this, you will use boundary conditions that customize the solution for the particular experiment of interest: 2. Use Fick s First Law to calculate J O (0, t) from C O (x, t): 3. Calculate the time-dependent diffusion-limited current: i = nfaj O (0, t)

4 now our solution is fully constrained and we need t back!! 626 inverse L.T. using Table A.1.1 in B&F =

5 What s efrc? Well, first of all, what s the error function: erf? 627

6 Now what s efrc? 628

7 Now what s efrc? 629 Gaussian distribution, with mean = 0 and std. dev. = 1/sqrt(2)

8 Does this make sense? 630 well, for large x, erfc = 0 (erf = 1) and so C(x, t) = C* Check! and for x = 0, erfc = 1 (erf = 0) and so C(x, t) = 0 Check! Let s plot it!

9 Hey, these look completely reasonable! s 0.01s 0.1s t = 1s C* = 1 x 10-6 M D = 1 x 10-5 cm 2 s -1

10 How large is the diffusion layer? Recall 632 D Δ*= *the rms displacement 1D 2Dt plane 2D 4Dt wire, line, tube 3D 6Dt point, sphere, disk cm 2 Δ = 2d Dt = s = cm s Δ = 2d Dt, where d is the dimension a characteristic "diffusion length" root mean square (rms) displacement (standard deviation)

11 Hey, these look completely reasonable! s 0.01s 0.1s t = 1s C* = 1 x 10-6 M D = 1 x 10-5 cm 2 s -1 2Dt = 14 µm 40 µm 4 µm use the geometric area for calculations

12 OK that s Step # Solve Fick s Second Law to get C O (x, t), and in the process of doing this, you will use boundary conditions that customize the solution for the particular experiment of interest: 2. Use Fick s First Law to calculate J O (0, t) from C O (x, t): 3. Calculate the time-dependent diffusion-limited current: i = nfaj O (0, t)

13 now Step #2 635 (Fick s First Law) but we just derived C O (x, t): and so we need to evaluate: J O x, t = D O x C erf x 2 D O t

14 now Step #2 J O x, t = D O x C erf x 2 Dt 636 we use the Liebnitz Rule, to get for d/dx(erf(x)): see B&F, pg. 780, for details

15 now Step #2 J O x, t = D O x C erf we use the Liebnitz Rule, to get for d/dx(erf(x)): x 2 Dt 637 see B&F, pg. 780, for details and using this in conjunction with the chain rule, we get: J O x, t = D O C 1 2 D O t 2π exp x2 4D O t

16 now Step #2 J O x, t = D O x C erf we use the Liebnitz Rule, to get for d/dx(erf(x)): x 2 Dt 638 see B&F, pg. 780, for details and using this in conjunction with the chain rule, we get: J O x, t = D O C 1 2 D O t 2π exp x2 4D O t and when x = 0, we have what we wanted J O 0, t = C D O πt

17 OK that s Steps #1 and Solve Fick s Second Law to get C O (x, t). In the process of doing this, you will use boundary conditions that customize the solution for the particular experiment of interest. 2. Use Fick s First Law to calculate J 0 (0, t) from C(x, t): 3. Calculate the time-dependent diffusion-limited current: i = nfaj(0, t)

18 and finally, Step #3 using Step #2 640 J O 0, t = C D O πt and with i = nfaj(0, t) the Cottrell Equation

19 and finally, Step #3 using Step #2 641 J O 0, t = C D O πt and with i = nfaj(0, t) the Cottrell Equation Frederick Gardner Cottrell, in 1920 b. January 10, 1877, Oakland, California, U.S.A. d. November 16, 1948, Berkeley, California, U.S.A. established Research Corp. in 1912 initial funding from profits on patents for the electrostatic precipitator, used to clear smokestacks of charged soot particles

20 Cottrell, then at UC Berkeley, invented the electrostatic precipitator used to clear smokestacks of charged soot particles 642 corona discharge capture plates neutral particles charged particles > 1 kv

21 643

22 OK, so what does it predict? 644 the Cottrell Equation D = 1.5 x 10-6 cm 2 s -1 D = 1 x 10-5 cm 2 s -1

23 OK, so what does it predict? 645 the Cottrell Equation D = 1 x 10-5 cm 2 s -1 D = 1.5 x 10-6 cm 2 s -1

24 OK, so what does it predict? 646 the Cottrell Equation slope = nfaπ -1/2 D 1/2 C*

25 OK, so what does it predict? 647 the Cottrell Equation????? slope = nfaπ -1/2 D 1/2 C*

26 use it to measure D! but what are the problems with this approach? 648 the Cottrell Equation 1. Huge initial currents compliance current! 2. Noise. 3. Roughness factor adds to current at early times. 4. RC time limitations yield a negative deviation at short times. 5. Adsorbed (electrolyzable) junk adds to current at short times. 6. Convection, and edge effects, impose a long time limit on an experiment.

27 use it to measure D! but what are the problems with this approach? 649 the Cottrell Equation 1. Huge initial currents compliance current! 2. Noise. 3. Roughness factor adds to current at early times. 4. RC time limitations yield a negative deviation at short times. 5. Adsorbed (electrolyzable) junk adds to current at short times. 6. Convection, and edge effects, impose a long time limit on an experiment. solution: Integrate it, with respect to time: the integrated Cottrell Equation

28 650 the integrated Cottrell Equation Anson, Anal. Chem., 1966, 38, 54

29 this is called an Anson plot. 651 What is this positive intercept? Anson, Anal. Chem., 1966, 38, 54

30 this is called an Anson plot. 652 with Γ O, the surface excess of O (mol cm -2 ) What is this positive intercept? Anson, Anal. Chem., 1966, 38, 54

31 653 Fred Anson Anson, Anal. Chem., 1966, 38, 54

32 reversal methods allow you to get data on both O and R, even if you start with just one in solution. 654 Anson, Anal. Chem., 1966, 38, 54

33 655 Q t > τ = 2nFAD 1/2 R CR τ 1/2 + t τ 1/2 t 1/2 π 1/2 + Q dl + nfaγ R

34 So, where is the non-ideal data in a Cottrell and Anson plot? 656 Short times Long times?

35 back to one potential step = one data set, from one experiment: 657 experiment observable governing equation chronoamperometry meas I(t) chronocoulometry meas Q(t)

36 how else can D be measured? how about not using electrochemistry, but NMR What?

37 how else can D be measured? how about not using electrochemistry, but NMR What?

38 So, back to the limitations what then physically happens at longer times? 660 t = 0.1 s planar diffusion t = 1 s mixed diffusion t = 10 s radial diffusion 3 mm insulator Au How is current affected, relative to the Cottrell prediction?

39 A typical electrode used in a laboratory electrochemistry experiment has an area of 0.05 cm 2 to 1 cm A = π = cm 2 7 mm 2

40 we have equations for the signal and the noise, so 662 we can calculate the signal-to-noise ratio for a potential step experiment signal noise i = E R exp t RC here we assume E = 0.6 V, R = 100 Ω

41 we have equations for the signal and the noise, so 663 we can calculate the signal-to-noise ratio for a potential step experiment signal S/N icottrell / icap noise i = E R exp t RC here we assume E = 0.6 V, R = 100 Ω for S/N 10, t > 100 µs

42 the RC time constant of the cell imposes a lower limit on the accessible time window (~100 µs) for a potential step experiment but what is the origin of the long time limit? 664 signal charging take data here edge effects noise i = E R exp t RC here we assume E = 0.6 V, R = 100 Ω

43 The Cottrell Equation can be used to describe the behavior during a potential step experiment as long as the thickness of the diffusion layer is small relative to the electrode dimension (and, of course, the boundary layer / stagnant layer / Nernst diffusion layer (δ)) 665 So, how long is that? t (s) 2Dt 2Dt r 0 D = 10-5 cm 2 /s r 0 = 0.15 cm Answer: < 1 s

44 When the diffusion layer approaches the dimensions of the electrode diameter, radial diffusion to the edges of the electrode cause the flux to be larger than predicted by the Cottrell Equation, and non-uniform. 666 t = 0.1 s linear diffusion t = 1 s mixed diffusion t = 10 s spherical diffusion 3 mm insulator Au How is current affected, relative to the Cottrell prediction?

45 so in a potential step experiment current changes continuously with time. 2. radial diffusion (AKA edge effects ) limit the data acquisition time window to ~1 s. 3. charging imposes a lower limit of µs on this data acquisition time window. 4. maximum current densities are > 60 ma cm -2 initially, but just 100 µa cm -2 at S/N 10. but, why do we care?

46 Why do we care? One reason 668 we need to push this up to perform meaningful measurements of the kinetics of fast reactions

47 but deleterious edge effects also suggest an opportunity: 669 What if instead of avoiding radial diffusion, we exploit it? t = 0.1 s linear diffusion t = 1 s mixed diffusion t = 10 s spherical diffusion 3 mm insulator Au How is current affected, relative to the Cottrell prediction?

48 Let s design an experiment in which we intentionally operate in this radial diffusion limit the entire time! 670 well we actually start in the linear regime, and then switch over quickly side view top view 10 nm 50 µm insulator Au, C, Pt called ultramicroelectrodes or UMEs

49 in any planar diffusion experiment, current changes continuously with time this means: 671 C x, t t = 0 = D 2 C x, t x 2 has no solution but recall from Section that the boundary layer / stagnant layer / Nernst diffusion layer (δ) ends up reaching a steady-state distance due to natural convection anyway, this doesn t help us simplify our experiment

50 the diffusion layer grows with time (indefinitely) s 0.01s 0.1s t = 1s C* = 1 x 10-6 M D = 1 x 10-5 cm 2 s -1 2Dt = 14 µm 40 µm 4 µm use the geometric area for calculations

51 and thus diffusion-controlled currents decay with time (indefinitely) 673 the Cottrell Equation D = 1.5 x 10-6 cm 2 s -1 D = 1 x 10-5 cm 2 s -1

52 but the same is not true for purely spherical diffusion: 674 C r, t t = 0 = D 2 C r, t r r C r, t r which has solutions: C r, t = B + A r for a spherical diffusion field: 2 C r, t r 2 = 2A r 3 C r, t t = A r 2 and so C r, t t = 0 = D 2A r r A r 2

53 1. semi-infinite boundary condition lim C r, t = r C = B + 0 B = C C r, t = B + A r 2. electrode surface (edge/circumference) boundary condition 675 C r 0, t = 0 0 = C + A r 0 A = C r 0 so Fick s 2 nd Law predicts that the steady-state concentration gradient is: C r, t = C C r 0 r = C 1 r 0 r notice that here we can reach a timeindependent condition! What?

54 the diffusion layer thickness is 10r 0, no matter how small r 0 is! nm 100 nm 1 µm C r, t = C 1 r 0 r r 0 = 10 µm

55 the diffusion layer thickness is 10r 0, no matter how small r 0 is! nm 100 nm 1 µm C r, t = C 1 r 0 r r 0 = 10 µm Recall, for transient linear diffusion Δ = 2d Dt = cm 2 s s = cm

56 the diffusion-limited current pre-factor depends on electrode geometry 678 i = "x"nfdc r 0 but not scan rate! geometry sphere hemisphere "x" 4π 2π disk 4 ring π 2 b + a r 0 ln 16 b + a b a a b

57 the UME: You can buy one from BASi 679

58 680 scanning is often steady-state at a UME steady-state occurs when v << RTD/(nFr 02 ) Walsh, Lovelock, & Licence, Chem. Soc. Rev., 2010, 39, 4185

59 681 scanning is often steady-state at a UME steady-state occurs when v << RTD/(nFr 02 ) v (mv s -1 ) << 26 mv x (D/r 02 ) Walsh, Lovelock, & Licence, Chem. Soc. Rev., 2010, 39, 4185

60 682 scanning is often steady-state at a UME steady-state occurs when v << RTD/(nFr 02 ) v (mv s -1 ) << 26 mv x (D/r 02 ) for a BASi UME with r 0 = 5 μm (1 x 10-5 cm 2 s -1 ) / (0.5 x 10-3 cm) 2 = 26 x 40 mv s -1 s Walsh, Lovelock, & Licence, Chem. Soc. Rev., 2010, 39, 4185

61 683 scanning is often steady-state at a UME steady-state occurs when v << RTD/(nFr 02 ) v (mv s -1 ) << 26 mv x (D/r 02 ) for a BASi UME with r 0 = 5 μm (1 x 10-5 cm 2 s -1 ) / (0.5 x 10-3 cm) 2 = 26 x 40 mv s -1 s v << 1 V s -1 Wow! Walsh, Lovelock, & Licence, Chem. Soc. Rev., 2010, 39, 4185

62 684 scanning is often steady-state at a UME steady-state occurs when v << RTD/(nFr 02 ) v (mv s -1 ) << 26 mv x (D/r 02 ) for a BASi UME with r 0 = 5 μm (1 x 10-5 cm 2 s -1 ) / (0.5 x 10-3 cm) 2 = 26 x 40 mv s -1 s v << 1 V s -1 Wow! Wightman, Anal. Chem., 1981, 53, 1125A

63 685 Ching, Dudek, & Tabet, J. Chem. Educ., 1994, 71, 602