The lab report below was submitted as part of the coursework for
CM3292 Advanced Experiements in Analytical and Physical Chemistry. Please do not plagiarise from it as
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AIM
(A) To determine the E0’ value and n for the FeIII(CN)63-/ FeII(CN)64- couple in 0.1M KNO3 from the 2mM cyclic voltammogram
(B) To determine the effect of scan rate on peak height of the cyclic voltammogram
(C) To determine the concentration of unknown K3Fe(CN)6 solution using a calibration graph of concentration vs. peak height
(D) To determine the effect of the supporting electrolyte on the cyclic voltammogram
(A) To determine the E0’ value and n for the FeIII(CN)63-/ FeII(CN)64- couple in 0.1M KNO3 from the 2mM cyclic voltammogram
(B) To determine the effect of scan rate on peak height of the cyclic voltammogram
(C) To determine the concentration of unknown K3Fe(CN)6 solution using a calibration graph of concentration vs. peak height
(D) To determine the effect of the supporting electrolyte on the cyclic voltammogram
RESULTS and CALCULATIONS
Preparation of varying concentrations of K3Fe(CN)6
solutions
Concentration /mM
|
Volume of 50mM K3Fe(CN)6
solution used / mL
|
Volume of 0.1M KNO3
solution used / mL
|
Final volume / mL
|
2
|
2.00
|
48.00
|
50.00
|
4
|
4.00
|
46.00
|
50.00
|
6
|
6.00
|
44.00
|
50.00
|
8
|
8.00
|
42.00
|
50.00
|
10
|
10.00
|
40.00
|
50.00
|
(A) Determine the E0’ value
and n
From the cyclic voltammogram of 2mM K3Fe(CN)6 in 0.1M KNO3, Epa = 0.268 V, Epc = 0.185 V.
From the cyclic voltammogram of 2mM K3Fe(CN)6 in 0.1M KNO3, Epa = 0.268 V, Epc = 0.185 V.
Therefore, E0’ = (Epa
+ Epc) ÷ 2 = (0.268 + 0.185) ÷ 2 = 0.2265 V ≈ 0.227 V
Percentage
difference between literature value and experimental value = (0.227-0.225) /
0.225 x 100% =0.667 %. This small percentage difference between the literature
and experimental value implied that the E0’
obtained experimentally has a rather high degree of accuracy.
ΔEp = Epa – Epc
= (0.268 – 0.185)≈ 0.059 ÷ n
n ≈ 0.059 ÷ 0.144 ≈ 0.410 ≈ 1
n ≈ 0.059 ÷ 0.144 ≈ 0.410 ≈ 1
Hence, the number of electrons
transferred in the reaction for FeIII(CN)63-/
FeII(CN)64-couple in
0.1M KNO3 is deduced to be 1.
(B)
Determine the effect of scan rate on peak height
4mM K3Fe(CN)6 in
0.1M KNO3 was used. From the cyclic voltammograms at different scan
rates,
Scan Rate, v (mV/s)
|
ipa (×10-4A)
|
ipc (×10-4A)
|
lipa/ipcl
|
v1/2 (mV/s)1/2
|
20
|
3.514
|
-2.821
|
1.26
|
4.47
|
50
|
5.279
|
-4.541
|
1.16
|
7.07
|
75
|
6.223
|
-5.440
|
1.14
|
8.66
|
100
|
6.970
|
-6.102
|
1.14
|
10
|
Table 1 – Results for Scan Rate and
Peak Height for 4mM K3Fe(CN)6 in 0.1M KNO3
From Table 1, the graphs of ipa and ipc vs. v1/2
are plotted:
Figure 1 – Graph of
ipa and ipc against v1/2
|
Both the graph of ipc and ipa against v1/2
gave a linear plot with R2 greater than 0.99. This indicates that
the ipc and ipa were directly proportional to the
square root of the scan rate. From the graph, it also shows that as the scan
rate increases, the peak current also increases.
|
(C) Determine the concentration of unknown K3Fe(CN)6
solution
From the cyclic voltammograms at
concentrations of K3Fe(CN)6,
[K3Fe(CN)6] (mM)
|
ipa (×10-4A)
|
ipc (×10-4A)
|
lipa/ipcl
|
2
|
1.784
|
-1.595
|
1.12
|
4
|
3.514
|
-2.821
|
1.25
|
6
|
5.299
|
- 4.764
|
1.11
|
8
|
6.820
|
-6.099
|
1.12
|
10
|
8.365
|
-7.506
|
1.11
|
Table 2 – Results for Peak Height at
varying concentrations of K3Fe(CN)6
From Table 2, the graphs of ipa and ipc vs. [K3Fe(CN)6] are plotted:
Figure 2 – Graph of ipa and ipc against [K3Fe(CN)6]
From the cyclic voltammogram of the
unknown solution, the peak heights are determined to be:
ipa
= 4.378 ×10-4A, ipc = -4.002 ×10-4A
Using ipa calibration
graph, the
equation of the best-fit line is determined to be y = 0.823x + 0.216
Substituting the ipa value for the unknown solution, the concentration of the unknown solution was determined to be 5.057mM.
Substituting the ipa value for the unknown solution, the concentration of the unknown solution was determined to be 5.057mM.
Using ipc calibration graph, the equation of the best-fit line is
determined to be y = -0.755x – 0.027
Substituting the ipc value for the unknown solution into the equation, the concentration of the unknown solution was determined to be 5.265mM.
Substituting the ipc value for the unknown solution into the equation, the concentration of the unknown solution was determined to be 5.265mM.
Thus, average value for the
concentration of the unknown K3Fe(CN)6 solution
= (5.265 + 5.057) ÷2 = 5.161mM
= (5.265 + 5.057) ÷2 = 5.161mM
(D) Determine the effect of the
supporting electrolyte
From
the cyclic voltammogram in different supporting electrolytes,
Solution
|
Scan Rate
(mV/s) |
Epa
(V) |
Epc
(V) |
ipa
(×10-4A) |
ipc
(×10-4A) |
E0’
(V) |
ΔEp
(V) |
4mM K3Fe(CN)6 in 0.1M KNO3
|
20
|
0.275
|
0.173
|
3.514
|
-2.821
|
0.224
|
0.102
|
4mM K3Fe(CN)6 in 0.1M KCl
|
20
|
0.278
|
0.182
|
3.424
|
-3.000
|
0.230
|
0.096
|
Table 3 – Results for the two
supporting electrolytes: KNO3 and KCl
From Table 3, it can be seen that the Epa
and Epc values for both electrolytes are similar. The peak height
values, ipa and ipc, are slightly lower for the KCl
electrolyte.
DATA ANALYSIS
(A)
Determine the E0’ value and n
The E0’ value from the experiment was determined to be 0.227V. The experimental value deviates slightly from the literature value of 0.225V (by 0.667%). This may be due errors in the extrapolation of the baseline in the cyclic voltammogram. The temperature at which the experiment was conducted could be different from the conditions used for the literature value. A difference in temperature can affect the current flow through the solution and result in a deviation. These random errors should be minimised by taking more readings.
The E0’ value from the experiment was determined to be 0.227V. The experimental value deviates slightly from the literature value of 0.225V (by 0.667%). This may be due errors in the extrapolation of the baseline in the cyclic voltammogram. The temperature at which the experiment was conducted could be different from the conditions used for the literature value. A difference in temperature can affect the current flow through the solution and result in a deviation. These random errors should be minimised by taking more readings.
(B)
Determine the effect of scan rate on peak height
Voltammetric currents depend on the concentration gradient that is established very near the electrode during electrolysis. This is called the Nernst diffusion layer. The relationship between the scan rate and the current peak height can be explained by considering the size of the diffusion layer and the time taken to record the scan[1].
Voltammetric currents depend on the concentration gradient that is established very near the electrode during electrolysis. This is called the Nernst diffusion layer. The relationship between the scan rate and the current peak height can be explained by considering the size of the diffusion layer and the time taken to record the scan[1].
The
cyclic voltammogram will take longer to record as the scan rate is decreased.
At a slow scan rate, the diffusion layer will grow much further from the
electrode as compared to a fast scan. This leads to a concentration gradient to
the electrode surface that is much lower as compared to a fast scan. Thus, the
magnitude of the current is directly proportional to the scan rate[2].
According to the Randles-Sevcik equation[3],
ip = 2.686 × 105n3/2AD1/2Cv1/2
or simply put, the peak current, ip is directly proportional to the square root of the scan rate, v1/2 when other factors are kept constant. This is confirmed from Figure 1 which shows that the experimental values conform closely to the linear relationship between ip and v1/2 (with R2 values above 0.99).
ip = 2.686 × 105n3/2AD1/2Cv1/2
or simply put, the peak current, ip is directly proportional to the square root of the scan rate, v1/2 when other factors are kept constant. This is confirmed from Figure 1 which shows that the experimental values conform closely to the linear relationship between ip and v1/2 (with R2 values above 0.99).
(C)
Determine the concentration of unknown K3Fe(CN)6 solution
The relationship between peak current and concentration can be explained by considering the concentration gradient at the diffusion layer. Just outside the diffusion layer, is the bulk solution being analysed. The concentration of the bulk solution is approximately the concentration of the solution and the concentration of the analyte at the surface of the electrode is close to zero (at the potential which peak current occurs). This leads to a concentration gradient and subsequently, the voltammetric current. Thus, a solution of higher concentration will lead to a greater concentration gradient as the difference between the concentration at the electrode surface and the bulk solution is higher. This leads to a higher peak current.
The relationship between peak current and concentration can be explained by considering the concentration gradient at the diffusion layer. Just outside the diffusion layer, is the bulk solution being analysed. The concentration of the bulk solution is approximately the concentration of the solution and the concentration of the analyte at the surface of the electrode is close to zero (at the potential which peak current occurs). This leads to a concentration gradient and subsequently, the voltammetric current. Thus, a solution of higher concentration will lead to a greater concentration gradient as the difference between the concentration at the electrode surface and the bulk solution is higher. This leads to a higher peak current.
From
the Randles-Sevcik equation, the peak current, ip is directly
proportional to the concentration of the solution, C when other factors are
kept constant. This allows us to construct a calibration graph from the anodic
peak current, ipa and the cathodic peak current, ipc to
determine the concentration of the unknown solution. Figure 2 confirms the linear relationship between concentration and
peak current. The experimental data conform to linearity with R2
values higher than 0.995. Thus, the concentration obtained for the unknown
solution is considered quite accurate and determined to be 5.161mM.
(D) Determine the effect of the
supporting electrolyte
A supporting electrolyte is added to reduce the effects of migration of the analyte in the solution. It is most commonly an alkali metal salt that is added in excess which does not react at the working electrode at the potentials being used. The current in the cell is primarily due to charges carried by the excess of ions from the supporting electrolyte instead of the analyte and this reduces the effects of migration. The supporting electrolye also reduces the resistance of the solution[3].
A supporting electrolyte is added to reduce the effects of migration of the analyte in the solution. It is most commonly an alkali metal salt that is added in excess which does not react at the working electrode at the potentials being used. The current in the cell is primarily due to charges carried by the excess of ions from the supporting electrolyte instead of the analyte and this reduces the effects of migration. The supporting electrolye also reduces the resistance of the solution[3].
The magnitude of the current is
also determined by the rate of mass transport of ions to the edge of the Nernst
diffusion layer by convection and also the rate of transport of ions from the
outer edge of the diffusion layer to the electrode surface[5]. Ions
of a lower mass should have greater mobility. Thus, the supporting electolyte
which has a lower molecular mass should give a higher current. This is
confirmed by the experimental results. The cation of the two electrolytes are
the same. They differ in their anions: Cl- with a molar mass of
35.5g/mol and NO3- with a molar mass of 63.0g/mol. Since
the charges on the anions are the same, the difference in their mobilities is thus
due to their molar masses. As expected and observed, the ipc peak
current for the solution with KCl as the supporting electrolyte is higher than that
with KNO3. However, the ipa peak current for the former
is unexpectedly lower than the latter. This erratic result may be due to temperature
fluctuations.
Precautions and Other comments
Before recording the cyclic voltammogram, each solution was swirled to ensure homogenity. However, during the recording of the cyclic voltammogram, it was ensured that the solution was not perturbed in any way to prevent inaccuracies. The electrodes were cleaned and blotted dry before changing the solution to be analysed in order to prevent contamination.
Before recording the cyclic voltammogram, each solution was swirled to ensure homogenity. However, during the recording of the cyclic voltammogram, it was ensured that the solution was not perturbed in any way to prevent inaccuracies. The electrodes were cleaned and blotted dry before changing the solution to be analysed in order to prevent contamination.
The ipa/ipc values of all the cyclic voltammograms
are close to 1. This shows that the FeIII(CN)63-/ FeII(CN)64-
reaction is a reversible electrode reaction with no other reactions involved. The
accuracy of the experiments may be improved by controlling the temperatures of
the solutions and reducing temperature fluctuations.
CONCLUSION
For the FeIII(CN)63-/ FeII(CN)64- couple in 0.1M KNO3 from the 2mM cyclic voltammogram, the E0’ value is determined to be 0.227V and n is determined to be 1.
For the FeIII(CN)63-/ FeII(CN)64- couple in 0.1M KNO3 from the 2mM cyclic voltammogram, the E0’ value is determined to be 0.227V and n is determined to be 1.
The
concentration of the unknown K3Fe(CN)6 solution is
determined to be 5.161mM.
The
current peak height varies linearly with the square root of the scan rate, v1/2.
The supporting electrolyte with the lower molar mass produces a higher peak
current.
REFERENCES
[1]
Skoog et. al., Fundamentals of
Analytical Chemistry, 8th Edition, 2004, Brooks/Cole.
[2]
P.T. Kissinger and W.E. Heineman, Laboratory Techniques in Electroanalytical
Chemistry, 2nd Edition, 1996, Dekker.
[3]
E.Gileadi, E. Kirowa-Eisner and J. Penciner, Interfacial Electrochemistry, QD
571 G54 1975.
Where did you find that literature value for the standard reduction potential you used in your calculations?
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