Hello, I am newly studying cyclic voltammetry and kind of stuck understanding the CV of a blank buffer.
The buffer is 1X TE (Tris-HCl+EDTA, pH 8.0). I am attaching the voltammogram as an image. No additives or solutes such as KCl should be added; I need to analyze the electrochemical reactions of my system where 1X TE buffer is the only component.
My questions are:
- Beyond +2 V and -2 V, I am repeatedly observing steep increases in the ionic current. This seems to represent the decomposition of water or anything very unstable. But the steepness of the ionic current looks similar in -1 V to 0 V region. Would this mean that the system is unstable in this voltage range also?
- What would the plateau near +2 V and - 2 V mean? Would this be identifiable as a peak in the CV?
- Why does the current increase and decrease in between -2 V and 2 V? Does it mean that there is some reaction going on? If there is no reaction possible, why would the current change?
- Why is the current at V=0 not zero? Is the ionic current somewhat path dependent?
Thanks in advance!
If you stress your system beyond the (legal) electrochemical potential stability of the electrolyte (including decomposition of the water), then you enter in a chaotic mode of your cell's operation, and it's path (and your CV graph) depends on the long[1] lifetime(s) of the new unstable and metastable "particles" (and bubbles) you create.
So, try to start your initial CVs (of a blank buffer) using a small and safe electrochemical potential window (VLow, VHigh); later on, you might want to increase, stepwise (say, by 50mV), these initial potential limits up to a maximum (safe) window, for your case study.
1. Long lifetime(s) compared with the the time duration of your CV.
According to my small knowledge if I am not wrong you are saying something about the increasing current level of two vertex,
1. If that's the case something is forming film on your WE, so it takes high current to go to the same point
2. I believe 1V is enough to oxidize the water/buffer
if the peak is very wide you can go for the maximum point but checking point by point and mention it as a centered value.
3. 3rd q same as 1
4. is not it a continuous process, so you might think about leakage current as well as the electrical diagram of the CV 3 electrode system. the voltage is zero there wrt RE, you know if it is really zero. Again please see the total electrical diagram of the 3 electrode system
I think that you do not have sufficient ions in your solution to see any real electrochemistry. Instead, you are basically showing Ohm's Law: E = IR by the cyclic voltammetry where the solution resistance is about 80,000 ohms. The dominant anodic oxidation and cathodic reduction of H2O is not even seen in your results.
Dr. Melvin H. Miles
There is a (quite) sufficient ions' density in this buffer[1] solution, even thought it has[1] a faint pH=8.
However,
1) their effective mobility is very low, and
2) the interfaces are modified,
due to the long[2] lifetime(s) of the new unstable and metastable "particles" (and bubbles), created in the phase of the extreme V-parts, whenever we try, uncommonly, to test-scan in a so wide potential window (VLow, VHigh).
The dominant anodic oxidation and cathodic reduction of the solute (H2O) is not shown, easily, due to the low (effective mobility) kinetics.
1. A partial quotation from Kidan Lee 's question: "The buffer is 1X TE (Tris-HCl+EDTA, pH 8.0)."
2. Long lifetime(s) compared with the the time duration of your CV.
Of course the nature of the electrode is also important (e.g. graphite, gold...). Does your application really needs such a large potential window? EDTA and Tris may also be irreversibly oxidized (reduced) and decomposes at such high potententials further decomposition products may be adsorbed and desorbed on the electrode which may look as an apparent oxidation or reduction wave try sweeping between e.g. -0,4 and 0.8 V initially and increase or decrease the potential window accordingly. (I see you already done that)
Best regards, Mike
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