电化学实验报告实验报告

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 K 3 [ Fe (CN)6] + 0.1M KCl

Lab report

1) Plot curves of LSV curve, and describe why current changes with sweeping voltage?

0.6

0.4

0.2

0.0

-0.2

-0.4

0.000000

0.000002

0.000004

0.000006

0.000008

0.000010

Potential/V

C u r r e n t /u A

Scan Rate: 20mV/s

Reason: V oltage 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;

0.6

0.4

0.2

0.0

-0.2

-0.4

-0.00004

-0.00002

0.00000

0.00002

0.00004

0.00006

Potential/V

C u r r e n t /A

A--- 20 mv/s

H---600mv/s B--- 50 mv/s C---100mv/s D---200mv/s E---300mv/s F---400mv/s G---500mv/s S c a n R a t e

A

B C D

E

F G H

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

20 50 100 200 300 400 500 600 Peak current (uA)

Ip c 8.336 13.17 18.50 25.96 31.54 36.17 40.23 43.95 Ip a

-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 voltag

E(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 ip c and decrease eventually. (2).when voltage 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 opposite