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1、Experimental class on “ Fuel Cell and ElectrochemistryExperiment setupEquipment: CHI760D electrochemical stationThree electrode system. WE: CE: RE: Saturated Calomel ElectrodeSolution: 1 .0 X 10-3mol/L K 3 Fe (CN) 6 + 0,1M KClLab reportPlot curves of LSV curve, and describe why current changes with

2、sweeping voltage?Reason: Voltage is a driving force to an electrode reactions, it is concerned with the equilibrium of electron transfer at electrode surface. As the altering of applied voltage, the Fermi-level is raised (or lowered), which changing the energy state of the electrons. Making the over

3、all barrier height (ie activation energy) alter as a function of the applied voltage. In this reaction, when voltage is 0.6V, there is no electron transfer, so the current is zero.With the voltage to the more reductive values, the current increases. When the diffusion layer has grown sufficiently ab

4、ove the electrode so that the flux of reactant to the electrode is not fast enough to satisfy that required by Nernst Equation. The peak is obtaining. When the reaction continued, it would get a situation that there will be a lower reactant concentration at the electrode than in bulk solution, that

5、is, the supply of fresh reactant to the surface decreased, so current decreases.Plot the curves of CV curves with different scan rate;0.00006H-600mv/s G-500mv/s F-400mv/s E-300mv/s D-200mv/s C-100mv/s B- 50 mv/s A- 20 mv/s0.000040.000020.00000-0.00002-0.000040.60.40.20.0-0.2-0.4araR nacs0.00006H-600

6、mv/s G-500mv/s F-400mv/s E-300mv/s D-200mv/s C-100mv/s B- 50 mv/s A- 20 mv/s0.000040.000020.00000-0.00002-0.000040.60.40.20.0-0.2-0.4araR nacsPotential/VScan rate (mV/s)2050100200300400500600Peak current (uA)Ipc8.33613.1718.5025.9631.5436.1740.2343.95Ipa-8.263-13.01-18.19-25.26-30.50-34.88-38.68-42.

7、12Ratio of Peak current1.0091.0121.0171.0281.0341.0371.0401.043Peak voltag E(V)V10.1710.1890.1910.1900.1870.1860.1830.183V20.2420.2550.2590.2620.2620.2620.2620.262Peak voltage difference (mV)7166687275767979From the CV curves, fill the table3)4) According to the result, describe why curves shows cer

8、tain trend, and how peak current and peak voltage difference change with scan rate?Answer: From above data and curve, we can obtain:At a fixed scan rate : (1).from initial positive voltage to more reductive values, the current beginto flow, then reach a peak ipc and decrease eventually. (2).when vol

9、tage moves back, the equilibrium positions gradually converting electrolysis product (Fe 2+) back to reactant (Fe 3+), the current flow is from the solution species back to the electrode and so occurs in the oppositesense to the forward. The process has another current peak ipa. It has same reason o

10、f linear sweep voltammetry.At different scan rate, the ratio of peak current ip c/ip a is about equal to 1 (1.009 1.043).At different scan rate, the position of peak voltage do not alter greatly C Ep is about a constant (66-79).mV/s)20501002003004005006001/2V(mV/s) 1/24.4727.0711014.14217.3212022.36

11、124.495Ip (uA)8.33613.1718.5025.9631.5436.1740.2343.95Ep(V)0.1710.1890.1910.1900.1870.1860.1830.183Ep (mV)7166687275767979 Ip-V1/2Figure 1: Peak current VS radical sign of scan rate Interpretation:It is apparent that the peak current is linear to radical sign of scan rate, whichsatisfy this equation

12、: ip = kv 1/2Co.Reason:This can be rationalised by considering the size of the diffusion layer and the time taken to record the scan. Clearly the linear sweep voltammogram will take longer to record as the scan rate is decreased. Therefore the size of the diffusion layer above the electrode surface

13、will be different depending upon the voltage scan rate used. In a slow voltage scan the diffusion layer will grow much further from the electrode in comparison to a fast scan. Consequently the flux to the electrode surface is considerably smaller at slow scan rates than it is at faster rates. As the

14、 current isproportional to the flux towards the electrode, the magnitude of the current will be lower at slow scan rates and higher at high rates.Ep-VFigure 2: Peak potential VS scan rateEp-v1601200 -|1|1|1|-|1-i0100200300400500600700v (mV/s)Figure 3: the change of Peak potential VS scan rateInterpretation: Figure 2: shows that the position of the peak current occurs at the same voltage.Figure 3: shows the Pea