CASE STUDIES IN CONTROLLED POTENTIAL
6.4. CASE III. STUDYING THE OXIDATION OF ORGANIC MOLECULES
Organic electrode reaction mechanisms can provide insight into biochemical reaction path- ways. This case study demonstrates the use of controlled potential methods to investigate possible intermediates in the decomposition of a common cellular constituent, uric acid.
Figure 6.6 shows the cyclic voltammogram for uric acid recorded at a moderately fast scan rate of 500 mV/s in an aqueous solution at a glassy carbon electrode [8]. The scan started by going from 0 V in the negative direction. There was no sign of reduction of the starting material, even out as far as−1.5 V. However, scanning in the positive direction produced a large oxidation peak around+0.4 V. Scanning back in the negative direction again yielded a small reduction peak paired with the huge oxidation peak. A second reduc- tion peak (that was not present in the first cathodic-going segment) was also observed. As the scan rate was increased the first reduction peak (Ic) grew with respect to the oxidation
0.5 0.0
Ia
Ic IIc
–0.5
EP(pHO-12) = [0.685–0.055pH]Volt versus SCE
Potential/Volt versus SCE 20 μA
Current
–1.0 –1.5
FIGURE 6.6 Cyclic voltammogram of 2 mM uric acid in pH 7.5 phosphate buffer recorded at 500 mV/s on a pyrolytic graphite electrode. Source: Reproduced with permission from Dryhurst et al. [8]. Copyright 1983, American Chemical Society.
k k peak height (data not shown) and the second reduction peak (IIc) decreased in size. This
behavior suggests that a chemical step follows the initial electron transfer process (associ- ated with peak Ia). This type of sequence is called an EC reaction mechanism (E=electron transfer, C=chemical step).
A⇄B+ne− (6.18)
B+C−→X+P (6.19)
A and B are a redox couple, but B is unstable. It reacts to form X, leaving little B avail- able near the electrode to be reduced to back to A on the cathodic scan. This scheme also includes the possibility that B decomposes on its own (that is, there is no second reactant, C and, perhaps no additional product, P). The new reduction peak, IIc, suggests that X can also be reduced.
What sort of information can one derive about the mechanism? Certainly, the iden- tity of the products associated with each peak is of interest. Determining the number of electrons and protons involved in each process can help in making the identification and outlining a possible mechanism. Another important question is whether there are inter- mediates that can be identified. That is, are there transient species that appear on the way to forming compound X? If so, could these intermediates react with other molecules under different conditions? The probability of such side reactions is related, in part, to the rates of reaction for various chemical steps. Consequently, measuring the lifetimes of species under these experimental conditions is of interest.
Answering the question of the number electrons and number of protons transferred in the first oxidation step can be useful evidence for identifying the product of the first oxida- tion peak, that is, B. The best way of finding the number of electrons,n, is by coulometry.
Holding the potential at+0.6 V (much more positive than the oxidation peak on the CV) and measuring the charge for oxidizing a known number of moles of reactant provide the data for calculating n from Faraday’s law:
Q=∫ idt=nFN (6.20)
n= Q
FN = ∫ idt
FN (6.21)
whereN=the number of moles reacting andF=96 485 C/mol.
One way of doing this is to electrolyze the entire quantity of a weighed amount of reactant added to the cell. This process usually means stirring the solution and using a working electrode with a large surface area (such as a Pt mesh) and recording the current until it drops to a tiny fraction of the original value. Such an experiment generally takes a few hours, but, in addition to providing a value forn, it also provides a supply of the final product, X, that can be identified from various forms of spectroscopy, including mass spectrometry. Dryhurst’s group performed a bulk electrolysis at+0.6 V for the oxidation of uric acid and calculated a value of 2 forn[8].
It is interesting to think about the analysis of the electrolysis product for a couple of reasons. Product isolation can be a challenge and the authors employed a clever strategy.
Furthermore, it was also an interesting example of electrochemical phenomena in action.
k k They freeze-dried the solution so that they could redissolve the material in a smaller
volume of water and separate the electrolysis product from the salt that made up the original supporting electrolyte using liquid chromatography with water as the mobile phase. The authors’ choice of column was a commercial gel permeation chromatography resin, Sephadex G-10™. It is a porous material in the form of beads made by cross-linking polysaccharide chains. Retention is based on molecular size. That material has a low molecular weight cut-off limit of only 700 Da. Molecules bigger than the cut-off limit do not penetrate the pores inside particles and are eluted by flushing the column with the volume of solvent equivalent to the space between particles. They are unretained. In contrast, the smallest solutes can explore the volume inside the channels of each particle and require more solvent to push them out of the column. They emerge last based on that mechanism. Curiously, the sodium phosphate, with a much smaller hydrated radius than uric acid, eluted first. Some mechanism other than size exclusion must have been operating. Earlier work with gel resins has demonstrated that the polysaccharide material contains a small fraction of carboxylic acid functional groups; some are naturally there in the polysaccharide and some are present because of residual oleic acid used in the man- ufacturing of the resin [9]. These acidic groups ionize even at slightly acidic pHs. Buffer solutions are most often used as the mobile phase with these resins and the ionic strength of these solutions are high enough to compress the electrical double layer on the surface of the resin (including within the channels inside the particles). However, in this work, pure water was the mobile phase. That means that the influence of the negative surface charge extended a much greater distance away from the surface. (Recall the Gouy–Chapman model discussed in Chapter 1.) This negative charge repels anions. Consequently, using the gel permeation column with pure water excluded the phosphate anions from the pores forcing the salt to elute well ahead of the organic molecules. This strategy worked well as long as the amount of salt in the sample was relatively small [9, 10].
Using mass spectrometry and infrared spectroscopy, the authors identified the elec- trolysis product, X, as allantoin [8]. Comparing the structures of uric acid and allantoin, it is apparent that a lot happens after two electrons are
Uric acid Allantoin
O
O O
HN
H N N N H H
H O O
O H2N
H N H N
N H
removed from uric acid. Allantoin represents X in Eq. (6.2). There must be several inter- mediate steps in going from A to X. Identifying intermediate species would help define the pathway.
By repeating the cyclic voltammetry experiment with uric acid at different pHs, the authors observed the oxidation peak, Ia, and its partner, Ic, shift in potential by 55 mV toward 0 V for every increase of one pH unit. That is, in going between the oxidized and reduced forms of this redox pair, hydrogen ions are transferred as well as electrons. This
k k reaction can be generalized as
Ox+ne−+mH+⇄Red (6.22)
The corresponding Nernst equation reveals how the formal potential depends on pH.
E=Eo− (2.303)RT nF log
{ aRed aOxamH+
}
= [
Eo−(2.303)RT nF log
{ 1 amH+
}]
− (2.303)RT nF log
{aRed aOx
}
= [
Eo−(2.303)RTm
nF pH
]
−(2.303)RT nF log
{aRed aOx
}
(6.23) where term in the brackets,
[
Eo−(2.303)RTmnF pH ]
, is the formal potential,Eo′, for the redox couple. The average of the peak potentials for the redox pair is a measure of the formal potential. Consequently, that average shifts with pH. The coefficient (2.303)RT/Fis equal to 59 mV at 25∘C. Then, in general the shift inEo′with pH is given by
𝜕Eo′
𝜕(pH) = 𝜕
𝜕(pH)
{−(2.303)RTm
nF pH
}
= −0.059 16(m n )
(6.24) So, for a shift of 0.059 V per pH unit,m= n. In this case, m= n= 2. Therefore, the oxidation process at Iais a two-electron, two-proton step [6].
A⇄B+2e−+2H+ (6.25)
A common structural change associated with the loss of two electrons and two hydro- gen ions in an organic molecule is the formation of a double bond. This idea suggested a quinoid structure [11]. One of the resonance forms might be drawn as this:
O
O O
HN N
N N H
Because this work was performed in buffers between pH 7 and 8, uric acid (pKa=5.75) would be in the anion form [11]. Therefore, the oxidation peak is more appropriately described with this first step:
O NH HN
HN
O N
O
O
HN N
N N
O
O + 2e– + 2H+
A B
–
–
(6.26)
k k
–180 0.2
Current function 0.0
–0.2 –0.4
–120 –60 (E–E1/2)n (mV)
(a) (b)
0 0.01 500 10
A B + ne– k X
k/a = 0.1, 0.01
0.1
60 –2.0
∣ip,back/ip,forward∣
0.0 0.2 0.4 0.6 0.8 1.0
–1.0 log (kτ)
0.0
FIGURE 6.7 (a) Simulated cyclic voltammograms of an electron transfer step followed by an irre- versible chemical reaction for different scan rates,𝜐, wherek/a=(k/𝜐)(RT/nF). The faster the scan rate compared to the rate of the chemical step (smaller values ofk/a), the higher the concentration of the intermediate, Red, available near the electrode during the return scan and the bigger the return current peak. The term,E1/2, on thex-axis represents the “half-wave” potential which is an estimate of the for- mal potential,Eo′. (b) A plot of the current ratio versus the log of the product of the rate constant,k, and the time,𝜏, required to scan between the formal potential and the switching potential. An estimate of k can be made, by finding the x-coordinate corresponding to the point on the curve associated with a measured ratio of peak currents. Source: Adapted with permission from Nicholson and Shain [12].
Copyright 1964, American Chemical Society.
What happens after the electron transfer? Apparently, structure B is unstable; an irre- versible chemical reaction converts B to some other species, C. Just how fast is that reac- tion? There are two well-established ways of measuring the rate constant for the following chemical step. Both methods will be examined here.
The first approach is to scan at faster rates until the return peak grows in. The under- lying idea is that the speed of the return scan must be comparable to the rate of the decom- position reaction in order to observe any current for the return peak. A small peak was observed for Icin Figure 6.6, indicating that the time required to scan back from the switch- ing potential was on the order of the lifetime of the intermediate, B. Figure 6.7a shows a set of computer-simulated current-voltage curves over a large range of scan rates.
This graph was adapted from a classic paper by Nicholson and Shain [12] and requires some explanation to make it digestible. (The curve for the anodic peak, Ia in Figure 6.6, points in the upward direction in this graph, but the oxidation step is shown here as the forward electron-transfer process to be consistent with the uric acid example.) Nicholson and Shain have tried to summarize a lot of information in one diagram for their read- ers. The current has been normalized (removing the square root dependence of the peak height on scan rate) and is represented here as a “current function” on the vertical axis.
That maneuver merely keeps all of the voltammograms on the same current scale. The rate constant for the chemical step is represented byk. Cyclic voltammograms were plotted for different scan rates. The scan rate is embedded in the term,a. That is,a=[(nF)/(RT)](𝜐t), wheretis the time in seconds from the start of the scan. [(RT)/(nF)] has the units of volts;
𝜐is the scan rate in V/s; andkis in s−1. Consequently,k/ais proportional tok/𝜐,the rate
k k constant of the chemical step compared to the voltage scan rate. The important idea to keep
in mind is that the value ofk/adecreases with faster scan rates. Another way of looking at it is that the larger values ofk/arepresent experiments at slower scan rates where the intermediate, B, has more time to decompose before the return scan starts. Consequently, there will be less B available on the return scan and a smaller return current peak will be observed. What is important about Figure 6.7a is the trend: as the scan rate increases (decreasingk/a), the shape of the CV approaches the ideal of a pair of peaks that are sym- metric about the formal potential (represented by theE1/2in thex-axis). The panel on the right side in Figure 6.7 predicts the ratio of the peak currents for a wide range of scan rates.
This plot can be used to calculate an estimate of the rate constant,k, for the chemical step.
An example calculation should help to make this analysis clear.
Figure 6.8 is a cyclic voltammogram taken at a scan rate of 1200 mV/s. The return peak is smaller than expected for an ideal reversible electron transfer process. It decreases in size with decreasing scan rate suggesting an EC mechanism. A separate scan in the same electrolyte solution, but without the analyte was made in order to determine the baseline. The formal potential can be estimated from the average of the peak potentials.Eo′
appears to be 0.125 V versus normal hydrogen electrode (NHE). The switching potential was=−0.270 V. The time,𝜏, required to scan between the formal potential,Eo′ and the switching potential is
𝜏= (Eo′−E𝜆)
𝜐 = 0.125 V− (−0.270 V)
1.2 V∕s =0.329s (6.27)
The peak height ratio can be calculated from the method [13] that was illustrated in Figure 5.13 as shown in Eq. (6.28).
ip,back
ip,forward = (ip,back)0
ip,forward+ 0.485(iλ)0
ip,forward +0.086= 5.5
10.4 +0.485(3.9)
10.4 +0.086=0.796 (6.28)
ibackground
ip,forward
(iλ)0
(ip,back)0
ip,forward ip,forward ip,forward
ip,back (ip,back)0 0.485(iλ)0
+ 0.086 +
=
E˚ʹ = 0.125 V
0.10 0.15 0.20 0.25 0.30 –0.05 0.0 0.05
Eλ = 0.270 V
ip,forward = 10.4 àA
(ip,back)0 = 5.5 àA
(iλ)0 = 3.9 àA 10.0
5.0
0
–5.0
–10.0
Current (àA)
Applied voltage, E(V) vs NHE
FIGURE 6.8 Example calculation for the rate constant for the following chemical step in an EC mech-
k k From the curve in Figure 6.7b, thex-coordinate corresponding to ay-value of 0.796 is
approximately equal to−0.70. That means log(k𝜏)= −0.70. Solving for the rate constant gives
log(k𝜏) = −0.70 or k= 1 𝜏
(
10−0.70)
= 1
0.329 s (
10−0.70)
=6.06s−1 (6.29) In summary, the pattern of a growing return peak (compared to the forward peak) with increasing scan rates suggested a mechanism of an electron transfer step followed by an irreversible chemical step. The scan rate study provided data for calculating the rate constant for the following chemical step. The average of the peak potentials provided an estimate of the formal potential. Table 6.1 is a list of other common electrode processes with coupled chemical steps that have been well studied (theoretically and experimentally) [12, 14] including some comments about distinguishing characteristics.
Scanning faster is not always practical. A second approach to finding rate constants for coupled chemical steps is to apply a double step chronoamperometry experiment [14, 16].
This is the method that Dryhurst and coworkers applied to the study of the uric acid oxida- tion. If the reaction is not too fast, one can determine whether the reaction with respect to the concentration of the product for the electron transfer step is first order (or pseudo-first order as might be the case, if the solvent or electrolyte were to react) or second order. One can also estimate the rate constant for the process of the chemical conversion of the inter- mediate to another compound. The fact that a small return peak (IC) can be observed in the cyclic voltammogram in Figure 6.6 suggests that this chemical step is on the right time scale for such an investigation [8].
The basic idea underlying the double potential step experiment is that the first step brings the system to a potential where the intermediate is generated and the second step is used to observe how the intermediate behaves [16]. Figure 6.9a is a general diagram showing the applied potential and the current response as a function of time. In this case, the voltage was stepped from 0 V, a value where no electrode reaction occurs, out to+0.6 V, where the product B in Eq. (6.8), is formed. The voltage was held for a predetermined time, 𝜏, and the current was measured. The current response to this “forward” step is an oxidation or anodic current, but for ease of reference it will be referred to here as the forward current,if. Next, the voltage was stepped back to 0 V again and the current was measured at a time, 2𝜏, or𝜏seconds from the start of the second potential step. The cur- rent response for this “back step” was for the reduction of compound B and was in the cathodic direction. It will be called the “back current,”ib, here for clarity. If the reaction in Eq. (6.8) was unperturbed by any following chemical step, the current following both the forward and back steps would follow the Cottrell equation (5.26) for the two species A and B, respectively. In an uncomplicated case (reversible electron transfer and no chemical reaction), the theory predicts that the ratio of the back current to the forward current, that isib(att=2𝜏)/if(att=𝜏), would be equal to 0.293 for all values of𝜏[16]. (The ratio is not unity because most of B diffuses away from the electrode. Only a fraction is reduced back to A.)
However, when B takes part in a chemical reaction, even less B is available for the back current. Consequently, the current ratio decreases with𝜏and with increasing values of the rate constant,k. (Of course, for this one reaction,kis fixed.) If one performs the double step experiment multiple times at different values of𝜏, a plot of the current ratio versus𝜏
k k TABLE 6.1 Electrode mechanisms
I. Reversible electron transfer
O+ne−→R CV characteristics: CV peak height ratio≈1;ΔEp≈59/n mV.
II. Irreversible electron transfer
O+ne−→R
CV characteristics: the current peak shifts in proportion to the square root of the scan rate.
Totally irreversible in the electrochemical sense may still show a return peak, but ΔEp>200 mV.
III. Reversible chemical reaction preceding a reversible electron transfer – CrErmechanism:
Z⇌O O+ne−⇌R where the equilibrium for the chemical step,K=k1/k−1.
IV. Reversible chemical reaction preceding an irreversible electron transfer – CrEimechanism:
Z⇌O O+ne−→R where the equilibrium for the chemical step,K=k1/k−1.
V. Reversible electron transfer followed by a reversible chemical reaction – ErCrmechanism:
O+ne−⇌R R⇌Z where the equilibrium for the chemical step,K=k1/k−1.
VI. Reversible electron transfer followed by an irreversible chemical reaction – ErCimechanism:
O+ne−⇌R R→k Z
(a) CV characteristics: small return peak for slow scan rates with increasing peak height ratio with increasing scan rates.
Chronoamperometry double step:k=0.406/𝜏1/2where𝜏1/2=value of time from start of step for normalized currentRI=0.5.
(Continued)
k k TABLE 6.1 (Continued)
(b) For product reacting with itself:
O+e−⇌R 2R→k Z
Chronoamperometry double step:k=0.830[O]/𝜏1/2where𝜏1/2=value of time from start of step for normalized currentRI=0.5 and [O]=original concentration of the starting material.
(c) For product reacting with starting material:
O+2e−⇌R−2 R−2+O→k Z
Chronoamperometry double step:k=0.922/𝜏1/2where𝜏1/2=value of time from start of step for normalized currentRI=0.5 and [O]=original concentration of the starting material.
VII. Reversible electron transfer followed by a regeneration of the starting material – catalytic mechanism:
O+ne−⇌R R+Z→k O
CVs vary in shape as a function of scan rate and concentrations of O and Z. See Case Study IV later in this chapter.
VIII. Irreversible electron transfer followed by a regeneration of the starting material – catalytic mechanism:
O+ne−→R R+Z→k O
IX. Multiple electron transfer with intervening chemical reaction(s) (a) ECE mechanism (first order):
O+n1e−⇌ R, R →k1 X X+n2e− ⇌ Y
For double step chronoamperometry:k=0.273/𝜏1/2where𝜏1/2=value of time from start of step for normalized currentRI=0.5.
(b) ECE mechanism (second order):
O+n1e−⇌ R, R+O →k1 X X+n2e− ⇌ Y,
When R reacts with O to form a dimer, double step chronoamperometry gives:
k=0.690/𝜏1/2where𝜏1/2=value of time from start of step for normalized currentRI=0.5.
Many systems have been reported in the literature for ECE and ECEC mechanisms. Their behavior is usually complicated but can be elucidated using digital simulation.
Source: Adapted with permission from Refs. [12], [14], and [15].