1298 Lecture #18 of 18

Transcription

1 Lecture #18 of

2 1299 Q: What s in this set of lectures? A: B&F Chapters 9, 10, and 6 main concepts: Sections : Sections : Rotating (Ring-)Disk Electrochemistry Electrochemical Impedance Spectroscopy Sections , 11.7, 14.3: Linear Sweep Voltammetry (LSV), Cyclic Voltammetry (CV), Thin-Layer Electrochemistry To really learn about your laboratory systems move beyond steady-state conditions!

3 Sloooooooooooowwwww masssssss traaaaaannnnssssport dooooooowwwwwwwwnnnnn, Dr. Butler and Dr. Volmer 1300 Goes through (0, 0) means both R and O are present Recall that curvature near (0, 0) implies activation overpotential and not concentration overpotential

4 Sloooooooooooowwwww masssssss traaaaaannnnssssport dooooooowwwwwwwwnnnnn, Dr. Butler and Dr. Volmer 1301

5 Sloooooooooooowwwww masssssss traaaaaannnnssssport dooooooowwwwwwwwnnnnn, Dr. Butler and Dr. Volmer 1302 Wait a minute All I need to do in order to observe a CV/LSV peak is stop stirring? Seriously? Why didn t anyone tell me sooner?

6 Sloooooooooooowwwww masssssss traaaaaannnnssssport dooooooowwwwwwwwnnnnn, Dr. Butler and Dr. Volmer 1303 Wait a minute All I need to do in order to observe a CV/LSV peak is stop stirring? Seriously? Why didn t anyone tell me sooner? WHO CAN EXPLAIN THIS?

7 with (a little) stirring, the diffusion layer is fully formed (i.e. time-independent) at each potential during the sweep 1304 FLASHBACK an unstirred solution will have δ cm after ~1 sec (Bockris, Reddy, and Gamboa-Aldeco, Modern EChem, Vol. 2A, 2002, pg. 1098) Bard & Faulkner, 2nd Ed., Figure stirring causes δ to become welldefined, time-invariant, and a short distance to conditions of the bulk

8 with (a little) stirring, the diffusion layer is fully formed (i.e. time-independent) at each potential during the sweep 1305 FLASHBACK an unstirred solution will have δ cm after ~1 sec (Bockris, Reddy, and Gamboa-Aldeco, Modern EChem, Vol. 2A, 2002, pg. 1098) Bard & Faulkner, 2nd Ed., Figure stirring causes δ to become welldefined, time-invariant, and a short distance to conditions of the bulk

9 without stirring, the diffusion layer grows over time and with a big (enough) potential step (or even bigger, or a little smaller again), the Cottrell equation results FLASHBACK s 0.01s 0.1s t = 1s C* = 1 x 10-6 M D = 1 x 10-5 M > 200 mv > 200 mv E eq

10 without stirring, the diffusion layer grows over time and with a big (enough) potential step (or even bigger, or a little smaller again), the Cottrell equation results FLASHBACK 1307 hybrid catalysis and diffusion > 200 mv > 200 mv E eq

11 without stirring, the diffusion layer grows over time and with a big (enough) potential step (or even bigger, or a little smaller again), the Cottrell equation results FLASHBACK 1308 mass-transfer limited (Cottrellian) hybrid catalysis and diffusion

12 without stirring, the diffusion layer grows over time and with a big (enough) potential step (or even bigger, or a little smaller again), the Cottrell equation results FLASHBACK peak occurs after E 1/ mass-transfer limited (Cottrellian) hybrid catalysis and diffusion

13 who invented linear sweep voltammetry? John E. B. Randles and A. Sevcik 1310 the Randles equivalent circuit approximation of an electrochemical cell same as in EIS!

14 1311

15 Randles Sevcik Equation (T = 298 K) 1312

16 1313 Randles Sevcik Equation (T = 298 K) What all LSV/CV ers should know: i p is proportional to the square root of the scan rate when the molecules are in solution and not stuck to the surface but when surface-adsorbed, i p is proportional to the scan rate

17 1314

18 this is an analog oscilloscope how do you capture these data? 1315

19 how do you capture this data? Answer: you photograph it! 1316 Tektronix C 59A Oscilloscope Camera f2.8.67mag w/ Back Film Pack

20 in the 1960 s-1980 s, X Y plotters were used to record ALL data 1317

21 and that plotter was connected to a voltammetric analyzer 1318 the digital instruments of today do not actually sweep and so are imperfect

22 1319 in the lab in 1956 Irving Shain Rich Nicholson in 1963

23 1320

24 1321

25 1322 the switching potential or pos/neg limit slope = v units, V s -1

26 the derivation of these equations is messy (involving the Laplace transform and numerical approximations) thus, we ll omit it 1323 but the key result from Nicholson and Shain is the following: σ = nf RT v the dimensionless current function

27 1324

28 1325 NOTE: is the maximum value for π 1/2 χ(σt) but not at E 1/2 Why?

29 is the maximum value for χ(σt), but not at 0 V vs E 1/2 Why? 1326 χ(σt) = mv

30 In this experiment, two things happen concurrently: 1) C(0, t) decreases, and 2) δ increases with t 1/2 1327

31 at least one thing about this I E trace makes some sense The behavior at large E E o is Cottrellian 1328

32 at least one thing about this I E trace makes some sense The behavior at large E E o is Cottrellian 1329

33 but is there justification for the pre-cottrellian peak being located at mv? 1330 Consider two limiting cases: 1) The reaction rate is diffusion-controlled, and the diffusion-layer thickness, δ, is independent of time, and is ~0.5 mm thick after ~1 sec in a solution that is not artificially stirred Bockris, Reddy, and Gamboa-Aldeco, Modern EChem, Vol. 2A, 2002, pg. 1098

34 but is there justification for the pre-cottrellian peak being located at mv? 1331 Consider two limiting cases: 1) The reaction rate is diffusion-controlled, and the diffusion-layer thickness, δ, is independent of time, and is ~0.5 mm thick after ~1 sec in a solution that is not artificially stirred 2) The reaction rate is activation-controlled. There is no diffusion layer no diffusion!

35 1) The reaction rate is diffusion-controlled, and the diffusionlayer thickness, δ, is independent of time. 1332

36 now, according to Fick s first law, the current will be proportional 1333 to the concentration gradient at x = 0 J i x = D i C i x x

37 now, according to Fick s first law, the current will be proportional 1334 to the concentration gradient at x = 0 the linearized version of which is: J i x = D i C i x x J i x = D i C o C o 0, t o first, consider a case where δ O is independent of time. In this case, J(0) will depend only on C O (0, t) and J max will correspond to C O (0, t) = 0.

38 now, according to Fick s first law, the current will be proportional to the concentration gradient at x = 0 J i x = D i C i x x 1335 the linearized version of which is: J i x = D i C o C o 0, t o first, consider a case where δ O is independent of time. In this case, J(0) will depend only on C O (0, t) and J max will correspond to C O (0, t) = 0. Based on this we get a sigmoidal I E curve (S-shaped), with a defined limiting current, which we ve seen many times in this course already and is obviously not what we see for CV s here. So the observed peaked response must derive from motion of δ O with time, convoluted with the potential dependence of C O (0, t)

39 we ve already seen this. There is no peak in the current. Question: How negative must one scan before obtaining i l? 1336 E = E 1/2 + RT nf ln i l i i

40 at 0 V, you have just 50% of i l so, to get 90% of i l, you need to apply ~ -55 mv past E eq 1337 i/il = 0.9 at -55 mv

41 okay, so what about the other limiting case? This one we have not seen before ) The reaction rate is diffusion-controlled, and the diffusion-layer thickness, δ, is independent of time. 2) The reaction rate is activation-controlled! There is no diffusion layer no diffusion!

42 let s imagine electrochemical systems for which diffusion does not control the rate of faradaic reactions 1339 example: redox chemistry of an adsorbed monolayer: 2H + + 2e 2Pt H UPD hydrogen HER Clavilier s papillon (butterfly pattern) OER UPD oxygen

43 let s imagine electrochemical systems for which diffusion does not control the rate of faradaic reactions 1340 example: redox chemistry with an ultra-thin Nafion film electrode Nafion [Ru II (bpy) 3 ] 2+ [Ru III (bpy) 3 ] 3+ d << (2Dt) 1/2

44 let s imagine electrochemical systems for which diffusion does not control the rate of faradaic reactions 1341 example: redox chemistry with an ultra-thin Nafion film electrode Nafion [Ru II (bpy) 3 ] 2+ [Ru III (bpy) 3 ] 3+ d << (2Dt) 1/2 tiny peak splitting may be due to ir u drop keep currents small

45 this is called: thin-layer (zero-gap) electrochemistry which we already discussed in the context of single-molecule EChem 1342

46 this is called: thin-layer (zero-gap) electrochemistry 1343 Question: what is a thin-layer cell? Answer: Any cell with a thickness: l 2Dt

47 the voltammetric response will therefore be proportional to the derivative of these curves more on this in a second 1344 R O dconc/de has a maximum at E o this capacitance times v, is current

48 what does B&F tell us about it? in Section 11.7! 1345 i = n2 F 2 vvc O RT exp nf RT 1 + exp nf RT E Eo E Eo 2 i p = n2 F 2 vvc O 4RT

49 so, thin-layer voltammetry has the following properties: 1346 i p = n2 F 2 vvc O 4RT ip V (the total volume of the thin-layer cell) and ip Co*. Taken together, this really means that... ip Γ (the coverage of the surface by electroactive molecules in units of moles cm -2 ) ip v 1 important this is how one recognizes & diagnoses thin-layer behavior experimentally. More on this later NOTE: no diffusion, so no D!

50 so, to sum up our observations about these two limiting cases: diffusion-controlled, static δ. Ep E o > 55 mv activation-controlled, NO δ. E p E o = 0 mv expanding δ using LSV/CV. E p E o = 28.5 mv mv

51 now let s take a closer look at thin-layer behavior, notably because it is highly relevant to molecular homogeneous electrocatalysis anyway, there are three types of thin-layer cells: nm to 50 µm 2 electrodes, E 1 = E 2 1 electrode, E 1 2 electrodes, E 1 E 2

52 1349 E 0 E 1 = E 2

53 3 cases of interest: electrode, E1. 2 electrodes, E1 E2

54 what s the current? 1351 consider the generic reaction: 1 electrode, E1. assuming the concentration everywhere in the cell follows C(x, t) = C(0, t), which means it is uniform (NOT as shown above): i = nfv dc O t dt 2 electrodes, E1 E2 = nf la dc O t dt

55 i = nfv dc O t dt = nf la dc O t dt 1352 j = nfl dc O t dt Note: i and j l = the cell thickness small l small V small j.

56 i = nfv dc O t dt = nf la dc O t dt 1353 j = nfl dc O t dt Note: i and j l = the cell thickness small l small V small j. Now, according to the Nernst Eq C O t = C O exp nf RT E Eo 1 this equation makes sense: if E = E 0, C O (t) = 0.5C O * E << E 0, C O (t) = 0 you re reducing as fast as possible E >> E 0, C O (t) = C O * you re doing nothing

57 i = nfv dc O t dt = nf la dc O t dt 1354 j = nfl dc O t dt Note: i and j l = the cell thickness small l small V small j. Now, according to the Nernst Eq C O t = C O exp nf RT E Eo 1 there is no explicit time dependence in this equation, but E is time dependent if we scan: E(t) = E i + vt. Substituting and differentiating

58 i = nfv dc O t dt = nf la dc O t dt 1355 j = nfl dc O t dt Note: i and j l = the cell thickness small l small V small j. Now, according to the Nernst Eq C O t = C O exp nf RT E Eo 1 there is no explicit time dependence in this equation, but E is time dependent if we scan: E(t) = E i + vt. Substituting and differentiating

59 i = n2 F 2 vvc O RT exp nf RT 1 + exp nf RT E Eo E Eo i p = n2 F 2 vvc O 4RT Now, according to the Nernst Eq C O t = C O exp nf RT E Eo 1 there is no explicit time dependence in this equation, but E is time dependent if we scan: E(t) = E i + vt. Substituting and differentiating

60 i = n2 F 2 vvc O RT exp nf RT 1 + exp nf RT E Eo E Eo i p = n2 F 2 vvc O 4RT as the volume of the cell is decreased, by reducing the cell thickness for example, i p falls. Also, i p depends on scan rate and is proportional to v 1! (Recall that for an LSV/CV, i p is proportional to v 1/2.)

61 again, here is the resulting i vs E thin-layer-cell voltammogram: 1358 i = n2 F 2 vvc O RT exp nf RT 1 + exp nf RT E Eo E Eo 2 i p = n2 F 2 vvc O 4RT

62 schematically what is happening is the following: imagine doing the experiment in many small potential steps current 1359 R O voltage

63 schematically what is happening is the following: imagine doing the experiment in many small potential steps current 1360 R O voltage

64 schematically what is happening is the following: imagine doing the experiment in many small potential steps current 1361 R O voltage

65 schematically what is happening is the following: imagine doing the experiment in many small potential steps current 1362 R O Q = nfvδc i = nfv ΔC Δt voltage

66 schematically what is happening is the following: imagine doing the experiment in many small potential steps current 1363 R O Q = nfvδc i = nfv ΔC Δt voltage

67 schematically what is happening is the following: imagine doing the experiment in many small potential steps current 1364 R O voltage

68 schematically what is happening is the following: imagine doing the experiment in many small potential steps current 1365 R O potential

69 Now this makes more sense 1366 R O dconc/de has a maximum at E o this capacitance times v, is current

70 Okay, so the two-electrode thin-layer cell (with E1 = E2) gives the peaked I E curve that we just calculated nm to 50 µm 2 electrodes, E1 E2

71 Now, what happens if you get rid of one electrode? nm to 50 µm 2 electrodes, E1 E2

72 Answer: Nothing! The I E curve is the same as the two-electrode case 1369

73 both electrodes are doing the same thing and the rate of diffusion is, by definition, negligible 1370

74 so, this equation, and these conclusions, apply both to one-electrode and two-electrode thin-layer cells (with E 1 = E 2 ) 1371 i p = n2 F 2 vvc O 4RT ip V (the total volume of the thin-layer cell) and ip Co*. Taken together, this really means that... ip Γ (the coverage of the surface by electroactive molecules in units of moles cm -2 ) ip v 1 important this is how one recognizes & diagnoses thin-layer behavior experimentally. More on this later NOTE: no diffusion, so no D!

75 Now what about the two-electrode E 1 E 2 case? nm to 50 µm 2 electrodes, E 1 E 2

76 Now what about the two-electrode E 1 E 2 case? Huh? 1373

77 in this instance, when O is consumed, it is simultaneously regenerated at the other electrode 1374 Much larger Observed for homogeneous electrocatalysis!

78 What s the current? Just Fick s first law 1375 J 0 = D C x x x = 0 and then linearize this as in Chapter 1 J 0 = D ΔC Δx = D C l C 0 l The numerator here will be bounded by C O *, so the limiting current is i l = nfad C O l

79 What s the current? 1376 i l = nfad C O l NOTE: We ve got D in this equation. That s because the current depends on the transport rate of molecules across the cell.

80 one can also calculate the gain imparted by the positive feedback produced by the second electrode gain = i l i p = nfad C O l n 2 F 2 v la C O 4RT = 4DRT nfvl

81 one can also calculate the gain imparted by the positive feedback produced by the second electrode gain = i l i p = nfad C O l n 2 F 2 v la C O 4RT = 4DRT nfvl example: for v = 100 mv s -1, and l = 10 µm we have: gain = n l 2 = = 1.0 x x 10 4 cm 2 = x 10 5 nl 2

82 one can also calculate the gain imparted by the positive feedback produced by the second electrode gain = i l i p = nfad C O l n 2 F 2 v la C O 4RT = 4DRT nfvl example: for v = 100 mv s -1, and l = 10 µm we have: gain = n l 2 = = 1.0 x x 10 4 cm 2 = x 10 5 nl 2 For l = 1 µm, we get 1000; for l = 100 nm we get 10 5 Wow! Amplified!

83 { 1380 effectively a thin-layer region example: for v = 100 mv s -1, and l = 10 µm we have: gain = n l 2 = 1.0 x 10 5 nl 2 = 1.0 x x 10 4 cm 2 = 10 For l = 1 µm, we get 1000; for l = 100 nm we get 10 5 Wow! Amplified!

84 ANYWAY in cyclic voltammetry you scan back and forth 1381

85 for a Nernstian, reversible reaction, ΔE p mv for n =

86 how do you measure ip? well, for the first forward scan, you just measure it 1383

87 how do you measure ip? well, for the first forward scan, you just measure it 1384 How can you convert the x axis from E to t?

88 how do you measure ip? well, for the first forward scan, you just measure it 1385 How can you convert the x axis from E to t?... Divide by v! /v = t (s)

89 how do you measure ip? well, for the first forward scan, you just measure it 1386 and now, what is this function? /v = t (s)

90 how do you measure ip? well, for the first forward scan, you just measure it 1387 and now, what is this function? A Cottrellian current transient /v = t (s)

91 so ip for the reverse scan must be measured from the decaying, Cottrellian current of the forward scan! 1388 i t 1/2

92 a similar problem exists if you have two processes in close proximity, in terms of potential here is a solution 1389 i p

93 How is this reversible CV affected by slow electron transfer kinetics (k 0, α)? 1390 χ(σ,t) = mv

94 answer: ΔE p increases from ~60 mv to larger values 1391

95 answer: ΔE p increases from ~60 mv to larger values 1392 Peaks can be: (a) Electrochemically reversible, Evans, et al., J. Chem. Educ. 1983, 60, 290

96 answer: ΔE p increases from ~60 mv to larger values 1393 Peaks can be: (a) Electrochemically reversible, (b) Quasi-reversible ( E pa E pc > 58/n mv at room temperature, after ir u and/or other potential corrections), or (c) Irreversible (i pc i pa, and they are not even close) Slow kinetics and ir u are indistinguishable by CV (thus, turn a knob, or keep ir u small)

97 1394 ir u E applied E attenuated... ir u drop looks a lot like slow catalysis minimize i and/or R u, or just correct for ir u

98 peak current increases with v 1/2 but, is faster always better? what about signal-to-noise ratio? 1395 signal: Randles Sevcik Equation (T = 298 K) noise: i c = C d Av Double-layer charging

99 peak current increases with v 1/2 but, is faster always better? what about signal-to-noise ratio? 1396 signal: Randles Sevcik Equation (T = 298 K) noise: i c = C d Av Double-layer charging ratio: S: N = 2.69 x 105 n 3/2 AD 1/2 C v 1/2 C d Av = 2.69 x 105 n 3/2 D 1/2 C C d v 1/2 Therefore, for better S:N, slow scan rates are best!

100 peak current increases with v 1/2 but, is faster always better? what about signal-to-noise ratio? 1397 signal: Randles Sevcik Equation (T = 298 K) noise: i c = C d Av Double-layer charging ratio: S: N = 2.69 x 105 n 3/2 AD 1/2 C v 1/2 C d Av = 2.69 x 105 n 3/2 D 1/2 C C d v 1/2 Therefore, for better S:N, slow scan rates are best!

101 but UMEs can help measure things like fast kinetics! 1398 measuring processes that occur in small spaces e.g. single cells, SECM, etc. (i is small ; j is large) measuring in highly resistive media (R u is small ) e.g. solvent glasses, no supporting electrolyte, nonpolar solvents, gas phase reactions observing & measuring the kinetics of fast reactions (C d is small; R u is small ) i = E R exp t RC

102 but UMEs can help measure things like fast kinetics! 1399 measuring processes that occur in small spaces e.g. single cells, SECM, etc. (i is small ; j is large) measuring in highly resistive media (R u is small ) e.g. solvent glasses, no supporting electrolyte, nonpolar solvents, gas phase reactions observing & measuring the kinetics of fast reactions (C d is small; R u is small ) i = E R exp t RC but wait Recall that for better S:N slow scan rates are best Uh oh!

103 Experimentally, we also observe an overpotential that is intrinsic to the electron transfer process e [Ru III (bpy) 3 ] 3+ [Ru II (bpy) 3 ] e anthracene anthracene as examples, these are two of the fastest known heterogeneous electron transfer rxns

104 1401 Science 1990, 250, 1118 r 0 = 4.7 µm 6.8 pa r 0 = 2.3 nm r 0 = 1.1 µm 1.9 pa r 0 = 1.1 nm

105 smaller is better! 1402

106 Recall 1403 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

107 1404

108 1405 anthracene + 1e anthracene scan rate is in megavolts s -1! That is MV s -1 wow!

109 Anthracene and [Ru II (bpy) 3 ] 2+ are not that similar, chemically So what do these two ultrafast reactions have in common? e [Ru III (bpy) 3 ] 3+ [Ru II (bpy) 3 ] e anthracene anthracene

110 1407 means Ru(II) means Ru(III)

111 1408

112 1409 Anthracene and [Ru II (bpy) 3 ] 2+ are not that similar, chemically So what do these two ultrafast reactions have in common? 1) reactant and products are almost structurally identical 2) electron transfer involves no bond-making or bond-breaking 3) in polar solvents, bigger is faster thus, exchange current (density) is large due to small reorganization energy!

113 RECALL without stirring, the diffusion layer grows over time and with a big (enough) potential step (or even bigger, or a little smaller again), the Cottrell equation results FLASHBACK 1410 mass-transfer limited (Cottrellian) hybrid catalysis and diffusion

114 Why are CVs useful when Tafel Plots seem ideal? 1411 CVs are not pure Butler Volmer behavior ever, unless you what?

115 Why are CVs useful when Tafel Plots seem ideal? 1412 CVs are not pure Butler Volmer behavior ever, unless you what? Why is Tafel (and/or with RDE) better than CV? Stir! Not limited by mass transfer and so one can measure interfacial faradaic charge-transfer kinetic parameters with ease (α, j 0, k 0 )

116 Why are CVs useful when Tafel Plots seem ideal? 1413 CVs are not pure Butler Volmer behavior ever, unless you what? Why is Tafel (and/or with RDE) better than CV? Stir! Not limited by mass transfer and so one can measure interfacial faradaic charge-transfer kinetic parameters with ease (α, j 0, k 0 ) Why is CV better than Tafel? Can tell if species is adsorbed to the surface by performing a facile scan-rate dependence (linear i p vs. v 1/2 = homo vs v 1 = hetero) Easy to compare 1ET and 2ET steps, as i p contains n. Randles Sevcik Equation (T = 298 K)

117 Why are CVs useful when Tafel Plots seem ideal? 1414 CVs are not pure Butler Volmer behavior ever, unless you what? Why is Tafel (and/or with RDE) better than CV? Stir! Not limited by mass transfer and so one can measure interfacial faradaic charge-transfer kinetic parameters with ease (α, j 0, k 0 ) Why is CV better than Tafel? Can tell if species is adsorbed to the surface by performing a facile scan-rate dependence (linear i p vs. v 1/2 = homo vs v 1 = hetero) Easy to compare 1ET and 2ET steps, as i p contains n. Can tease out kinetics from shape, if you are a computer.

118 Why are CVs useful when Tafel Plots seem ideal? 1415 CV is not pure Butler Volmer region ever, unless you what? Stir Why is Tafel (and/or with RDE) better than CV? Not limited by mass transfer and so one can measure interfacial faradaic charge-transfer kinetic parameters with ease (α, j 0, k 0 ) Why is CV better than Tafel? Can tell if species is adsorbed to the surface by performing a facile scan-rate dependence (linear i p vs. v 1/2 = homo vs v 1 = hetero) Easy to compare 1ET and 2ET steps, as i p contains n. Can tease out kinetics from shape, if you are a computer. What is the midpoint potential for Tafel plots versus CV plots? E eq E 0, if D O = D R

119 RECALL: Course goal, i.e. best 2-hr-long final-exam question ever! 1416 Q: Explain cyclic voltammetry. From syllabus WE DID IT! Course philosophy Theory/Experiments versus Technologies (me vs you) I will teach the theory, history, and experimental specifics, and you will teach the technologies, and real-world and academic state-of-the-art Evans,, Kelly, J. Chem. Educ. 1983, 60, 290

120 1417 Q: What was in this set of lectures? A: B&F Chapters 9, 10, and 6 main concepts: Sections : Sections : Rotating (Ring-)Disk Electrochemistry Electrochemical Impedance Spectroscopy Sections , 11.7, 14.3: Linear Sweep Voltammetry (LSV), Cyclic Voltammetry (CV), Thin-Layer Electrochemistry To really learn about your laboratory systems move beyond steady-state conditions!