Transcript Slide 1
Hydrodynamic Methods Lecture 17.1 Rotated Disk Electrode Voltammetry RDEV r1 2f in s-1, so f in rps revolutions per second “dead” or diffusion layer Laminar flow occurs up to a point, at too high , we find that turbulent flow occurs. This is when the value r1 2 exceeds the Reynold’s number for that particular fluid with a given kinematic viscosity, n, in cm2 s-1. n g cm1s 1 (Poise) v v d g cm3 vH 2O is ~ 0.01cm2 s 1 at 20 C (*what is vCH 3CN at 20 C ??*) Look in Table 9.2.1 So, should be ~ 2 ×105 n/r2, but other limitations actually mean < 1000 s-1 or f 10,000 rpm. On the low side, must rotate fast enough to establish constant, homogeneous supply of material to electrode surface. > 10 s-1 Hydrodynamic Methods Lecture 17.2 If one applies a potential which is that needed to obtain mass-transport limited conditions, then what is i ? consider: - hydrodynamics - diffusion Why? Co C Real profile 0 x d Recall that d is F(1/). So, C( o ,t ) i nFADo x i nFAmo Co ; mo Do d How solve? As we did before except incorporate hydrodynamics. Also: Co ( x,0) Co Co (, t ) Co Co (0, t ) F ( E ) Two Cases: 1. Reversible use q expression (Nernst) 2. Before reach MT limit and - irrev. - quasi-rev. ET rxns. Hydrodynamic Methods Lecture 17.3 no iDL effects. Case 1: Levich Equation ilim 0.62nFADo2 / 3n 1/ 61/ 2Co Know: d Know: 1.6 D1/ 3n 1/ 6 1/ 2 ; Levich Layer Levich plot If reaction is D–C, then ilim vs. 1/2 is linear with zero intercept. Also if ET reversible: 0.059 DR o EE log n Do 2/3 il ,c i 0.059 log n i E1/2 Then plot of Eapp vs. log Case 2A: il i 0.059 V will be straight with slope i n Totally irreversible; O only i nFA kf Co kbCR Zero But, kf is F(E) , so we denote this kf(E). We call this current iK and it is: This is the Kinetic current. iK nFAkf E Co na F E E o kf k exp RT So, at high enough -h, we should get kf ?? NO. Hydrodynamic Methods Lecture 17.4 We have no ET effects at -h, Irrev. Rev. for i, so we merely get ilim B-V / No MT ic MT effects + - E vs Ref ET effects So, if we could vary E and measure iK, we could get k ?? ia Yes! How? 1 1 1 ; iK nFAkf E Co i iK ilim Turns out we have: i F 1/ 2 Koutecký – Levich or Inverse Levich plot 1 1/ 2 vs. at a given Eapp. i So make plot of E1 1 i Slope is turbulence 1 0.62nFADo2 / 3v 1/ 6Co intercept is 1 iK 1 i i Same slopes In each. vary Eapp -1/2 iK E1 E2 E3 E4(on i lim) -1/2 Levich line (for E>E4) 1/2 E1>E2>E3>E4 More - Hydrodynamic Methods Lecture 17.5 1 at Eapp . We know So, we get intercept nFAk ( E )C f o n, F , A, Co . Get k f at Ei . P lot ln k f vs.h or ( E E o ) and get interceptof ln k o . Also, slope is na F RT Why? k f ( E ) k o exp na Fh / RT Case 2B: Quasi – Reversible for O and R MT? 1 b 1 b b1 i 1 b i o il , c il , a Fnc(E) ET b k exp nF E E o / RT Now we have both kf and kb a function of n. Thus, the Koutecký – Levich plots do not have same slope for various Potentials (h). Problems! Minimize errors by using small potential range near the foot of the wave where i is not changing so drastically. Hydrodynamic Methods Lecture 17.6 RRDE dR r1 (R) Ring r3 (D) Disk dD r2 dR = dD r1 disk radius r2 – r1 gap r3 – r2 width of ring The collection Efficiency, N, is defined as iR N iD It is a Function of electrode geometry but is independent of , Co , Do , DR , etc. if R is stable. kchem If R Z occurs, then Nexptl < Ntheo and N = F(). Hydrodynamic Methods Lecture 17.7 For RRDE Collection Experiments: 1. ERing is held positive enough so as to oxidize any R. 2. No bulk R, CR 0 3. EDisk is scanned. 4. iDisk is measured. 5. iRing is measured. O + ne iD,c ERing R iD,lim + - EDisk vs. Ref iR,lim R O + ne iR,a N iR F ( ) if R is st able and reactionis D C. iD For D C , iD 1/ 2 and iR 1/ 2 Review: i st able R : DPSCA r 0.293; CV iF i p ,a i p ,c RRDE N F ( ) 1