2025年电化学实验研究报告探索电极反应机制

Experimentalclasson"FuelCellandElectrochemistry"

Experimentsetup

Equipment:CHI760Delectrochemicalstation

Threeelectrodesystem.WE:CE:RE:SaturatedCalomelElectrode

Solution:1.0×10³mol/LK₃[Fe(CN)6]+0.1MKCl

Labreport

1)PlotcurvesofLSVcurve,anddescribewhycurrentchangeswithsweepingvoltage?

Current/uA

Potential/V

Reason:Voltageisadrivingforcetoanelectrodereactions,itisconcernedwiththeequilibriumofelectrontransferatelectrodesurface.Asthealteringofappliedvoltage,theFermi-levelisraised(orlowered),whichchangingtheenergystateoftheelectrons.Makingtheoverallbarrierheight(ieactivationenergy)alterasafunctionoftheappliedvoltage.

(1).Inthisreaction,whenvoltageis0.6V,thereisnoelectrontransfer,sothecurrentiszero.Withthevoltagetothemorereductivevalues,thecurrentincreases.

(2).WhenthediffusionlayerhasgrownsufficientlyabovetheelectrodesothatthefluxofreactanttotheelectrodeisnotfastenoughtosatisfythatrequiredbyNernstEquation.Thepeakisobtaining.

(3).Whenthereactioncontinued,itwouldgetasituationthattherewillbealowerreactantconcentrationattheelectrodethaninbulksolution,thatis,thesupplyoffreshreactanttothesurfacedecreased,socurrent

decreases.

2)PlotthecurvesofCVcurveswithdifferentscanrate;

79).

Current/A

Potential/V

3)FromtheCVcurves,fillthetable

Scanrate(mV/s)

20

50

100

200

300

400

500

600

Peak

current(

uA)

lpe

8.336

13.17

18.50

25.96

31.54

36.17

40.23

43.95

lpa

-8.263

-13.01

-18.19

-25.26

-30.50

-34.88

-38.68

-42.12

RatioofPeakcurrent

1.009

1.012

1.017

1.028

1.034

1.037

1.040

1.043

Peak

voltagE

(V)

V1

0.171

0.189

0.191

0.190

0.187

0.186

0.183

0.183

V2

0.242

0.255

0.259

0.262

0.262

0.262

0.262

0.262

Peakvoltage

difference(mV)

71

66

68

72

75

76

79

79

4)Accordingtotheresult,describewhycurvesshowscertaintrend,andhowpeakcurrentandpeakvoltagedifferencechangewithscanrate?

Answer:Fromabovedataandcurve,wecanobtain:

A.Atafixedscanrate:(1).frominitialpositivevoltagetomorereductivevalues,thecurrentbegintoflow,thenreachapeakipeanddecreaseeventually.(2).whenvoltagemovesback,theequilibriumpositionsgraduallyconvertingelectrolysisproduct(Fe²+)backtoreactant(Fe³+),thecurrentflowisfromthesolutionspeciesbacktotheelectrodeandsooccursintheoppositesensetotheforward.Theprocesshasanothercurrentpeakipa.Ithassamereasonoflinearsweepvoltammetry.

B.Atdifferentscanrate,theratioofpeakcurrentip/ipaisaboutequalto1(1.009-1.043).

C.Atdifferentscanrate,thepositionofpeakvoltagedonotaltergreatly△Episaboutaconstant(66--

v(mV/s)

20

50

100

200

300

400

500

600

v1/2(mV/s)¹/2

4.472

7.071

10

14.142

17.321

20

22.361

24.495

Ip(uA)

8.336

13.17

18.50

25.96

31.54

36.17

40.23

43.95

Ep(V)

0.171

0.189

0.191

0.190

0.187

0.186

0.183

0.183

Ep'(mV)

71

66

68

72

75

76

79

79

(1)Ip--V1/2

vn/dI

v¹/2/(mV/s)¹/2

Figure1:PeakcurrentVSradicalsignofscanrate

Interpretation:Itisapparentthatthepeakcurrentislineartoradicalsignofscanrate,whichsatisfythisequation:ip=kv¹/2Co.

Reason:Thiscanberationalisedbyconsideringthesizeofthediffusionlayerandthetimetakentorecordthescan.Clearlythelinearsweepvoltammogramwilltakelongertorecordasthescanrateisdecreased.Thereforethesizeofthediffusionlayerabovetheelectrodesurfacewillbedifferentdependinguponthevoltagescanrateused.Inaslowvoltagescanthediffusionlayerwillgrowmuchfurtherfromtheelectrodeincomparisontoafastscan.Consequentlythefluxtotheelectrodesurfaceisconsiderablysmalleratslowscanratesthanitisatfasterrates.Asthecurrentisproportionaltothefluxtowardstheelectrode,themagnitudeofthecurrentwillbeloweratslowscanratesandhigherathighrates.

(2)E₀--V

Ep(V)

v(mV/s)

Figure2:PeakpotentialVSscanrate

(3)Ep'--v

Ep'(mV)

v(mV/s)

Figure3:thechangeofPeakpotentialVSscanrate

Interpretation:Figure2:showsthatthepositionofthepeakcurrentoccursatthesamevoltage.Figure3:showsthePe