Lecture #4 of 17 4/10/2019. Q: What s in this set of lectures? A: B&F Chapters 1, 15 & 4 main concepts:

Transcription

1 206 Lecure #4 of Q: Wha s in his se of lecures? A: B&F Chapers 1, 15 & 4 main conceps: Secion 1.1: Redox reacions Chaper 15: Elecrochemical insrumenaion Secion 1.2: Charging inerfaces Secion 1.3: Overview of elecrochemical experimens Secion 1.4: Mass ransfer and Semi-empirical reamen of elecrochemical observaions Chaper 4: Mass ransfer 208 Looking forward Secion 1.1 (and some of Chaper 15) 2-elecrode versus 3-elecrode measuremens Reference elecrodes Poeniosas Compliance volage/curren J E and I E curves Kineic overpoenial Faradaic reacions 1

2 FROM LAST TIME: now, as menioned earlier, unforunaely, real reference elecrodes can do none of hese hings perfecly 209 Specifically, we would really like o have a reference elecrode ha has he following aribues: 1. I has a well-defined and invarian poenial. Tha is, no maer how much curren we draw from his elecrode, is poenial mus no vary. 2. I has zero impedance. Tha is, i imposes no resisive load on our cell. 3. I does no conaminae our soluion. Tha is, i is no a source of undesired ions in our elecrochemical cell. so we resor o a 3-elecrode poeniosa 210 WE = working elecrode RE = reference elecrode CE = couner (or auxiliary) elecrode Ou of sigh, ou of mind is a bad moo! hp:// invened in 1937 by Hickling 211 Hickling, Trans. Faraday Soc., 1942, 38, 27 2

3 invened in 1937 by Hickling 212 Hickling, Trans. Faraday Soc., 1942, 38, 27 invened in 1937 by Hickling 213 his is a vacuum ube! Hickling, Trans. Faraday Soc., 1942, 38, 27 invened in 1937 by Hickling 214 in fac, i's his vacuum ube! hps:// Hickling, Trans. Faraday Soc., 1942, 38, 27 3

4 his is how (many famous elecrochemiss) do his oday 215 meaning wih somewha old insrumens like described in B&F an op-amp hp://en.wikipedia.org/wiki/operaional_amplifier hp:// his is how (many famous elecrochemiss) do his oday 216 meaning wih somewha old insrumens like described in B&F an op-amp 20 ransisors, 11 resisors, and 1 capacior; Wow! hp://en.wikipedia.org/wiki/operaional_amplifier hp:// his is how (many famous elecrochemiss) do his oday 217 meaning wih somewha old insrumens like described in B&F an op-amp 20 ransisors, 11 resisors, and 1 capacior; Wow! hp://en.wikipedia.org/wiki/operaional_amplifier hp:// fixed values (vs. ground) 4

5 his is how (many famous elecrochemiss) do his oday 218 meaning wih somewha old insrumens like described in B&F an op-amp Op-amp Golden Rules Ideal rules ha are close o correc in pracice. For an op-amp wih feedback (which we have), (1) The Volage Rule: The oupu (V OUT ) aemps o do whaever is necessary o make he poenial difference beween he inpus (IN+ and IN-) zero (because V and V + are fixed). (2) The Curren Rule: The inpus o IN+ and INdraw no ne curren. Thus, by Ohm s law, he impedance is essenially infinie! Horowiz and Hill, The Ar of Elecronics, Cambridge Universiy Press, 1980 hp://hyperphysics.phy-asr.gsu.edu/hbase/elecronic/opampi.hml#c2 his is how (many famous elecrochemiss) do his oday 219 meaning wih somewha old insrumens like described in B&F an op-amp Op-amp Golden Rules Ideal rules ha are close o correc in pracice. For an op-amp wih feedback (which we have), (1) The Volage Rule: The oupu (V OUT ) aemps o do whaever is necessary o make he poenial difference beween he inpus (IN+ and IN-) zero (because V and V + are fixed). (2) The Curren Rule: The inpus o IN+ and INdraw no ne curren. Thus, by Ohm s law, he impedance is essenially infinie! How does he oupu hen pass curren? hp:// brain-in-profile-clipar.hml hp://shujaabbas.hubpages.com/h ub/caroon-boxing-champion Horowiz and Hill, The Ar of Elecronics, Cambridge Universiy Press, 1980 hp://hyperphysics.phy-asr.gsu.edu/hbase/elecronic/opampi.hml#c2 his is how (many famous elecrochemiss) do his oday hree op-amps! 220 an op-amp vs. ground hp://elecronicdesign.com/conen/14978/59899_fig_02.jpg 5

6 his is how (many famous elecrochemiss) do his oday hree op-amps! 221 a curren-o-volage converer (curren follower) wih IN + held a ground, and hus IN a virual ground; I in (WE) = V ou /R, where R is ermed he feedback resisor vs. ground hp://elecronicdesign.com/conen/14978/59899_fig_02.jpg his is how (many famous elecrochemiss) do his oday hree op-amps! 222 a curren-o-volage converer (curren follower) wih IN + held a ground, and hus IN a virual ground; I in (WE) = V ou /R, where R is ermed he feedback resisor vs. ground hp://elecronicdesign.com/conen/14978/59899_fig_02.jpg his is how (many famous elecrochemiss) do his oday hree op-amps! 223 a curren-o-volage converer (curren follower) wih IN + held a ground, and hus IN a virual ground; I in (WE) = V ou /R, where R is ermed he feedback resisor a volage follower wih uniy gain due o he wo 10kΩ resisors, and so V in = V ou, and even if I ou is large, i does no draw from V in (sable), bu raher from he muscle (leads no shown) vs. ground hp://elecronicdesign.com/conen/14978/59899_fig_02.jpg 6

7 his is how (many famous elecrochemiss) do his oday V IN+ = V IN I IN+ = I IN 0 hree op-amps! V IN+ = V in(re) + I IN+R V IN = V ou + I IN R 224 Thus, V in(re) = V ou a curren-o-volage converer (curren follower) wih IN + held a ground, and hus IN a virual ground; I in (WE) = V ou /R, where R is ermed he feedback resisor a volage follower wih uniy gain due o he wo 10kΩ resisors, and so V in = V ou, and even if I ou is large, i does no draw from V in (sable), bu raher from he muscle (leads no shown) vs. ground hp://elecronicdesign.com/conen/14978/59899_fig_02.jpg his is how (many famous elecrochemiss) do his oday hree op-amps! 225 f a volage-o-curren amplifier (I ou = V in /R f ) supplies curren beween he CE (and WE via he ground) in order o mainain he difference in poenial beween he WE/ground and RE (E app = V in ); gain = R f /R a curren-o-volage converer (curren follower) wih IN + held a ground, and hus IN a virual ground; I in (WE) = V ou /R, where R is ermed he feedback resisor a volage follower wih uniy gain due o he wo 10kΩ resisors, and so V in = V ou, and even if I ou is large, i does no draw from V in (sable), bu raher from he muscle (leads no shown) vs. ground hp://elecronicdesign.com/conen/14978/59899_fig_02.jpg his is how (many famous elecrochemiss) do his oday I in = I ou hree op-amps! I in = V in/r V in/r = V ou/r f I ou = V ou/r f Thus, V ou = V in R f/r f 226 a volage-o-curren amplifier (I ou = V in /R f ) supplies curren beween he CE (and WE via he ground) in order o mainain he difference in poenial beween he WE/ground and RE (E app = V in ); gain = R f /R a curren-o-volage converer (curren follower) wih IN + held a ground, and hus IN a virual ground; I in (WE) = V ou /R, where R is ermed he feedback resisor a volage follower wih uniy gain due o he wo 10kΩ resisors, and so V in = V ou, and even if I ou is large, i does no draw from V in (sable), bu raher from he muscle (leads no shown) vs. ground hp://elecronicdesign.com/conen/14978/59899_fig_02.jpg 7

8 If we wan o be able o adjus he volage on he WE, we inroduce E app here 227 E app f a volage-o-curren amplifier (I ou = V in /R f ) supplies curren beween he CE (and WE via he ground) in order o mainain he difference in poenial beween he WE/ground and RE (E app = V in ); gain = R f /R a curren-o-volage converer (curren follower) wih IN + held a ground, and hus IN a virual ground; I in (WE) = V ou /R, where R is ermed he feedback resisor a volage follower wih uniy gain due o he wo 10kΩ resisors, and so V in = V ou, and even if I ou is large, i does no draw from V in (sable), bu raher from he muscle (leads no shown) vs. ground hp://elecronicdesign.com/conen/14978/59899_fig_02.jpg 228 Noe especially he following for older (simpler) poeniosas: 1. The working elecrode (WE) is a (virual) ground and has a very low impedance, Z = R + ix. You canno ge an elecrical shock a his elecrode or a his inpu o he poeniosa. 2. Amplifier U3 akes he curren a he WE and convers i ino a poenial so i can be recorded. V = IR a he oupu of U3. 3. The reference elecrode (RE), conneced o he non-invering inpu (+) of he op-amp U2, is asked o source a minue amoun of curren (~3 fa for his paricular op-amp; 0 fa is he ideal case). 4. Op-amp U1 produces, a he couner elecrode (CE), an oupu curren, Iou, ha is proporional o he poenial difference beween RE and WE (i.e. ground). Cauion: You CAN ge a lehal shock a his elecrode. However, his power is no infinie (your wall sockes have a limied power hey can supply). The poeniosa limis are ermed he compliance volage and compliance curren Beware of compliance volage issues (maximum volage o CE) and compliance curren oo! x x mM ferrocene, 0.1M [NE 4 ][BF 4 ], MeCN flow rae = 1ml/hr 4.0x x10-5 I /A x x x x mv/s E /V hp:// 8

9 Acive I/E Converer versus Passive I/E Converer meaning older 230 hp:// Acive I/E Converer versus Passive I/E Converer meaning newer 231 Many modern poeniosas use he archiecure shown here, ye i is rarely discussed a lengh in exbooks! This scheme has been used by Gamry, PAR, Solarron, and perhaps ohers [and likely Bio-Logic]. Unlike he acive I/E converer design, his I/E converer is passive curren only flows hrough passive circui elemens (e.g. R, C, no op-amp) he working elecrode is NOT a (virual) (earh) ground he elecromeer is differenial beween he RE and he WE sense (RE #2) hp:// Acive I/E Converer versus Passive I/E Converer meaning newer 232 hp:// 9

10 Poeniosa summary for non-ee majors * The poeniosa does no conrol he poenial of he working elecrode! * The poeniosa conrols he poenial of he couner elecrode only (relaive o he working elecrode) * The couner elecrode is he mos imporan elecrode, followed by he reference elecrode * Compliance volage and compliance curren limis are very imporan in he choice of he poeniosa / applicaion * Wih a few componens you can build your own poeniosa for < $100! * Passive poeniosas do no hold he WE a earh ground, bu can measure poenials across elecrolye inerfaces 233 Rowe,..., Plaxco, PLoS One, 2011, 6, e23783 Mo,..., Sykes, J. Chem. Educ., 2014, 91, 1028 and ha is why we use a 3-elecrode poeniosa 234 WE = working elecrode CE = couner (or auxiliary) elecrode RE = reference elecrode Ou of sigh, ou of mind is a bad moo! hp:// 235 An example of wo RE scales You re welcome! B&F back inside 10

11 ... okay, now le s measure he curren ha flows as we change he poenial of he plainum working elecrode (imagine ha we have a poeniosa here ha allows us o do ha). 236 Le s suppose ha his WE is plainum, and ha all hree elecrodes are immersed in 1.0 M HCl hp:// okay, now le s measure he curren ha flows as we change he poenial of 237 he plainum working elecrode (imagine ha we have a poeniosa here ha allows us o do ha). Wha is he saring poenial for his experimen? Tha is, wha is he open-circui poenial? an oxidaion: 2Cl (aq) Cl 2 (g) + 2e in 1.0 M HCl SHE a reducion: 2H + (aq) + 2e H 2 (g)... okay, now le s measure he curren ha flows as we change he poenial of 238 he plainum working elecrode (imagine ha we have a poeniosa here ha allows us o do ha). Wha is he saring poenial for his experimen? Tha is, wha is he open-circui poenial (E oc )? an oxidaion: 2Cl (aq) Cl 2 (g) + 2e a reducion: 2H + (aq) + 2e H 2 (g) in 1.0 M HCl Somewhere in here bu no sure where i is no clear which halfreacion is dominan SHE alhough my guess i ha E oc is se by impuriies like Fe(III) and Fe(II) 11

12 ... okay, now le s measure he curren ha flows as we change he poenial of he plainum working elecrode (imagine ha we have a poeniosa here ha allows us o do ha). 239 i Curren flow is proporional o rae, so le s wrie his like chemiss: an oxidaion: 2Cl (aq) Cl 2 (g) + 2e Rae mol s = dn d = i nf bu for elecrochemiss, his is less useful because i depends on he experimenal se-up (i.e. elecrode area)! SHE a reducion: 2H + (aq) + 2e H 2 (g)... okay, now le s measure he curren ha flows as we change he poenial of he plainum working elecrode (imagine ha we have a poeniosa here ha allows us o do ha). 240 i Curren flow is proporional o rae, so le s wrie his like chemiss: an oxidaion: 2Cl (aq) Cl 2 (g) + 2e Rae mol s = dn d = i nf bu for elecrochemiss, his is less useful because i depends on he experimenal se-up (i.e. elecrode area)! SHE a reducion: 2H + (aq) + 2e H 2 (g) Rae mol s cm 2 = i nfa = j nf his is much beer! Can we define each of hese variables by name and uni?... okay, now le s measure he curren ha flows as we change he poenial of he plainum working elecrode (imagine ha we have a poeniosa here ha allows us o do ha). i an oxidaion: 2Cl (aq) Cl 2 (g) + 2e 241 Given ha he reacions shown here occur in a single cell (no sal bridge), when his E is applied o he WE and curren flows a his I, wha curren flows a a large P CE and where on his plo would you pu a poin o show is response? SHE a reducion: 2H + (aq) + 2e H 2 (g) 12

13 ... okay, now le s measure he curren ha flows as we change he poenial of he plainum working elecrode (imagine ha we have a poeniosa here ha allows us o do ha). an oxidaion: 2Cl (aq) Cl 2 (g) + 2e 242 Given ha he reacions shown here occur in a single cell (no sal bridge), when his E is applied o he WE and curren flows a his I, wha curren flows a a large P CE and where on his plo would you pu a poin o show is response? SHE Pssss! Don forge me! I m sill here! a reducion: 2H + (aq) + 2e H 2 (g)... okay, now le s measure he curren ha flows as we change he poenial of he plainum working elecrode (imagine ha we have a poeniosa here ha allows us o do ha). an oxidaion: 2Cl (aq) Cl 2 (g) + 2e 243 Given ha he reacions shown here occur in a single cell (no sal bridge), when his E is applied o he WE and curren flows a his I, wha curren flows a a large P CE and where on his plo would you pu a poin o show is response? SHE a reducion: 2H + (aq) + 2e H 2 (g) Curren maching equal and opposie he curren a he WE is he same as ha a he CE Kirchhoff s Curren Law... okay, now le s measure he curren ha flows as we change he poenial of he plainum working elecrode (imagine ha we have a poeniosa here ha allows us o do ha). an oxidaion: 2Cl (aq) Cl 2 (g) + 2e 244 Given ha he reacions shown here occur in a single cell (no sal bridge), when his E is applied o he WE and curren flows a his I, wha curren flows a a large P CE and where on his plo would you pu a poin o show is response? SHE a reducion: 2H + (aq) + 2e H 2 (g) Curren maching equal and opposie he curren a he WE is he same as ha a he CE Kirchhoff s Curren Law okay, bu wha if proons could no be reduced so easily? An overpoenial is required! 13

14 Now, preend his experimenal I E curve (from my labs; in fac) was measured when he P WE was swiched wih a Hg WE why does lile curren flow unil ~ -1 V? 245 vs. SHE Now, preend his experimenal I E curve was measured when he P WE was swiched wih a Hg WE why does lile curren flow unil ~ -1 V? Overpoenial! which is presen due o kineic/rae/curren limiaions 246 η = E app E Eq vs. SHE Now, preend his experimenal I E curve was measured when he P WE was swiched wih a Hg WE why does lile curren flow unil ~ -1 V? Overpoenial! which is presen due o kineic/rae/curren limiaions 247 η = E app E Eq Wha if you dump in Cd 2+, whose E 0 (Cd 2+ /Cd 0 ) -0.4 V vs. SHE? vs. SHE 14

15 Now, preend his experimenal I E curve was measured when he P WE was swiched wih a Hg WE why does lile curren flow unil ~ -1 V? Overpoenial! which is presen due o kineic/rae/curren limiaions 248 η = E app E Eq Wha if you dump in Cd 2+, whose E 0 (Cd 2+ /Cd 0 ) -0.4 V vs. SHE? Curren response will be kineically deermined and curren will sar o pass a abou -0.4 V vs. SHE vs. SHE Now, preend his experimenal I E curve was measured when he P WE was swiched wih a Hg WE why does lile curren flow unil ~ -1 V? Overpoenial! which is presen due o kineic/rae/curren limiaions 249 η = E app E Eq Wha if you dump in Cd 2+, whose E 0 (Cd 2+ /Cd 0 ) -0.4 V vs. SHE? Curren response will be kineically deermined and curren will sar o pass a abou -0.4 V vs. SHE vs. SHE Also, don forge abou possible compliance volage issues (maximum volage o CE) and compliance curren oo! under condiions of seady-sae curren flow, we are concerned wih mached (equal and opposie) currens a he WE and CE 250 WE vs. RE CE vs. RE vs. SHE 15

16 more erminology 251 supporing elecrolye an iner sal added o impar ionic conduciviy o he soluion (e.g. 1 M HCl, in his case) background limis he wo poenial limis a which he pure solven + supporing elecrolye begin o reac a he working elecrode 252 Elecrochemical window (Poenial range) * Red arrow enries are all measured in 1 M acid (1) If you waned an aqueous baery wih a large volage, which elecrode is bes? (2) Beween aqueous and non-aqueous baeries, which can generae he larges poenial? B&F back inside cover 253 Elecrochemical window (Poenial range) * Red arrow enries are all measured in 1 M acid (1) If you waned an aqueous baery wih a large volage, which elecrode is bes? Hg (in 0.1 M E 4 NOH) or C (in 0.1 M KCl) (2) Beween aqueous and non-aqueous baeries, which can generae he larges poenial? Non-aqueous!... much larger solven window B&F back inside cover 16

17 more erminology supporing elecrolye an iner sal added o impar ionic conduciviy o he soluion (e.g. 1 M HCl, in his case) 254 background limis he wo poenial limis a which he pure solven + supporing elecrolye begin o reac a he working elecrode polarizable elecrode an elecrode operaed wihin a poenial range in which no Faradaic elecrochemisry occurs Scienis Faraday s law The amoun of chemical reacion caused by he flow of curren is proporional o he amoun of elecriciy passed. (B&F) an oxidaion: 2Cl (aq) Cl 2 (g) + 2e solven window SHE Michael Faraday ( ) from Wiki a reducion: 2H + (aq) + 2e H 2 (g) more erminology supporing elecrolye an iner sal added o impar ionic conduciviy o he soluion (e.g. 1 M HCl, in his case) 255 background limis he wo poenial limis a which he pure solven + supporing elecrolye begin o reac a he working elecrode polarizable elecrode an elecrode operaed wihin a poenial range in which no Faradaic elecrochemisry occurs Scienis Faraday s law The amoun of chemical reacion caused by he flow of curren is proporional o he amoun of elecriciy passed. (B&F) an oxidaion: 2Cl (aq) Cl 2 (g) + 2e solven window SHE Michael Faraday ( ) from Wiki Typically, chemical reacion is measured by mass (g) and elecriciy passed is measured by charge (C) don forge z in he mah! a reducion: 2H + (aq) + 2e H 2 (g) and, even more erminology Faradaic elecrochemisry elecrochemisry characerized by he flow of curren o/from elecron donor/accepor species presen a he elecrode surface 256 Non-Faradaic elecrochemisry elecrochemisry characerized by he flow of curren o/from an elecrode surface in he absence of donor/accepor species, ypically dominaed by capaciive charging of he elecrode or adsorpion/desorpion phenomena Scienis Faraday s law The amoun of chemical reacion caused by he flow of curren is proporional o he amoun of elecriciy passed. (B&F) an oxidaion: 2Cl (aq) Cl 2 (g) + 2e solven window SHE Michael Faraday ( ) from Wiki Typically, chemical reacion is measured by mass (g) and elecriciy passed is measured by charge (C) don forge z in he mah! a reducion: 2H + (aq) + 2e H 2 (g) 17

18 and lasly, ypical WE ranges for EChem experimens/echnologies 257 J 3E-y 2E-y 1E-y -2.0E-x -1.5E-x -1.0E-x -0.5E-x -1E-y -2E-y 0.5E-x 1.0E-x 1.5E-x 2.0E-x E, vs. SHE -3E-y Cyclic volammogram: x = 1, y = 4 5 ΔE = 500 mv J = ±100 μa/cm 2 Nanopore: x = -1, y = 9 E = ±10 V J = ±1 na/cm 2 Phooelecrochemisry: x = 0 1, y = 2 E = E oc = ±700 mv J = J sc = ±30 ma/cm 2 Fuel Cell / Baery: x = 0, y = 0 E = 1 3 V J = 1 2 A/cm A review of Secion 1.1 (and some of Chaper 15) 2-elecrode versus 3-elecrode measuremens Reference elecrodes Poeniosas Compliance volage/curren J E and I E curves Kineic overpoenial Faradaic reacions 259 Q: Wha s in his se of lecures? A: B&F Chapers 1, 15 & 4 main conceps: Secion 1.1: Redox reacions Chaper 15: Elecrochemical insrumenaion Secion 1.2: Charging inerfaces Secion 1.3: Overview of elecrochemical experimens Secion 1.4: Mass ransfer and Semi-empirical reamen of elecrochemical observaions Chaper 4: Mass ransfer 18

19 260 Looking forward Secions 1.2 and 1.3 RC circuis (~90% of slides) Elecrochemically acive surface area Uncompensaed resisance Placemen of elecrodes, and oher properies 261 our P WE is polarizable wihin his poenial range SHE a polarizable elecrode ha has been polarized imposes a background elecric response an added curren in a volammeric experimen, for example (no observable on his scale as i is small) ha is ransien (e.g. a blip) and is observed in all elecrochemical experimens We need o undersand his background curren! The elecrical response of a polarizable elecrode is approximaed by a series resisor and capacior (a series RC circui) Power E Supply (T)oal 262 resisor = soluion R capacior = WE inerface C E T = E R + E cap + 19

20 The elecrical response of a polarizable elecrode is approximaed by a series resisor and capacior (a series RC circui) Power Supply R = he soluion resisance (beween he WE and RE) 263 resisor = soluion capacior = WE inerface R C The elecrical response of a polarizable elecrode is approximaed by a series resisor and capacior (a series RC circui) Power Supply R = he soluion resisance (beween he WE and RE) 264 C = he ne capaciance (of he WE and he CE), C(T)oal 1 = C T C 1 C 2 resisor = soluion capacior = WE inerface R C o measure C 2(WE), make C 1(CE) large or use a hree-elecrode seup and a psa Firs, wha are approximae values for R (S) and C (d)? 265 Power Supply R = he soluion resisance (beween he WE and RE) In aqueous soluions conaining 0.1 M supporing elecrolye, R = a few ohms; for non-aq., R > 100 Ω resisor = soluion capacior = WE inerface C = he ne capaciance (of he WE and he RE), C(T)oal ~20 µf/cm 2 of elecrode area for gold or plainum; 2 5 µf/cm 2 for carbon, ypically bu hese change slighly wih poenial as we will see laer (Farad is C/V, where V is J/C) R C 20

21 266 i = E R exp RC Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure i = E R exp RC Could here be a problem wih an insananeous 6 V poenial sep, for example? Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure i = E R exp RC Could here be a problem wih an insananeous 6 V poenial sep, for example? Compliance curren! (a = 0, E = ir (Ohm s law)) Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure

22 269 i = E R exp RC Wha porion of E app is acually fel by he WE a = 0? Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure i = E R exp RC Wha porion of E app is acually fel by he WE a = 0? Lile of i! Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure NOTE: Elecronics can limi he observaion of rapid chemical kineics (i.e. he rae-deermining sep is charging and no elecron ransfer) E E WE vs. RE Wha porion of E app is acually fel by he WE a = 0? Lile of i! Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure

23 272 i = E R exp RC bu where did his equaion for curren come from? who s comforable wih me jus giving you his equaion? Le s manipulae unis! Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure B&F eqn. (1.2.8) E = E R + E C = ir s + q C d R C E T = E R + E cap B&F eqn. (1.2.8) E = E R + E C = ir s + q C d Unis: C Unis: C/V R C 23

24 275 B&F eqn. (1.2.8) E = E R + E C = ir s + q C d B&F eqn. (1.2.9) i = 1 E q = dq d Unis: C Unis: C/V R C Unis: C/s 276 B&F eqn. (1.2.8) E = E R + E C = ir s + q C d B&F eqn. (1.2.9) i = 1 E q = dq d Unis: C Unis: C/V R C Unis: C/s Need o inegrae! 277 B&F eqn. (1.2.8) E = E R + E C = ir s + q C d B&F eqn. (1.2.9) i = 1 E q = dq d Unis: C/s Need o inegrae! 1 d = 1 R s E q dq = C d C d dq EC d +q Unis: C Unis: C/V R C 24

25 278 B&F eqn. (1.2.8) E = E R + E C = ir s + q C d Unis: C B&F eqn. (1.2.9) i = 1 E q = dq Unis: C/V R C d Unis: C/s Need o inegrae! 1 d = 1 R s E q dq = C d C d dq EC d +q 1 = ln EC d + q ln EC d (assuming ha a = 0, q = 0) = ln EC d + q EC d Inegraed! = ln EC d + q ln EC d = ln EC d + q EC d Inegraed! = ln EC d + q ln EC d EC d e = ECd + q = ln EC d + q EC d Inegraed! 25

26 281 1 = ln EC d + q ln EC d EC d e = ECd + q = ln EC d + q EC d Inegraed! q = EC d 1 e q B&F eqn. (1.2.10) = ln EC d + q ln EC d EC d e = ECd + q = ln EC d + q EC d Inegraed! q = EC d 1 e B&F eqn. (1.2.10) Need o differeniae! dq d = EC d 1 e = ln EC d + q ln EC d EC d e = ECd + q = ln EC d + q EC d Inegraed! q = EC d 1 e Need o differeniae! dq d = EC d 1 e = E e = I R s Done! B&F eqn. (1.2.10) B&F eqn. (1.2.6) 26

27 284 i = E R exp RC Wha are he unis of RC? R (Ω) x Cap (F) R (V / C/s) x Cap (C/V) R Cap (V-s/C x C/V) R Cap (s)! Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure i = E R exp RC Wha are he unis of RC? R (Ω) x Cap (F) R (V / C/s) x Cap (C/V) R Cap (V-s/C x C/V) R Cap (s)! Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure i = E R exp RC Wha are he unis of RC? R (Ω) x Cap (F) R (V / C/s) x Cap (C/V) R Cap (V-s/C x C/V) R Cap (s)! Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure

28 287 Why is 37% of he iniial signal noeworhy? i = E R exp RC Wha are he unis of RC? R (Ω) x Cap (F) R (V / C/s) x Cap (C/V) R Cap (V-s/C x C/V) R Cap (s)! Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure Why is 37% of he iniial signal noeworhy? Plug in = RC! Ah ha! i = E R exp RC Wha are he unis of RC? R (Ω) x Cap (F) R (V / C/s) x Cap (C/V) R Cap (V-s/C x C/V) R Cap (s)! Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure Example: Consider he case where R = 1 Ω and C = 20 µf/cm 2. How long will i ake o charge C o 95% of is maximum capaciy? 28

29 290 Example: Consider he case where R = 1 Ω and C = 20 µf/cm 2. How long will i ake o charge C o 95% of is maximum capaciy? q = EC d 1 e B&F eqn. (1.2.10) q 291 Example: Consider he case where R = 1 Ω and C = 20 µf/cm 2. How long will i ake o charge C o 95% of is maximum capaciy? q = EC d 1 e q = EC d B&F eqn. (1.2.10) q Example: Consider he case where R = 1 Ω and C = 20 µf/cm 2. How long will i ake o charge C o 95% of is maximum capaciy? 292 q = EC d 1 e q = EC d B&F eqn. (1.2.10) q q = 1 e q 29

30 Example: Consider he case where R = 1 Ω and C = 20 µf/cm 2. How long will i ake o charge C o 95% of is maximum capaciy? 293 q = EC d 1 e q = EC d B&F eqn. (1.2.10) q q = 1 e q 0.95 = 1 e = e 0.95 Example: Consider he case where R = 1 Ω and C = 20 µf/cm 2. How long will i ake o charge C o 95% of is maximum capaciy? 294 q = EC d 1 e q = EC d B&F eqn. (1.2.10) q q = 1 e q 0.95 = 1 e = e 0.95 ln 0.05 = 0.95 Example: Consider he case where R = 1 Ω and C = 20 µf/cm 2. How long will i ake o charge C o 95% of is maximum capaciy? 295 q = EC d 1 e q = EC d B&F eqn. (1.2.10) q q = 1 e q 0.95 = 1 e = e 0.95 assuming 1 cm 2 ln 0.05 = = 1Ω 20μF ln = 60μs 30

31 Example: Consider he case where R = 1 Ω and C = 20 µf/cm 2. How long will i ake o charge C o 95% of is maximum capaciy? 296 q = EC d 1 e q = EC d q = 1 e q 0.95 = 1 e = e 0.95 ln 0.05 = 0.95 B&F eqn. (1.2.10) assuming 1 cm = 1Ω = 60μs As above, assuming a 6 V poenial sep, now wha is he average curren ha flows up o 0.95? 20μF ln Example: Consider he case where R = 1 Ω and C = 20 µf/cm 2. How long will i ake o charge C o 95% of is maximum capaciy? 297 q = EC d 1 e q = EC d q = 1 e q 0.95 = 1 e = e 0.95 ln 0.05 = 0.95 B&F eqn. (1.2.10) assuming 1 cm = 1Ω = 60μs As above, assuming a 6 V poenial sep, now wha is he average curren ha flows up o 0.95? I avg = C / s = (C/V x V) / s 20μF ln Example: Consider he case where R = 1 Ω and C = 20 µf/cm 2. How long will i ake o charge C o 95% of is maximum capaciy? 298 q = EC d 1 e q = EC d q = 1 e q 0.95 = 1 e = e 0.95 ln 0.05 = 0.95 B&F eqn. (1.2.10) assuming 1 cm = 1Ω = 60μs As above, assuming a 6 V poenial sep, now wha is he average curren ha flows up o 0.95? I avg = C / s = (C/V x V) / s = 20 μf x 6 V / 60 μs 20μF ln

32 Example: Consider he case where R = 1 Ω and C = 20 µf/cm 2. How long will i ake o charge C o 95% of is maximum capaciy? 299 q = EC d 1 e q = EC d q = 1 e q 0.95 = 1 e = e 0.95 ln 0.05 = 0.95 B&F eqn. (1.2.10) assuming 1 cm = 1Ω = 60μs As above, assuming a 6 V poenial sep, now wha is he average curren ha flows up o 0.95? I avg = C / s = (C/V x V) / s = 20 μf x 6 V / 60 μs = 120 μc / 60 μs 20μF ln Example: Consider he case where R = 1 Ω and C = 20 µf/cm 2. How long will i ake o charge C o 95% of is maximum capaciy? 300 q = EC d 1 e q = EC d q = 1 e q 0.95 = 1 e = e 0.95 ln 0.05 = 0.95 B&F eqn. (1.2.10) assuming 1 cm = 1Ω = 60μs As above, assuming a 6 V poenial sep, now wha is he average curren ha flows up o 0.95? I avg = C / s = (C/V x V) / s = 20 μf x 6 V / 60 μs = 120 μc / 60 μs = 2 A! Compliance? 20μF ln Curren-sep galvanosaic chronopoeniomery Curren sep (ha is, incremen he curren by an amoun, i): E = i R + C B&F eqn. (1.2.12) Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure

33 Curren-sep galvanosaic chronopoeniomery Curren sep (ha is, incremen he curren by an amoun, i): E = i R + C B&F eqn. (1.2.12) So, a consan applied curren resuls in a linear sweep of he poenial hus, wha if we insead applied he poenial sweep? Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure Linear-sweep volammery cyclic volammery Poenial scan (ha is, ramp he applied poenial, E() = v for one direcion): scan rae i = vc d 1 exp R S C d B&F eqn. (1.2.15) Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure Linear-sweep volammery cyclic volammery Poenial scan (ha is, ramp he applied poenial, E() = v for one direcion): scan rae i = vc d 1 exp R S C d B&F eqn. (1.2.15) ASIDE: Recall, for a poenial sep, he same shape bu for charge (q) q q = EC d 1 e B&F eqn. (1.2.10) Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure

34 Linear-sweep volammery cyclic volammery Poenial scan (ha is, ramp he applied poenial, E() = v for one direcion): scan rae i = vc d 1 exp R S C d B&F eqn. (1.2.15) So he oal curren envelope a any poenial ha is well-removed from he swiching poenial will be: i = 2Cv, wih v s unis being V/s i = 2Cv Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure Linear-sweep volammery cyclic volammery This is an example of a cyclic volammogram wih obvious RC charging hp:// /hesuieworld.com*wp-includes*heme-compa*dallas-exas-scenery-5417.jpg/ Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure i = 2Cv 34