Student name:_________________________________________

CHEM 1079 Analytical Science

Practical: Voltammetry Simulation

Due date:

The program Polar41c has been downloaded onto the computers in the Department's computer lab. You can download a copy yourself (as a zip file) from http://www.DrHuang.com, unzip and install it on your own computer. Note that the program is too large to fit on a floppy disc.

In this exercise you will perform a number of experiments on a voltammetry simulator. You may not have covered this material in lectures and should refer to your CHEM 1079 lecture notes and textbook (Harris, Chapter 18).

DC Voltammetry

In voltammetry, an inert electrode (eg., Pt, C, Hg) is placed in a solution containing electroactive species. (Polarography is a subset of voltammetry that uses Hg electrodes.) The potential of the electrode is changed as a function of time:

If an electrochemical reaction occurs, electrons are transferred and a current flows. The current is measured as a function of the potential applied to the electrode. A voltammogram is a plot of the current, i, on the vertical axis versus the potential, E, on the horizontal axis. Cathodic (reduction) currents are shown in the positive y-direction and negative potentials are shown in the positive x-direction.

Explain the purposes of the three electrodes used in voltammetry.

In this series of experiments we are looking at the two-electron reduction of Cu²⁺ to Cu. The E^o of the reduction is +0.34 V relative to the standard hydrogen electrode, or +0.10 V relative to the saturated calomel electrode, SCE, (which is more commonly used as a reference in voltammetry).

If we apply a potential greater (more positive) than 0.10 V vs. SCE, the reaction

Cu²⁺ + 2e ® Cu

will not occur. In voltammetry of metal ions, we start with an applied potential well in excess (more positive) of the E^o of the metal ion present and scan the potential in the negative direction as a linear function of time. As the potential approaches 0.10 V, the reaction above begins to occur and a current flows. As the potential is brought closer to 0.1 V, this current increases until, a little way past 0.1 V, it flattens out. See for yourself:

Load the program from Windows:

· Start/ Programs/ Polar/ Polar

· Maximise

Now run the simulation:

· Input/ Techniques ® DC voltammetry

· Input/ Mechanism ® A + 2e®B

· Run/ Simulate

Sketch the polarogram (i versus E):

Note the gradual increase of current with voltage around the E^o. Current (coulombs per second) is a measure of the rate of the electrochemical reaction. The reaction starts slowly at first but is driven faster as the voltage is increased. Note also the levelling off of the current to form a plateau.

What causes this plateau? As Cu²⁺ ions near the surface of the electrode react to form Cu (that is deposited on to the electrode), the region close to the electrode is

depleted of Cu²⁺ ions. We have

to wait for more Cu²⁺ ions to diffuse from the bulk solution (where they are in high concentration) to the electrode surface (where they are in low concentration) before the reaction can continue. The rate of reaction, and hence the current, becomes limited by the rate of diffusion of ions from the bulk of the solution to the electrode surface, and the current plateaus.

There are a couple of parameters that can be obtained from the voltammogram that are of interest to us. The diffusion current , i_d, is the (diffusion) limited current at the top of the voltammogram. The half-wave potential, E_1/2, is the potential at i_d/2.

Find the values of i_d and E_1/2:

· Analysis/ Find Halfwave E

You can put gridlines in if you want:

· Display/ Option ® Grid

Show i_d and E_1/2, on the voltammogram you sketched on the previous page.

Let’s now look at the effect of changing the concentration of the Cu²⁺ ions will have on the voltammogram.

Run the simulation:

· Display/ Option ® Overlap (this enables the two voltammograms to be displayed on the same plot)

· Input/ Chemicals ® C_anal(A) = 2e-3

· Run/ Simulate

What has happened to i_d and E_1/2? Find out:

· Analysis/ Find Halfwave E

Sketch the voltammograms for the 1 x 10^-3 M and 2 x 10^-3 M solutions on the same axes.

You may well have expected to see a change in E_1/2 with concentration since this is what is seen in potentiometry (as described by the Nernst equation). In voltammetry, E_1/2 is unchanged by concentration, and it is i_d that is concentration dependent.

Try another couple of concentrations, tabulating the data (to 3 sig. figs. only) below:

[Cu^2+] / M

id / A

Now plot i_d versus concentration using the data you collected in the simulations in the space below. What do you notice?

The voltammogram for a copper solution of unknown concentration run under the same conditions has i_d = 7 x 10^-4 A. Using the calibration graph above, determine the concentration of the solution.

The diffusion-limited current is concentration dependent and so can be used, as above, as an analytical probe to determine the concentration of an unknown copper solution. To ensure the concentration dependence of the limiting current, we have to ensure that the Cu²⁺ ions move by diffusion (under the influence of a concentration gradient) and not by convection (stirring) or migration (under the influence of an electric field). Since we are applying a potential across the electrodes, the ions will move by migration – a positively charged ion will be attracted to a negatively charged electrode. To minimise the migration of the electroactive species, we add a large excess of electroinactive supporting electrode to the solution. The electroinactive ions move by migration, but the electroactive Cu²⁺ ions move predominantly by diffusion so the Faradaic current is proportional to concentration.

The linear relationship between i_d and concentration can be used as an analytical tool to determine the concentration of an electroactive species in solution. But how do we know what electroactive species is/are present? It's the value of E_1/2 that can indicate this. As you know, redox couples have distinct reduction potentials. The half-wave potential, E_1/2 is very closely related to the E^o values.

Let’s look at the voltammogram for the reduction of Pb²⁺ (E^o = - 0.40 V vs. SCE).

Run the simulation

· Display/ Option ® Overlap off

· Input/ Chemicals ® C_anal(A) = 1e-3

· Input/ Instrument ® E_end = -0.6 V

· Run/ Simulate

· Display/ Option ® Overlap on

· Input/ Chemicals E^o = - 0.40

· Run/ Simulate

· Analysis/ Find Halfwave E

Draw voltammograms for Cu²⁺ and Pb²⁺ on the same axes. E_1/2 can be used for identification.

Next we look at the effect of changing the number of electrons (n) involved in the reduction process. What do you think will happen to the voltammogram as you increase the number of electrons? Will the current that flows increase, decrease, or stay the same? Will the diffusion limited current, i_d, increase, decrease, or stay the same? Will there be any effect on E_1/2?

Do a series of experiments to investigate the effect of n:

· Display/ Option ® Overlap off

· Input/ Techniques ® DC voltammetry

· Input/ Chemicals ® E^o = 0.1

· Input/ Instrument ® E_end = - 0.3 V

· Input/ Mechanism ® A + 1e®B (default setting)

· Run/ Simulate

· Display/ Option ® Overlap on

· Input/ Mechanism ® A + 2e®B

· Run/ Simulate

· Input/ Mechanism ® A + 3e®B

· Run/ Simulate

· Analysis/ Find Halfwave E

Draw a series of voltammograms for n = 1, 2 and 3.

The diffusion limited current, i_d, increases linearly with n. The shape of the voltammogram is given by the equation:

E = E_1/2 + (0.0592/n)*log [(i_d - i)/i]

So, as n is increased the voltammetric wave (i versus E) becomes steeper.

Normal Pulse Voltammetry

Normal pulse (NP) voltammetry replaces the DC voltage ramp (the linear change of potential with time) with voltage pulses of increasing amplitude.

The current is sampled near the end of the pulse when the Faradaic current (due to the reduction of Cu²⁺) is large compared to the charging current (due to the migration of the electroinactive supporting electrolyte ions in response to the charging of the electrode) – ie., the signal-to-noise ratio is largest.

Run the simulation:

· Display/ Option ® Overlap off

· Input/ Chemicals ® D = 1e-5

· Run/ Simulate

· Display/ Option ® Overlap on

· Input/ Techniques ® Normal pulse voltammetry

· Run/ Simulate

· Analysis/ Find Halfwave E

Draw the DC and NP voltammograms on the same scale.

Were you expecting a similar shape wave? Has E_1/2 moved? How do the i_d's compare? Because the current is sampled near the end of the pulse when the Faradaic current (due to the reduction of Cu²⁺) is large compared to the charging current (due to the migration of ions in response to the charging of the electrode), the signal is increased (as you can see). But also the noise is decreased and hence the signal-to-noise ratio is improved by a factor of about 10. The detection limit of NP voltammetry is 10^-6 M, compared with 10^-5 M for DC.

Differential Pulse Voltammetry

In differential pulse (DP) voltammetry, small pulses are superimposed on a linear voltage ramp.

Current is measured before and after the application of the pulse and the difference in current is recorded as a function of the applied voltage. The resulting voltammogram is nearly the derivative of a DC voltammogram. A peak is produced.

Run the simulation:

· Display/ Option ® Overlap off

· Input/ Techniques ® DC voltammetry

· Run/ Simulate

· Display/ Option ® Overlap on

· Input/ Techniques ® Differential pulse voltammetry

· Run/ Simulate

· Analysis/ Find Peak

What peak parameter do you think is most likely to be used in place of E_1/2 to identify an electroactive species? What peak parameter do you expect to be related to the concentration of the electroactive species? Indicate these parameters on a sketch of a DP voltammogram (on the next page).

Run a series of experiments to test that the peak current, i_p, and the peak area are proportional to concentration, c. Collect data from the simulation

[Cu^2+] / M

ip / A

Peak Area

Now graph i_p and peak area versus c:

The magnitude of the pulse that is superimposed on the DC ramp in DP voltammetry can be changed. What effect might increasing the size of this voltage pulse have on driving the reaction and hence on the current? Might any other peak parameters be affected?

Well, try it and see:

· Display/ Option ® Overlap off

· Input/ Chemicals ® C_anal = 1e-3

· Input/ Instrument ® E_pulse = - 0.05 (default setting)

· Run/ Simulate

· Display/ Option ® Overlap on

· Input/ Instrument ® E_pulse = - 0.02

· Run/ Simulate

· Input/ Instrument ® E_pulse = - 0.01

· Run/ Simulate

· Analysis/ Find Peak

Sketch some voltammograms:

You were probably expecting to see an increase in the signal, but what else do you notice? Estimate the peak width (in volts) at half-height. To help you do this, you could insert gridlines and/or print the voltammograms and measure the peak widths at half height with a ruler.

What do you notice about the peak widths? What implications does this have for resolution of neighbouring peaks?

DP voltammetry allows an even greater increase in the signal to noise ratio and the detection limit is of the order of 10^-7 M. Because it is easier to distinguish partially overlapping peaks than partially overlapping waves, resolution is improved (provided too large a pulse is not used) from 0.2 V for DC and NP to 0.05 V for DP voltammetry.

Square Wave Voltammetry

In square wave (SQ) voltammetry, an alternating (positive, negative, positive, negative….) wave that is square in shape is superimposed on a voltage staircase.

The reverse (anodic) pulse causes re-oxidation of the product (Cu(s)) of each forward (cathodic) pulse. The voltammetric signal is the difference between these two signals and hence will only be recorded when significant amounts of the oxidised (Cu²⁺) and reduced forms (Cu(s)) are present.

Run the simulation:

· Display/ Option ® Overlap off

· Input/ Techniques ® Square Wave voltammetry

· Run/ Simulate

Sketch the shape of the SQ voltammogram:

Square wave voltammetry is about 5 times more sensitive than differential pulse. It is also about 100 times faster - an entire square wave voltammogram can be obtained in less than one second. It can be used for the detection of compounds emerging from a liquid chromatograph. Compounds are distinguished by their retention times and E_1/2 and are quantified by peak height or area.

Stripping Analysis

In stripping analysis, Cu²⁺ is concentrated onto the electrode (as Cu) by reduction at a potential more negative (why?) than E_1/2. The copper metal is then stripped off the electrode by reversing the voltage - ie., sweeping in a positive direction - driving the oxidation reaction. The current measured during the oxidation is related to the amount of Cu initially deposited which, under a given set of experimental conditions, is related to the concentration of Cu²

Run the simulation:

· Display/ Option ® Overlap off

· Input/ Techniques ® DC voltammetry

· Run/ Simulate

· Display/ Option ® Overlap on

· Input/ Instrument ® Pre-concentration on. Note that the potential will be held at - 0.3 V for 600 s before being reversed and swept in the positive direction.

· Run/ Simulate

· Analysis/ Find Halfwave E

Sketch the shape of the DC stripping voltammogram.

How does the current measured compare with that for a normal DC voltammogram? Think about both the signs and magnitudes of the currents.

Stripping is the most sensitive of voltammetric techniques (detection limits are ca. 10^-10 M) because the analyte is concentrated from dilute solution. The longer the period of concentration, the more sensitive the analysis. Confirm this:

· Input/ Instrument ® t_pre = 1200 s

· Run/ Simulate

· Analysis/ Find Halfwave E