2023年电化学实验报告实验报告.doc

Experimental class on“Fuel Cell and Electrochemistry”
Experiment setup
Equipment: CHI760D electrochemical station
Three electrode system. WE: CE: RE: Saturated Calomel Electrode
Solution: 1.0 ×10-3mol/L K3 [ Fe (CN)6] + 0.1M KCl

Lab report
1) Plot curves of LSV curve, and describe why current changes with 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 overall barrier 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 sufficiently 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 solution, that is, the supply of fresh reactant to the surface decreased, so current decreases.

2) Plot the curves of CV curves with different scan rate;

3) From the CV curves, fill the table
Scan rate (mV/s)

20
50
100
200
300
400
500
600
Peak current(uA)
Ipc
8.336
13.17
18.50
25.96
31.54
36.17
40.23
43.95
Ipa
-8.263
-13.01
-18.19
-25.26
-30.50
-34.88
-38.68
-42.12
Ratio of Peak current
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
Peak voltage difference (mV)
71
66
68
72
75
76
79
79

4) According to the result, describe why curves shows certain trend, and how peak current and peak voltage difference change with scan rate?

Answer: From above data and curve, we can obtain:
A. 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, 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 voltammetry.
B. At different scan rate, the ratio of peak current ipc/ipa is about equal to 1 (1.009—1.043).
C. At different scan rate, the position of peak voltage do not alter greatly ΔEp is about a constant (66--79).

ν(mV/s)
20
50
100
200
300
400
500
600
ν1/2 (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

Figure 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, which satisfy this equation: ip = kv1/2C0.
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 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 current is proportional 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 rate

Interpretation: Figure 2: shows that the position of the peak current occurs at the same voltage.
Figure 3: shows the Peak voltage difference is constant in different scan rate.

Reason: The characteristic of this electrode reaction has rapid electron transfer kinetics and there is no charge transfer in double layer, so this reaction is reversible.