2022年电化学实验报告实验报告

1、Experimental class on“Fuel Cell and Electrochemistry”Experiment setupEquipment: CHI760D electrochemical stationThree electrode system. WE: CE: RE: Saturated Calomel ElectrodeSolution: 1.0 10-3mol/L K3 Fe (CN)6 + 0.1M KClLab report Plot curves of LSV curve, and describe why current changes with sweep

2、ing 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 overall b

3、arrier height (ie activation energy) alter as a function of the applied voltage. (1). 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. (2). When the diffusion layer has grown sufficien

4、tly above 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.(3). When the reaction continued, it would get a situation that there will be a lower reactant concentration at the electrode than in bulk solut

5、ion, that is, the supply of fresh reactant to the surface decreased, so current decreases.Plot the curves of CV curves with different scan rate;From the CV curves, fill the tableScan rate (mV/s)2050100200300400500600Peak current(uA)Ipc8.33613.1718.5025.9631.5436.1740.2343.95Ipa-8.263-13.01-18.19-25.

6、26-30.50-34.88-38.68-42.12Ratio of Peak current1.0091.0121.0171.0281.0341.0371.0401.043Peak voltagE(V)V10.1710.1890.1910.1900.1870.1860.1830.183V20.2420.2550.2590.2620.2620.2620.2620.262Peak voltage difference (mV)7166687275767979According to the result, describe why curves shows certain trend, and

7、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 begin to flow, then reach a peak ipc and decrease eventually. (2).when voltage moves back

8、, the equilibrium positions gradually converting electrolysis product (Fe2+) back to reactant (Fe3+), the current flow is from the solution species back to the electrode and so occurs in the opposite sense to the forward. The process has another current peak ipa. It has same reason of linear sweep v

9、oltammetry.At different scan rate, the ratio of peak current ipc/ipa is about equal to 1 (1.0091.043).At different scan rate, the position of peak voltage do not alter greatly Ep is about a constant (66-79).(mV/s)20501002003004005006001/2 (mV/s) 1/24.4727.0711014.14217.3212022.36124.495Ip (uA)8.3361

10、3.1718.5025.9631.5436.1740.2343.95Ep(V)0.1710.1890.1910.1900.1870.1860.1830.183Ep(mV)7166687275767979(1) Ip-V1/2Figure 1: Peak current VS radical sign of scan rateInterpretation: It is apparent that the peak current is linear to radical sign of scan rate, which satisfy this equation: ip = kv1/2C0. R

11、eason: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 will be different

12、 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 current is propo

13、rtional to the flux towards the electrode, the magnitude of the current will be lower at slow scan rates and higher at high rates.(2) Ep-V Figure 2: Peak potential VS scan rate(3) Ep-v 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 Peak voltage differen