206 Lecture #4 of 17

Transcription

1 Lecture #4 of

2 207 Q: What s in this set of lectures? A: B&F Chapters 1, 15 & 4 main concepts: Section 1.1: Redox reactions Chapter 15: Electrochemical instrumentation Section 1.2: Charging interfaces Section 1.3: Overview of electrochemical experiments Section 1.4: Mass transfer and Semi-empirical treatment of electrochemical observations Chapter 4: Mass transfer

3 208 Looking forward Section 1.1 (and some of Chapter 15) 2-electrode versus 3-electrode measurements Reference electrodes Potentiostats Compliance voltage/current J E and I E curves Kinetic overpotential Faradaic reactions

4 FROM LAST TIME: now, as mentioned earlier, unfortunately, real reference electrodes can do none of these things perfectly 209 Specifically, we would really like to have a reference electrode that has the following attributes: 1. It has a well-defined and invariant potential. That is, no matter how much current we draw from this electrode, its potential must not vary. 2. It has zero impedance. That is, it imposes no resistive load on our cell. 3. It does not contaminate our solution. That is, it is not a source of undesired ions in our electrochemical cell.

5 so we resort to a 3-electrode potentiostat 210 WE = working electrode CE = counter (or auxiliary) electrode RE = reference electrode Out of sight, out of mind is a bad motto!

6 invented in 1937 by Hickling 211 Hickling, Trans. Faraday Soc., 1942, 38, 27

7 invented in 1937 by Hickling 212 Hickling, Trans. Faraday Soc., 1942, 38, 27

8 invented in 1937 by Hickling 213 this is a vacuum tube! Hickling, Trans. Faraday Soc., 1942, 38, 27

9 invented in 1937 by Hickling 214 in fact, it's this vacuum tube! Hickling, Trans. Faraday Soc., 1942, 38, 27

10 this is how (many famous electrochemists) do this today meaning with somewhat old instruments like described in B&F 215 an op-amp

11 this is how (many famous electrochemists) do this today meaning with somewhat old instruments like described in B&F 216 an op-amp 20 transistors, 11 resistors, and 1 capacitor; Wow!

12 this is how (many famous electrochemists) do this today meaning with somewhat old instruments like described in B&F 217 an op-amp 20 transistors, 11 resistors, and 1 capacitor; Wow! fixed values (vs. ground)

13 this is how (many famous electrochemists) do this today meaning with somewhat old instruments like described in B&F Op-amp Golden Rules Ideal rules that are close to correct in practice. For an op-amp with feedback (which we have), (1) The Voltage Rule: The output (V OUT ) attempts to do whatever is necessary to make the potential difference between the inputs (IN+ and IN-) zero (because V and V + are fixed). (2) The Current Rule: The inputs to IN+ and INdraw no net current. Thus, by Ohm s law, the impedance is essentially infinite! 218 an op-amp Horowitz and Hill, The Art of Electronics, Cambridge University Press,

14 this is how (many famous electrochemists) do this today meaning with somewhat old instruments like described in B&F Op-amp Golden Rules Ideal rules that are close to correct in practice. For an op-amp with feedback (which we have), (1) The Voltage Rule: The output (V OUT ) attempts to do whatever is necessary to make the potential difference between the inputs (IN+ and IN-) zero (because V and V + are fixed). 219 an op-amp (2) The Current Rule: The inputs to IN+ and INdraw no net current. Thus, by Ohm s law, the impedance is essentially infinite! How does the output then pass current? brain-in-profile-clipart.html ub/cartoon-boxing-champion Horowitz and Hill, The Art of Electronics, Cambridge University Press,

15 this is how (many famous electrochemists) do this today three op-amps! 220 an op-amp vs. ground

16 this is how (many famous electrochemists) do this today three op-amps! 221 a current-to-voltage converter (current follower) with IN + held at ground, and thus IN at virtual ground; I in (WE) = V out /R, where R is termed the feedback resistor vs. ground

17 this is how (many famous electrochemists) do this today three op-amps! 222 a current-to-voltage converter (current follower) with IN + held at ground, and thus IN at virtual ground; I in (WE) = V out /R, where R is termed the feedback resistor vs. ground

18 this is how (many famous electrochemists) do this today three op-amps! 223 a current-to-voltage converter (current follower) with IN + held at ground, and thus IN at virtual ground; I in (WE) = V out /R, where R is termed the feedback resistor a voltage follower with unity gain due to the two 10kΩ resistors, and so V in = V out, and even if I out is large, it does not draw from V in (stable), but rather from the muscle (leads not shown) vs. ground

19 this is how (many famous electrochemists) do this today V IN+ = V IN I IN+ = I IN 0 three op-amps! V IN+ = V in (RE) + I IN+ R V IN = V out + I IN R 224 Thus, V in (RE) = V out a current-to-voltage converter (current follower) with IN + held at ground, and thus IN at virtual ground; I in (WE) = V out /R, where R is termed the feedback resistor a voltage follower with unity gain due to the two 10kΩ resistors, and so V in = V out, and even if I out is large, it does not draw from V in (stable), but rather from the muscle (leads not shown) vs. ground

20 this is how (many famous electrochemists) do this today three op-amps! 225 f a voltage-to-current amplifier (I out = V in /R f ) supplies current between the CE (and WE via the ground) in order to maintain the difference in potential between the WE/ground and RE (E app = V in ); gain = R f /R a current-to-voltage converter (current follower) with IN + held at ground, and thus IN at virtual ground; I in (WE) = V out /R, where R is termed the feedback resistor a voltage follower with unity gain due to the two 10kΩ resistors, and so V in = V out, and even if I out is large, it does not draw from V in (stable), but rather from the muscle (leads not shown) vs. ground

21 this is how (many famous electrochemists) do this today I in = I out three op-amps! I in = V in /R V in /R = V out /R f I out = V out/r f Thus, V out = V in R f /R f 226 a voltage-to-current amplifier (I out = V in /R f ) supplies current between the CE (and WE via the ground) in order to maintain the difference in potential between the WE/ground and RE (E app = V in ); gain = R f /R a current-to-voltage converter (current follower) with IN + held at ground, and thus IN at virtual ground; I in (WE) = V out /R, where R is termed the feedback resistor a voltage follower with unity gain due to the two 10kΩ resistors, and so V in = V out, and even if I out is large, it does not draw from V in (stable), but rather from the muscle (leads not shown) vs. ground

22 If we want to be able to adjust the voltage on the WE, we introduce E app here 227 E app f a voltage-to-current amplifier (I out = V in /R f ) supplies current between the CE (and WE via the ground) in order to maintain the difference in potential between the WE/ground and RE (E app = V in ); gain = R f /R a current-to-voltage converter (current follower) with IN + held at ground, and thus IN at virtual ground; I in (WE) = V out /R, where R is termed the feedback resistor a voltage follower with unity gain due to the two 10kΩ resistors, and so V in = V out, and even if I out is large, it does not draw from V in (stable), but rather from the muscle (leads not shown) vs. ground

23 Note especially the following for older (simpler) potentiostats: The working electrode (WE) is at (virtual) ground and has a very low impedance, Z = R + ix. You cannot get an electrical shock at this electrode or at this input to the potentiostat. 2. Amplifier U3 takes the current at the WE and converts it into a potential so it can be recorded. V = IR at the output of U3. 3. The reference electrode (RE), connected to the non-inverting input (+) of the op-amp U2, is asked to source a minute amount of current (~3 fa for this particular op-amp; 0 fa is the ideal case). 4. Op-amp U1 produces, at the counter electrode (CE), an output current, Iout, that is proportional to the potential difference between RE and WE (i.e. ground). Caution: You CAN get a lethal shock at this electrode. However, this power is not infinite (your wall sockets have a limited power they can supply). The potentiostat limits are termed the compliance voltage and compliance current

24 Beware of compliance voltage issues (maximum voltage to CE) and compliance current too! x x mM ferrocene, 0.1M [NEt 4 ][BF 4 ], MeCN flow rate = 1ml/hr 4.0x x10-5 I /A x x x x mv/s E /V

25 Active I/E Converter versus Passive I/E Converter meaning older 230

26 Active I/E Converter versus Passive I/E Converter meaning newer 231 Many modern potentiostats use the architecture shown here, yet it is rarely discussed at length in textbooks! This scheme has been used by Gamry, PAR, Solartron, and perhaps others [and likely Bio-Logic]. Unlike the active I/E converter design, this I/E converter is passive current only flows through passive circuit elements (e.g. R, C, not op-amp) the working electrode is NOT at (virtual) (earth) ground the electrometer is differential between the RE and the WE sense (RE #2)

27 Active I/E Converter versus Passive I/E Converter meaning newer 232

28 Potentiostat summary for non-ee majors 233 * The potentiostat does not control the potential of the working electrode! * The potentiostat controls the potential of the counter electrode only (relative to the working electrode) * The counter electrode is the most important electrode, followed by the reference electrode * Compliance voltage and compliance current limits are very important in the choice of the potentiostat / application * With a few components you can build your own potentiostat for < $100! * Passive potentiostats do not hold the WE at earth ground, but can measure potentials across electrolyte interfaces Rowe,..., Plaxco, PLoS One, 2011, 6, e23783 Mott,..., Sykes, J. Chem. Educ., 2014, 91, 1028

29 and that is why we use a 3-electrode potentiostat 234 WE = working electrode CE = counter (or auxiliary) electrode RE = reference electrode Out of sight, out of mind is a bad motto!

30 235 An example of two RE scales You re welcome! B&F back inside

31 ... okay, now let s measure the current that flows as we change the potential of the platinum working electrode (imagine that we have a potentiostat here that allows us to do that). 236 Let s suppose that this WE is platinum, and that all three electrodes are immersed in 1.0 M HCl

32 ... okay, now let s measure the current that flows as we change the potential of 237 the platinum working electrode (imagine that we have a potentiostat here that allows us to do that). What is the starting potential for this experiment? That is, what is the open-circuit potential? an oxidation: 2Cl (aq) Cl 2 (g) + 2e in 1.0 M HCl SHE a reduction: 2H + (aq) + 2e H 2 (g)

33 ... okay, now let s measure the current that flows as we change the potential of 238 the platinum working electrode (imagine that we have a potentiostat here that allows us to do that). What is the starting potential for this experiment? That is, what is the open-circuit potential (E oc )? an oxidation: 2Cl (aq) Cl 2 (g) + 2e a reduction: 2H + (aq) + 2e H 2 (g) SHE in 1.0 M HCl Somewhere in here but not sure where it is not clear which halfreaction is dominant although my guess it that E oc is set by impurities like Fe(III) and Fe(II)

34 ... okay, now let s measure the current that flows as we change the potential of the platinum working electrode (imagine that we have a potentiostat here that allows us to do that). i Current flow is proportional to rate, so let s write this like chemists: an oxidation: 2Cl (aq) Cl 2 (g) + 2e Rate mol s SHE = dn dt = i nf 239 but for electrochemists, this is less useful because it depends on the experimental set-up (i.e. electrode area)! a reduction: 2H + (aq) + 2e H 2 (g)

35 ... okay, now let s measure the current that flows as we change the potential of the platinum working electrode (imagine that we have a potentiostat here that allows us to do that). i Current flow is proportional to rate, so let s write this like chemists: an oxidation: 2Cl (aq) Cl 2 (g) + 2e Rate mol s SHE = dn dt = i nf 240 but for electrochemists, this is less useful because it depends on the experimental set-up (i.e. electrode area)! a reduction: 2H + (aq) + 2e H 2 (g) Rate mol s cm 2 = i nfa = j nf this is much better! Can we define each of these variables by name and unit?

36 ... okay, now let s measure the current that flows as we change the potential of the platinum working electrode (imagine that we have a potentiostat here that allows us to do that). i an oxidation: 2Cl (aq) Cl 2 (g) + 2e SHE 241 Given that the reactions shown here occur in a single cell (no salt bridge), when this E is applied to the WE and current flows at this I, what current flows at a large Pt CE and where on this plot would you put a point to show its response? a reduction: 2H + (aq) + 2e H 2 (g)

37 ... okay, now let s measure the current that flows as we change the potential of the platinum working electrode (imagine that we have a potentiostat here that allows us to do that). an oxidation: 2Cl (aq) Cl 2 (g) + 2e 242 Given that the reactions shown here occur in a single cell (no salt bridge), when this E is applied to the WE and current flows at this I, what current flows at a large Pt CE and where on this plot would you put a point to show its response? SHE Psssst! Don t forget me! I m still here! a reduction: 2H + (aq) + 2e H 2 (g)

38 ... okay, now let s measure the current that flows as we change the potential of the platinum working electrode (imagine that we have a potentiostat here that allows us to do that). an oxidation: 2Cl (aq) Cl 2 (g) + 2e 243 Given that the reactions shown here occur in a single cell (no salt bridge), when this E is applied to the WE and current flows at this I, what current flows at a large Pt CE and where on this plot would you put a point to show its response? SHE a reduction: 2H + (aq) + 2e H 2 (g) Current matching equal and opposite the current at the WE is the same as that at the CE Kirchhoff s Current Law

39 ... okay, now let s measure the current that flows as we change the potential of the platinum working electrode (imagine that we have a potentiostat here that allows us to do that). an oxidation: 2Cl (aq) Cl 2 (g) + 2e 244 Given that the reactions shown here occur in a single cell (no salt bridge), when this E is applied to the WE and current flows at this I, what current flows at a large Pt CE and where on this plot would you put a point to show its response? SHE a reduction: 2H + (aq) + 2e H 2 (g) Current matching equal and opposite the current at the WE is the same as that at the CE Kirchhoff s Current Law okay, but what if protons could not be reduced so easily? An overpotential is required!

40 Now, pretend this experimental I E curve (from my labs; in fact) was measured when the Pt WE was switched with a Hg WE why does little current flow until ~ -1 V? 245 vs. SHE

41 Now, pretend this experimental I E curve was measured when the Pt WE was switched with a Hg WE why does little current flow until ~ -1 V? 246 Overpotential! which is present due to kinetic/rate/current limitations η = E app E Eq vs. SHE

42 Now, pretend this experimental I E curve was measured when the Pt WE was switched with a Hg WE why does little current flow until ~ -1 V? 247 Overpotential! which is present due to kinetic/rate/current limitations η = E app E Eq What if you dump in Cd 2+, whose E 0 (Cd 2+ /Cd 0 ) -0.4 V vs. SHE? vs. SHE

43 Now, pretend this experimental I E curve was measured when the Pt WE was switched with a Hg WE why does little current flow until ~ -1 V? 248 Overpotential! which is present due to kinetic/rate/current limitations η = E app E Eq What if you dump in Cd 2+, whose E 0 (Cd 2+ /Cd 0 ) -0.4 V vs. SHE? Current response will be kinetically determined and current will start to pass at about -0.4 V vs. SHE vs. SHE

44 Now, pretend this experimental I E curve was measured when the Pt WE was switched with a Hg WE why does little current flow until ~ -1 V? 249 Overpotential! which is present due to kinetic/rate/current limitations η = E app E Eq What if you dump in Cd 2+, whose E 0 (Cd 2+ /Cd 0 ) -0.4 V vs. SHE? Current response will be kinetically determined and current will start to pass at about -0.4 V vs. SHE vs. SHE

45 Also, don t forget about possible compliance voltage issues (maximum voltage to CE) and compliance current too! under conditions of steady-state current flow, we are concerned with matched (equal and opposite) currents at the WE and CE 250 WE vs. RE CE vs. RE vs. SHE

46 more terminology 251 supporting electrolyte an inert salt added to impart ionic conductivity to the solution (e.g. 1 M HCl, in this case) background limits the two potential limits at which the pure solvent + supporting electrolyte begin to react at the working electrode

47 252 Electrochemical window (Potential range) * Red arrow entries are all measured in 1 M acid (1) If you wanted an aqueous battery with a large voltage, which electrode is best? (2) Between aqueous and non-aqueous batteries, which can generate the largest potential? B&F back inside cover

48 253 Electrochemical window (Potential range) * Red arrow entries are all measured in 1 M acid (1) If you wanted an aqueous battery with a large voltage, which electrode is best? Hg (in 0.1 M Et 4 NOH) or C (in 0.1 M KCl) (2) Between aqueous and non-aqueous batteries, which can generate the largest potential? Non-aqueous!... much larger solvent window B&F back inside cover

49 more terminology 254 supporting electrolyte an inert salt added to impart ionic conductivity to the solution (e.g. 1 M HCl, in this case) background limits the two potential limits at which the pure solvent + supporting electrolyte begin to react at the working electrode polarizable electrode an electrode operated within a potential range in which no Faradaic electrochemistry occurs Scientist Faraday s law The amount of chemical reaction caused by the flow of current is proportional to the amount of electricity passed. (B&F) an oxidation: 2Cl (aq) Cl 2 (g) + 2e solvent window SHE Michael Faraday ( ) from Wiki a reduction: 2H + (aq) + 2e H 2 (g)

50 more terminology 255 supporting electrolyte an inert salt added to impart ionic conductivity to the solution (e.g. 1 M HCl, in this case) background limits the two potential limits at which the pure solvent + supporting electrolyte begin to react at the working electrode polarizable electrode an electrode operated within a potential range in which no Faradaic electrochemistry occurs Scientist Faraday s law The amount of chemical reaction caused by the flow of current is proportional to the amount of electricity passed. (B&F) an oxidation: 2Cl (aq) Cl 2 (g) + 2e solvent window SHE Michael Faraday ( ) from Wiki Typically, chemical reaction is measured by mass (g) and electricity passed is measured by charge (C) don t forget z in the math! a reduction: 2H + (aq) + 2e H 2 (g)

51 and, even more terminology 256 Faradaic electrochemistry electrochemistry characterized by the flow of current to/from electron donor/acceptor species present at the electrode surface Non-Faradaic electrochemistry electrochemistry characterized by the flow of current to/from an electrode surface in the absence of donor/acceptor species, typically dominated by capacitive charging of the electrode or adsorption/desorption phenomena Scientist Faraday s law The amount of chemical reaction caused by the flow of current is proportional to the amount of electricity passed. (B&F) an oxidation: 2Cl (aq) Cl 2 (g) + 2e solvent window SHE Michael Faraday ( ) from Wiki Typically, chemical reaction is measured by mass (g) and electricity passed is measured by charge (C) don t forget z in the math! a reduction: 2H + (aq) + 2e H 2 (g)

52 and lastly, typical WE ranges for EChem experiments/technologies 257 J 3E-y 2E-y 1E-y -2.0E-x -1.5E-x -1.0E-x -0.5E-x -1E-y 0.5E-x 1.0E-x 1.5E-x 2.0E-x E, vs. SHE -2E-y -3E-y Cyclic voltammogram: 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 Photoelectrochemistry: x = 0 1, y = 2 E = E oc = ±700 mv J = J sc = ±30 ma/cm 2 Fuel Cell / Battery: x = 0, y = 0 E = 1 3 V J = 1 2 A/cm 2

53 258 A review of Section 1.1 (and some of Chapter 15) 2-electrode versus 3-electrode measurements Reference electrodes Potentiostats Compliance voltage/current J E and I E curves Kinetic overpotential Faradaic reactions

54 259 Q: What s in this set of lectures? A: B&F Chapters 1, 15 & 4 main concepts: Section 1.1: Redox reactions Chapter 15: Electrochemical instrumentation Section 1.2: Charging interfaces Section 1.3: Overview of electrochemical experiments Section 1.4: Mass transfer and Semi-empirical treatment of electrochemical observations Chapter 4: Mass transfer

55 260 Looking forward Sections 1.2 and 1.3 RC circuits (~90% of slides) Electrochemically active surface area Uncompensated resistance Placement of electrodes, and other properties

56 261 our Pt WE is polarizable within this potential range SHE a polarizable electrode that has been polarized imposes a background electric response an added current in a voltammetric experiment, for example (not observable on this scale as it is small) that is transient (e.g. a blip) and is observed in all electrochemical experiments We need to understand this background current!

57 The electrical response of a polarizable electrode is approximated by a series resistor and capacitor (a series RC circuit) 262 Power Supply E (T)otal resistor = solution R capacitor = WE interface C E T = E R + E cap +

58 The electrical response of a polarizable electrode is approximated by a series resistor and capacitor (a series RC circuit) 263 Power Supply R = the solution resistance (between the WE and RE) resistor = solution capacitor = WE interface R C

59 The electrical response of a polarizable electrode is approximated by a series resistor and capacitor (a series RC circuit) 264 Power Supply R = the solution resistance (between the WE and RE) C = the net capacitance (of the WE and the CE), C(T)otal 1 C T = 1 C C 2 resistor = solution capacitor = WE interface R C to measure C 2(WE), make C 1(CE) large or use a three-electrode setup and a pstat

60 First, what are approximate values for R (S) and C (d)? 265 Power Supply R = the solution resistance (between the WE and RE) In aqueous solutions containing 0.1 M supporting electrolyte, R = a few ohms; for non-aq., R > 100 Ω C = the net capacitance (of the WE and the RE), C(T)otal ~20 µf/cm 2 of electrode area for gold or platinum; 2 5 µf/cm 2 for carbon, typically but these change slightly with potential as we will see later resistor = solution capacitor = WE interface (Farad is C/V, where V is J/C) R C

61 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): i = E R exp t RC Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure 1.2.7

62 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): i = E R exp t RC Could there be a problem with an instantaneous 6 V potential step, for example? Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure 1.2.7

63 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): i = E R exp t RC Could there be a problem with an instantaneous 6 V potential step, for example? Compliance current! (at t = 0, E = ir (Ohm s law)) Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure 1.2.7

64 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): i = E R exp t RC What portion of E app is actually felt by the WE at t = 0? Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure 1.2.7

65 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): i = E R exp t RC What portion of E app is actually felt by the WE at t = 0? Little of it! Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure 1.2.7

66 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): NOTE: Electronics can limit the observation of rapid chemical kinetics (i.e. the rate-determining step is charging and not electron transfer) E E WE vs. RE t What portion of E app is actually felt by the WE at t = 0? Little of it! Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure 1.2.7

67 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): i = E R exp t RC but where did this equation for current come from? who s comfortable with me just giving you this equation? Let s manipulate units! Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure 1.2.7

68 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): B&F eqn. (1.2.8) E = E R + E C = ir s + q C d R E T = E R + E cap + C

69 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): B&F eqn. (1.2.8) E = E R + E C = ir s + q C d Units: C Units: C/V R C

70 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): B&F eqn. (1.2.8) E = E R + E C = ir s + q C d B&F eqn. (1.2.9) i = 1 R s E q C d = dq dt Units: C Units: C/V R C Units: C/s

71 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): B&F eqn. (1.2.8) E = E R + E C = ir s + q C d B&F eqn. (1.2.9) i = 1 R s E q C d = dq dt Units: C Units: C/V R C Units: C/s Need to integrate!

72 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): B&F eqn. (1.2.8) E = E R + E C = ir s + q C d B&F eqn. (1.2.9) i = 1 R s E q C d = dq dt Units: C Units: C/V R C Units: C/s Need to integrate! 1 dt = 1 R s E q dq = C d C d EC d +q dq

73 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): B&F eqn. (1.2.8) E = E R + E C = ir s + q C d B&F eqn. (1.2.9) Units: C/s i = 1 R s E q C d = dq dt Need to integrate! 1 dt = 1 R s E q dq = C d C d EC d +q dq Units: C Units: C/V 1 t = ln EC d + q ln EC d (assuming that at t = 0, q = 0) R C = ln EC d + q EC d Integrated!

74 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): 1 t = ln EC d + q ln EC d = ln EC d + q EC d Integrated!

75 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): 1 t = ln EC d + q ln EC d = ln EC d + q EC d Integrated! t EC d e = EC d + q

76 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): 1 t = ln EC d + q ln EC d = ln EC d + q EC d Integrated! t EC d e = EC d + q q = EC d 1 e t q B&F eqn. (1.2.10) t

77 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): 1 t = ln EC d + q ln EC d = ln EC d + q EC d Integrated! t EC d e = EC d + q q = EC d 1 e t B&F eqn. (1.2.10) Need to differentiate! dq dt = EC d 1 e t

78 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): 1 t = ln EC d + q ln EC d = ln EC d + q EC d Integrated! t EC d e = EC d + q q = EC d 1 e t Need to differentiate! dq dt = EC d 1 e t = E e t = I R s Done! B&F eqn. (1.2.10) B&F eqn. (1.2.6)

79 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): i = E R exp t RC What are the units 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 1.2.7

80 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): i = E R exp t RC What are the units 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 1.2.7

81 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): i = E R exp t RC What are the units 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 1.2.7

82 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): i = E R exp t RC Why is 37% of the initial signal noteworthy? What are the units 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 1.2.7

83 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Voltage step (that is, increment the potential by an amount, E): i = E R exp t RC Why is 37% of the initial signal noteworthy? Plug in t = RC! Ah ha! What are the units 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 1.2.7

84 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Example: Consider the case where R = 1 Ω and C = 20 µf/cm 2. How long will it take to charge C to 95% of its maximum capacity?

85 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Example: Consider the case where R = 1 Ω and C = 20 µf/cm 2. How long will it take to charge C to 95% of its maximum capacity? q = EC d 1 e t B&F eqn. (1.2.10) q t

86 Now, what response is obtained for various inputs to this circuit? Potential-step potentiostatic chronoamperometry (chronocoulometry) Example: Consider the case where R = 1 Ω and C = 20 µf/cm 2. How long will it take to charge C to 95% of its maximum capacity? q = EC d 1 e t q t = EC d B&F eqn. (1.2.10) q t

87 Example: Consider the case where R = 1 Ω and C = 20 µf/cm 2. How long will it take to charge C to 95% of its maximum capacity? 292 q = EC d 1 e t q t = EC d B&F eqn. (1.2.10) q q = 1 e q t t t

88 Example: Consider the case where R = 1 Ω and C = 20 µf/cm 2. How long will it take to charge C to 95% of its maximum capacity? 293 q = EC d 1 e t q t = EC d B&F eqn. (1.2.10) q q = 1 e q t t 0.95 = 1 e t = e t 0.95 t

89 Example: Consider the case where R = 1 Ω and C = 20 µf/cm 2. How long will it take to charge C to 95% of its maximum capacity? 294 q = EC d 1 e t q t = EC d B&F eqn. (1.2.10) q q = 1 e q t t 0.95 = 1 e t = e t 0.95 ln 0.05 = t 0.95 t

90 Example: Consider the case where R = 1 Ω and C = 20 µf/cm 2. How long will it take to charge C to 95% of its maximum capacity? 295 q = EC d 1 e t q t = EC d B&F eqn. (1.2.10) q q = 1 e q t t 0.95 = 1 e t = e t 0.95 assuming 1 cm 2 ln 0.05 = t 0.95 t 0.95 = 1Ω 20μF ln = 60μs t

91 Example: Consider the case where R = 1 Ω and C = 20 µf/cm 2. How long will it take to charge C to 95% of its maximum capacity? 296 q = EC d 1 e t q t = EC d q = 1 e q t t 0.95 = 1 e t = e t 0.95 ln 0.05 = t 0.95 B&F eqn. (1.2.10) assuming 1 cm 2 t 0.95 = 1Ω = 60μs As above, assuming a 6 V potential step, now what is the average current that flows up to t 0.95? 20μF ln 0. 05

92 Example: Consider the case where R = 1 Ω and C = 20 µf/cm 2. How long will it take to charge C to 95% of its maximum capacity? 297 q = EC d 1 e t q t = EC d q = 1 e q t t 0.95 = 1 e t = e t 0.95 ln 0.05 = t 0.95 B&F eqn. (1.2.10) assuming 1 cm 2 t 0.95 = 1Ω = 60μs As above, assuming a 6 V potential step, now what is the average current that flows up to t 0.95? I avg = C / s = (C/V x V) / s 20μF ln 0. 05

93 Example: Consider the case where R = 1 Ω and C = 20 µf/cm 2. How long will it take to charge C to 95% of its maximum capacity? 298 q = EC d 1 e t q t = EC d q = 1 e q t t 0.95 = 1 e t = e t 0.95 ln 0.05 = t 0.95 B&F eqn. (1.2.10) assuming 1 cm 2 t 0.95 = 1Ω = 60μs As above, assuming a 6 V potential step, now what is the average current that flows up to t 0.95? I avg = C / s = (C/V x V) / s = 20 μf x 6 V / 60 μs 20μF ln 0. 05

94 Example: Consider the case where R = 1 Ω and C = 20 µf/cm 2. How long will it take to charge C to 95% of its maximum capacity? 299 q = EC d 1 e t q t = EC d q = 1 e q t t 0.95 = 1 e t = e t 0.95 ln 0.05 = t 0.95 B&F eqn. (1.2.10) assuming 1 cm 2 t 0.95 = 1Ω = 60μs As above, assuming a 6 V potential step, now what is the average current that flows up to t 0.95? I avg = C / s = (C/V x V) / s = 20 μf x 6 V / 60 μs = 120 μc / 60 μs 20μF ln 0. 05

95 Example: Consider the case where R = 1 Ω and C = 20 µf/cm 2. How long will it take to charge C to 95% of its maximum capacity? 300 q = EC d 1 e t q t = EC d q = 1 e q t t 0.95 = 1 e t = e t 0.95 ln 0.05 = t 0.95 B&F eqn. (1.2.10) assuming 1 cm 2 t 0.95 = 1Ω = 60μs As above, assuming a 6 V potential step, now what is the average current that flows up to t 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 0. 05

96 Now, what response is obtained for various inputs to this circuit? Current-step galvanostatic chronopotentiometry Current step (that is, increment the current by an amount, i): E = i R + t C B&F eqn. (1.2.12) Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure 1.2.9

97 Now, what response is obtained for various inputs to this circuit? Current-step galvanostatic chronopotentiometry Current step (that is, increment the current by an amount, i): E = i R + t C B&F eqn. (1.2.12) So, a constant applied current results in a linear sweep of the potential thus, what if we instead applied the potential sweep? Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure 1.2.9

98 Now, what response is obtained for various inputs to this circuit? Linear-sweep voltammetry cyclic voltammetry Potential scan (that is, ramp the applied potential, E(t) = vt for one direction): i = vc d scan rate 1 exp t R S C d B&F eqn. (1.2.15) Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure

99 Now, what response is obtained for various inputs to this circuit? Linear-sweep voltammetry cyclic voltammetry Potential scan (that is, ramp the applied potential, E(t) = vt for one direction): i = vc d scan rate 1 exp t R S C d B&F eqn. (1.2.15) ASIDE: Recall, for a potential step, the same shape but for charge (q) q = EC d 1 e t B&F eqn. (1.2.10) q Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure t

100 Now, what response is obtained for various inputs to this circuit? Linear-sweep voltammetry cyclic voltammetry Potential scan (that is, ramp the applied potential, E(t) = vt for one direction): i = vc d scan rate 1 exp t R S C d B&F eqn. (1.2.15) So the total current envelope at any potential that is well-removed from the switching potential will be: i = 2Cv, with v s units being V/s i = 2Cv Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure

101 Now, what response is obtained for various inputs to this circuit? Linear-sweep voltammetry cyclic voltammetry This is an example of a cyclic voltammogram with obvious RC charging /thesuiteworld.com*wp-includes*theme-compat*dallas-texas-scenery-5417.jpg/ Bard & Faulkner, 2nd Ed., Wiley, 2001, Figure i = 2Cv