I am using two aqueous electrolytes (K2SO4 and KCl respectively) for my electrochemical cell and i wanted to check potential stability of both electrodes. I have attached CV obtained at 50mV scan rates for both electrolytes. What i can conclude from this about potential window of both electrolytes. I have charged my cell with electrodes with 1.5 V battery. potential range from the CV graph here is same for both.
can u suggest me what to do to get the potential window of these electrolytes from CV experiments.
So,
First, I would suggest you to find a reference electrode. Without such, nobody can't explain you for sure what happens on yours electrodes during the voltammetry. another recommendation is to decrease your sweep rate; 50mV/s can be too much if your want to detect the start of the O2 or H2 evolution for electrodes with high specific surface (graphenes, activated carbons, nano-materials) where the capacitive currents arriving from the double layers can be very high.
Pankaj,
you should offer more informations: which reference electrode are you using? What is your working electrode and counter electrode. As Artur and Denys said in aqueous solutions you are limited by water decomposition: at lower potential --> by hydrogen evolution, of higher potential --> oxygen evolution. In case you have chlorine ions --> Cl2 evolution will usually come earlier then oxygen evolution, that narrows your potential window. This potential window is strongly dependent of your working electrode material and pH. In case for Pt group metals, the potential window at pH = 1 is between 0 V - 1.6/1.7 V vs. RHE, because Pt has a negligible overpotential for hydrogen evolution and relatively small (compared to other materials) for oxygen evolution. For example dropped Hg electrode has a very high overpotential for hydrogen evolution (i think it was something around 1 V), which extremely widen your potential range.
Your CVs are very different due to the higher currents in case you using KCl. This current seems to be a faradaic current and i possibly could answer your question if you tell me the working electrode material, pH and the temperature!
I assume that your two CVs were recorded using the symmetrical two electrode cell (supercapacitor). If so, you should have NOT scanned your voltage to the negative range. By doing so, you have basically charged the cell (supercapacitor) twice, but in opposite directions (polarities). This is why your CVs show a voltage range of 2 V, which is NOT the same as potential window of your electrolyte.
Secondly, these CVs from the two-electrode cell are not suitable for determining the potential windows of the two electrolytes you had used. It is better to use the three-electrode cell with an inert working electrode (e.g. Pt or graphite) to record the CV, in which case the electrode potential (NOT the cell voltage) should be extended to as positive (or negative) as possible to bring about a continuously raising positive (or negative) current. The potential range between these positive and negative currents is the potential window of the electrolyte.
Nevertheless, from the two electrode CVs of yours, you can still say that the potential windows of both electrolyte is greater than 1 V, but this is not good enough for you.
Please note that the potential window of an electrolyte can be expressed in two ways. The first way is simply to say that the potential window is 1.8 V which means the width of the window. It is also often expressed as, for example, from -1.0 V to 0.8 V vs. reference (e.g. SCE or Ag/AgCl). Also, voltage is used for cell, and potential for electrode. Do not use voltage for electrode, nor potential for cell.
Run the CV with various bare working electrode with different limits (you'll see the proton reduction at negative potentials and water oxidation at positive ones depend on your using bare working electrodes). The potential limits for cyclic voltammetry in aqueous electrolytes also depends on your working electrode.
Thanks Everyone.
Dear Prof. George, So CV of three electrode configuration is the way i can get my correct potential window. The cell i use can be used only for 2 electrodes. Is there no way to get correct potential window with two electrode cell ?
It is not ideal to use a two electrode cell to measure the electrolyte potential window (EPW), but it is not incorrect to do so, depending on the cell structure and electrode properties.
The main challenge is that when the cell voltage is at zero, the potential is the same for both the positive and negative electrodes (called Potential of Zero Voltage or PZV), but it may not be at the centre of the EPW. When you increase the voltage of the cell, the potentials of the two electrodes will move away from the PZV in opposite directions. The speed of the potential change on each electrode is determined largely by the capacitance of the electrode. (The larger is the capacitance, the slower is the potential change.) In a symmetrical capacitor, the capacitance is the same for both the positive and negative electrodes. Thus, if the PZV is away from the centre of the EPW, one of the electrodes would reach its potential limit earlier than the other, giving rise to a maximum cell voltage that is lower than the potential window of the electrolyte.
If you have two electrodes system it is Voltage window of 1 V. To determine that properly you should scan very slowly 5 mV s-1 and follow the water splitting. To do that properly hold the voltage at the maximum voltage (Chronoapemerometry) you believe is the right one and compare the residual current to the one at the middle of voltammogram. It should match. I suggest you perform EIS at both voltage as well (let us say 0.5 and 1V ) If voltage is right Nyquist in both experiments should match..
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