I know that when extracting the peak anodic/cathodic currents, it is typical to extract a baseline. I presume to account for non-faradaic processes (i.e. double layer charging), however I have not found any references that actually delve into what processes give rise to this current.
Dear Hunter,
If the electrochemical couple is not fully reversible (i.e. the ratio of activities of the oxidised and reduced forms does not satisfy the Nernst equation for the applied potential) then yes, there may be a residual current coming from electroactive species at potentials significantly more negative than IPf or more positive than IPr. If not, the baseline current will be largely due to EDL charging. I am drawing this conclusion from the strong dependence of baseline currents on the scan rate for a reversible couple (see examples of cyclic voltammograms shown at different scan rates). Such a dependence would be anticipated from the charging (capacitance) current since Ic = dq/dt where q is the capacitor's charge, q = Cdl A (E - Epzc), Cdl is the capacitance per unit area and Epzc is the potential of zero charge (i.e. where the sign of the electrode charge reverses). Assuming the electrode area does not change, then
Ic = dq/dt = Cdl A dE/dt + A (E - Epzc) dCdl/dt
The first term is indeed proportional to the potential scan rate while the second term may become significant when adsorption processes change the double layer capacitance.
Dear Hunter,
The reverse peak current is more often problematic, with baselines affected by the location of the switching potential, among other things. I will send a Figure in an attachment for you. Proper extrapolation of baselines for peak current measurement in the absence of complications is shown at it. Many computer-controlled potentiostats have data systems that can accurately and quickly make these types of measurements. For situations where baseline assignment is more complicated, the current ratio can sometimes be conveniently calculated using the empirical method of Nicholson , which requires the measurement of the forward (ip,f) and reverse peak currents (designated (ip,r)0 to differentiate it from that used previously) from the same baseline, along with a third current, iλ, which is the current measured at the switching potential.
I hope this helps, kind regards, Fahimeh
i guess you are talking about double layer charging. it is because when you apply a particular potential to balance this charge opposite cahrged ions come near the electrode which gives rise to this current.
Fahimeh,
Yes I understand that how to draw the baseline properly, I am more so asking “what currents am i actually removing when using the baseline vs peak current measured from i=0?”
Zahid,
Is it solely EDL charging though? redox molecules are still being converted at the extreme potentials right? So even on the reverse sweep, at large positive potentials we are still reducing our redox species. Correct?
Dear Hunter,
If the electrochemical couple is not fully reversible (i.e. the ratio of activities of the oxidised and reduced forms does not satisfy the Nernst equation for the applied potential) then yes, there may be a residual current coming from electroactive species at potentials significantly more negative than IPf or more positive than IPr. If not, the baseline current will be largely due to EDL charging. I am drawing this conclusion from the strong dependence of baseline currents on the scan rate for a reversible couple (see examples of cyclic voltammograms shown at different scan rates). Such a dependence would be anticipated from the charging (capacitance) current since Ic = dq/dt where q is the capacitor's charge, q = Cdl A (E - Epzc), Cdl is the capacitance per unit area and Epzc is the potential of zero charge (i.e. where the sign of the electrode charge reverses). Assuming the electrode area does not change, then
Ic = dq/dt = Cdl A dE/dt + A (E - Epzc) dCdl/dt
The first term is indeed proportional to the potential scan rate while the second term may become significant when adsorption processes change the double layer capacitance.
What about Ohmic drop? Is the slope of the base line important to measure ip? Why do not take a 90 degree line over the base line to measure the ip?
- Badges
- Science topic
- Similar topics
- Phytochemistry
- Extracts