Differnetial Pulse Voltammetry of Ferricyanide PDF

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Introduction: There are three common types of pulse voltammetry. These include Normal Pulse Voltammetry (NVP), Differential Pulse Voltammetry (DVO) and Square Wave Voltammetry (SWV). In this lab, we performed Differential Pulse Voltammetry of Ferricyanide. Differential pulse voltammetry is the third type of voltammetry we have done so far with ferricyanide. The purpose of this lab was to determine the ½ cell potential for the ferri/ferrocyanide reaction and compare it to the value stated in literature and the values we obtained in the other two voltammetry labs, cyclic and hydrodynamic, as well as comparing the different types of pulse voltammetry previously mentioned. All pulse voltammetry is the difference in the rate of decay of the nonfaradiac current, also called the charging current, and the faradic current following a pulse. In voltammetry, we use a solid electrode. In all pulse techniques, there are four important parameters: pulse amplitude, pulse width, sample period, and pulse period. In normal pulse voltammetry, the potential returns to the initial value after a series of pulses of increasing amplitude.

Figure 1: Normal pulse waveform

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Figure 2: Normal pulse voltammogram

In differential pulse voltammetry, the potential is scanned with a series of pulses like in normal pulse voltammetry, but each potential is of a fixed small amplitude unlike normal pulse voltammetry.

Figure 3: Differential pulse waveform

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DPV of Ferricyanide

Current (A)

1.2E-04 1.0E-04

2 mM

8.0E-05

4 mM 6 mM

6.0E-05

8 mM 4.0E-05

10 mM

2.0E-05 0.0E+00 300.00 250.00 200.00 150.00 100.00

50.00

0.00

-50.00

Potential (V)

Figure 4: Differential pulse voltammograms

Lastly square wave voltammetry consists of a symmetrical square wave pulse of amplitude. The potential is superimposed on a staircase waveform. The peak height is directly proportional to the concentration.

Figure 5: Square waveform

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Figure 6: Square wave voltammogram Equation 1 gives the height of the peak which is proportional to the concentration of the electroactive species. 1

nFAD 2 C ip = √ πt m

Equation 1

In equation 1 ip is the peak current (A), F is Faraday’s constant, A, is the working area of the electrode (cm2), D is the diffusion coefficient of the analyte (cm2/sec), C is the concentration of the analyte (mol/cm3), and tm is the time after the application of the potential when the current is sampled (sec) To find the ½ cell potential equation 2 is used E1/2=(Ep + Δ E )/2

Equation 2

E is the pulse amplitude (mV), Ep is the cell potential, and E1/2 is the half-cell potential. In this lab, the half-cell potential was found to be 232mV. To find the concentration of the unknown ferricyanide equation 3 is used

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Unknown Concentration =

i p unknown +b m

Equation 3

The components of this equation are found using the calibration curve plot, and the unknown concentration value for this lab was 5.94mM.

Procedure: There were no changes to the procedure, it followed the lab handout: DPV CHEM 2410L fall 2018. Instrumentation:

Working electrode Reference electrode Auxiliary electrode potentiostat Stirring unit

Model Number MF-2012 MF-2052 MW-1033 RDE-2 57945-016

Company BASi BASi BASi BASi Van Waters & Krogers Mangestir

Reagents: chemicals

supplier

Lot number

Nitrogen

Airgas Nitrogen Company Fisher company Fisher company

Potassium ferricyanide Potassium nitrate

Expiration date

UN1066

Molecular weight (g/mol) NA

16211

101.10

NA

770521

329.25

NA

NA

Calculation 1: Determine gramps of KNO3 needed to make 1.0M stock solution 1.0 mol/L * 1.0 L * 101.10 g/mol = 101.10 g KNO 3 needed

6 Amounts of reagents used: Chemical K3Fe(CN)6 KNO3 Unknown K3Fe(CN)6

Amount Used (g) 0.8136 101.1034 Prepared by TA

Calculation 2: Determine the volume of K3Fe(CN)6 stock solution needed to make diluted solutions of 2, 4, 6, 8, and 10mM K3Fe(CN)6 (10mM * 50mL)/10mM = 50mL K3Fe(CN)6 needed to make a 10mM solution (8mM * 50mL)/10mM = 40mL K3Fe(CN)6 needed to make an 8mM solution (6mM * 50mL)/10mM = 30mL K3Fe(CN)6 needed to make a 6mM solution (4mM * 50mL)/10mM = 20mL K3Fe(CN)6 needed to make a 4mM solution (2mM * 50mL)/10mM = 10mL K3Fe(CN)6 needed to make a 2mM solution

Preparation of Standards: Concentration (mM) 2 4 6 8 10

Volume of K3Fe(CN)6 (mL) 10 20 30 40 50

Data Results and Discussion:

Volume of KNO3 (mL) 40 30 20 10 0

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2mM Differential

Pulse Voltammetry of Ferricyanide

4mM 6mM

0

8mM 10mM

0

Current (A)

0 0 0 0 0 0 -4.00E-01 -2.00E-01 0.00E+00 2.00E-01 4.00E-01 6.00E-01 8.00E-01 1.00E+00

Potential (mV)

Figure 7: Differential Pulse Voltammograms of 2, 4, 6, 8, and 10 mM Ferricyanide, working electrode was glassy carbon, reference electrode Ag|AgCl, and a platinum wire auxiliary electrode Table 1. Cathodic and anodic peak currents for 2, 4, 6, 8, and 10 mM Ferricyanide

Concentration (mM) 2 4 6 8 10

Minimum current (A) 1.19E-06 1.28E-06 1.30E-06 1.25E-06 1.37E-06

Maximum current (A) 3.35E-05 6.06E-05 8.57E-05 1.13E-04 1.30E-04

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Calibration Curve 1.40E-04 f(x) = 0 x + 0 R² = 0.99

1.20E-04

Current (A)

1.00E-04 8.00E-05 6.00E-05 4.00E-05 2.00E-05 0.00E+00 0.00E+00 2.00E+00 4.00E+00 6.00E+00 8.00E+00 1.00E+01 1.20E+01

Concentration (mM)

Figure 8: Calibration curve of maximum peak current percent concentration of ferricyanide

DPV of Unknown Concentration of Ferricyanide 8.00E-05 7.00E-05

Current (A)

6.00E-05 5.00E-05 4.00E-05 3.00E-05 2.00E-05 1.00E-05 0.00E+00 -4.00E-01-2.00E-010.00E+00 2.00E-01 4.00E-01 6.00E-01 8.00E-01 1.00E+00

Potential m(V)

Figure 9: Differential Pulse voltammogram of unknown solution of ferricyanide, working electrode was glassy carbon, reference electrode Ag|AgCl, and a platinum wire auxiliary electrode Table 2. Data for determining the concentration of the unknown ferricyanide solution

unknown

Peak Current (A) 6.94E-05

Calibration Line y=1.00E-05x+1.00E-05

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Calculation 3: Determination of the unknown concentration of ferricyanide: y=mx+b  x=(y-b)/m  x=(6.94E-05-1.00E-05)/(1.00E-05)  x=5.94mM The concentration of ferricyanide in the unknown calculated form the anodic peak current was found to be 5.94mM The unknown solution that the TA prepared was about 5.00mM so this is a reasonable calculation. Calculation 4: Determination of the half-call potential using equation two E1/2=(335mV+130mV) /2

E1/2=232mV

Calculation 5: Finding the percent difference from CV, HDV, and the literature CV % difference = ½ * |CV-DPV|/CV+DPV)*100 CV % difference = ½*|192mV-232mV|/ (192mV+232mV)*100 CV % difference = 4.72%

HDV % difference = ½ * |HDV-DPV|/CV+DPV)*100 HDV % difference = ½*|165mV232mV|/(165mV+232mV)*100 HDV % difference = 8.34% The percent difference from the literature value (225mV) can be found using the following equation: (experimental – actual)/actual * 100  (232mV-225mV)/225mV*100 = 3.11%

Conclusion and Opinion: The biggest issue we faced in this lab is the same as the previous two labs which is that the lab group was very big and not everyone got to use the instrumentation. A solution for this issue would be to have smaller lab groups, but that would be unlikely due to the fact that there is

10 only one instrument and a large number of students that take this course.

References: 1. Experiment 3Differential Pulse Voltammetry of Ferricyanide, Laboratory Handout, Chemistry 2410L, Fall Semester 2018...