Note that the "peak" actually has two sides. When you sweep the potential with CV, the cathodic peak ($E_{pc}$) and the anodic peak ($E_{ac}$) won't exactly line up. If you have a well-behaved, reversible redox species, the two peaks will be separated by $E_{pa}-E_{pc}=\frac{56.5\text{ mV}}{n}$ for $n$ electrons.
Do I measure to the onset of the oxidation/reduction curve or the peak
You do neither. Instead, you take $E^0 = (E_{pa}-E_{pc})/2$ - the average of the anodic and cathodic peaks. This is the standard potential for that redox reaction.
You mention that you're including ferrocene as an in-situ reference potential, so you would compare the $E^0$ potential of your redox couple to the $E^0$ of ferrocene.
What justification do I use to convert the voltage to an energy? How can I say that a difference in meV corresponds to eV?
In principal, you're monitoring an oxidation or reduction. So you measure the oxidation potential relative to ferrocene.
The correspondence to the HOMO energy is based on an interpretation of Koopmans' theorem. In short, the oxidation potential involves removing an electron from the molecule, and the lowest ionization potential should come from the HOMO.
So there should be a correlation between the oxidation potentials from electrochemistry and the orbital energies. Some people take this as fact, but it's good that you're skeptical. I see many papers that do this "conversion" without any warning that these are different things.
You asked about units - as you guessed, the conversion between potential (V) and eV (for orbitals) occurs, because in most cases, one electron is transferred, and so the charge is known. In the general case, this is certainly a cause for concern.
Maybe this might make sense if it was a one electron process, but how do I know that?
Heh. Generally, it's one. That's a good question in electrochemistry. There are multiple ways of determination. If you integrate the peaks, the area should correspond to the number of charges transferred. But usually, you don't oxidize or reduce the whole solution.
You can estimate the number of electrons based on the peak spacing, as indicated above $\frac{56.5 mV}{n}$.
Note that the "peak" actually has two sides. When you sweep the potential with CV, the cathodic peak ($E_{pc}$) and the anodic peak ($E_{ac}$) won't exactly line up. If you have a well-behaved, reversible redox species, the two peaks will be separated by $E_{pa}-E_{pc}=\frac{56.5\text{ mV}}{n}$ for $n$ electrons.
Do I measure to the onset of the oxidation/reduction curve or the peak
You do neither. Instead, you take $E^0 = (E_{pa}-E_{pc})/2$ - the average of the anodic and cathodic peaks. This is the standard potential for that redox reaction.
You mention that you're including ferrocene as an in-situ reference potential, so you would compare the $E^0$ potential of your redox couple to the $E^0$ of ferrocene.
What justification do I use to convert the voltage to an energy? How can I say that a difference in meV corresponds to eV?
In principal, you're monitoring an oxidation or reduction. So you measure the oxidation potential relative to ferrocene.
The correspondence to the HOMO energy is based on an interpretation of Koopmans' theorem. In short, the oxidation potential involves removing an electron from the molecule, and the lowest ionization potential should come from the HOMO.
So there should be a correlation between the oxidation potentials from electrochemistry and the orbital energies. Some people take this as fact, but it's good that you're skeptical. I see many papers that do this "conversion" without any warning that these are different things.
You asked about units - as you guessed, the conversion between potential (V) and eV (for orbitals) occurs, because in most cases, one electron is transferred, and so the charge is known. In the general case, this is certainly a cause for concern.
Maybe this might make sense if it was a one electron process, but how do I know that?
Heh. Generally, it's one. That's a good question in electrochemistry. There are multiple ways of determination. If you integrate the peaks, the area should correspond to the number of charges transferred. But usually, you don't oxidize or reduce the whole solution.
You can estimate the number of electrons based on the peak spacing, as indicated above $\frac{56.5 mV}{n}$.
Actually, do you not think it would be best to measure the onset of the oxidation/reduction peak? - The HOMO according to the theory you quoted should be the "first" electron allowed into the "band", so it would make sense that we consider the onset. That would match literature "Therefore, the oxidation and reduction potentials are estimated by the onset potential, defined as the potential at which the initial injection of holes or electrons to the HOMO and LUMO..." Buzan et al. More
Okay - Thank you for your detailed answer, that seriously helps to answer most of my questions. Apart from one regarding the dimensions (sorry for the physics): Im measuring volts [kg $m^2$ $s^−3$ $A^−1$] but without (rigorous?) justification, Im taking these volts for energy peaks and saying they correlate to an energy [kg $m^2$ $s^-2$] in eV. I.e. Im measuring the More
Yes, you can "convert" this way, but you're correct to be skeptical.
Let's start with interpreting the cyclic voltammetry curves themselves.
(figure from Wikipedia)
Note that the "peak" actually has two sides. When you sweep the potential with CV, the cathodic peak ($E_{pc}$) and the anodic peak ($E_{ac}$) won't exactly line up. If you have a well-behaved, reversible redox species, the two peaks will be separated by $E_{pa}-E_{pc}=\frac{56.5\text{ mV}}{n}$ for $n$ electrons.
You do neither. Instead, you take $E^0 = (E_{pa}-E_{pc})/2$ - the average of the anodic and cathodic peaks. This is the standard potential for that redox reaction.
You mention that you're including ferrocene as an in-situ reference potential, so you would compare the $E^0$ potential of your redox couple to the $E^0$ of ferrocene.
In principal, you're monitoring an oxidation or reduction. So you measure the oxidation potential relative to ferrocene.
The correspondence to the HOMO energy is based on an interpretation of Koopmans' theorem. In short, the oxidation potential involves removing an electron from the molecule, and the lowest ionization potential should come from the HOMO.
So there should be a correlation between the oxidation potentials from electrochemistry and the orbital energies. Some people take this as fact, but it's good that you're skeptical. I see many papers that do this "conversion" without any warning that these are different things.
You asked about units - as you guessed, the conversion between potential (V) and eV (for orbitals) occurs, because in most cases, one electron is transferred, and so the charge is known. In the general case, this is certainly a cause for concern.
Heh. Generally, it's one. That's a good question in electrochemistry. There are multiple ways of determination. If you integrate the peaks, the area should correspond to the number of charges transferred. But usually, you don't oxidize or reduce the whole solution.
You can estimate the number of electrons based on the peak spacing, as indicated above $\frac{56.5 mV}{n}$.
You can change the scan rate and use the Randles-Sevcik equation
$$i_p = 268,600 \ n^{\frac{3}{2}} AD^{\frac{1}{2}} Cv^{\frac{1}{2}}$$
They are not. There should be a correlation between oxidation and reduction potentials and HOMO and LUMO energies.
But step back and think - in one case, you have electrochemistry in the electrolyte solution at an electrode surface.
Yes, you can "convert" this way, but you're correct to be skeptical.
Let's start with interpreting the cyclic voltammetry curves themselves.
(figure from Wikipedia)
Note that the "peak" actually has two sides. When you sweep the potential with CV, the cathodic peak ($E_{pc}$) and the anodic peak ($E_{ac}$) won't exactly line up. If you have a well-behaved, reversible redox species, the two peaks will be separated by $E_{pa}-E_{pc}=\frac{56.5\text{ mV}}{n}$ for $n$ electrons.
You do neither. Instead, you take $E^0 = (E_{pa}-E_{pc})/2$ - the average of the anodic and cathodic peaks. This is the standard potential for that redox reaction.
You mention that you're including ferrocene as an in-situ reference potential, so you would compare the $E^0$ potential of your redox couple to the $E^0$ of ferrocene.
In principal, you're monitoring an oxidation or reduction. So you measure the oxidation potential relative to ferrocene.
The correspondence to the HOMO energy is based on an interpretation of Koopmans' theorem. In short, the oxidation potential involves removing an electron from the molecule, and the lowest ionization potential should come from the HOMO.
So there should be a correlation between the oxidation potentials from electrochemistry and the orbital energies. Some people take this as fact, but it's good that you're skeptical. I see many papers that do this "conversion" without any warning that these are different things.
You asked about units - as you guessed, the conversion between potential (V) and eV (for orbitals) occurs, because in most cases, one electron is transferred, and so the charge is known. In the general case, this is certainly a cause for concern.
Heh. Generally, it's one. That's a good question in electrochemistry. There are multiple ways of determination. If you integrate the peaks, the area should correspond to the number of charges transferred. But usually, you don't oxidize or reduce the whole solution.
You can estimate the number of electrons based on the peak spacing, as indicated above $\frac{56.5 mV}{n}$.
You can change the scan rate and use the Randles-Sevcik equation
$$i_p = 268,600 \ n^{\frac{3}{2}} AD^{\frac{1}{2}} Cv^{\frac{1}{2}}$$
They are not. There should be a correlation between oxidation and reduction potentials and HOMO and LUMO energies.
But step back and think - in one case, you have electrochemistry in the electrolyte solution at an electrode surface.
More
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