# Why does an ideal capacitor give rise to a rectangular cyclic voltammogram (CV)?

This question is sort of a sequel to my previous question about cyclic voltammetry (CV). One of the responses made reference to the fact that an ideal capacitor gives rise to a rectangular cyclic voltammogram. Can you please help me understand why this is the case? In other words, why does an ideal capacitor reach a constant current $I$ as soon as a voltage $V$ is applied?

I indeed see nearly ideal CVs in many literature articles (CVs which are rather rectangular with rounded corners). In other figures, though, I see relative deviation from "rectangles with rounded corners," in that I see abrupt peaks, spikes, or valleys.

For example, below I have plotted two figures from Khomenko, Electrochimica Acta 2005, 50, 2499-2506. Just very roughly and "hand wavy," what might be the qualitative reason for the "rectangle with rounded corners" behavior of Figure 8 (left) and the "abrupt peaks" behavior of Figure 4 (right)? Could it be that the sample in Figure 8 (left) is relatively unreactive toward apllied potential, whereas the sample in Figure 4 (right) undergoes redox (Faradaic) reactions -- indicating the presence of so-called pseudocapacitance -- when an external potential is applied?

Please know that I am not looking for an answer specific to the article to which I linked. I am only asking this question in the context of basic, qualitative aspects of cyclic voltammetry.

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The kind of analysis truly required is outside the scope of chemistry and would be best handled in the careful hands of an electrical engineer. I'll give a brief attempt that exploits an easy-to-use circuit simulator.

We have to first think of an equivalent circuit that we can use as a model to determine its behavior. I suggest the following:

where my variable is R1's value for its resistance, which I shall set to 0, 10 and 100 ohm. If we were to watch this happen in time then we would see the following:

Quickly converting these to Current vs Voltage and running two other simulations at different resistances and we get:

These results are due to setting up differential equations and solving them appropriately.

You can play with the circuit I made for yourself here.

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This is veering into physics, but the reason for your initial observation - that an ideal capacitor has a square cyclic voltammogram, as shown in Chris's diagram - is that the charge on an ideal capacitor is proportional to the voltage across it, with the constant of proportionality being the capacitance: $q = CV$. Differentiating with respect to time, we see that the current across the capacitor is $I = \frac{\mathrm{d}q}{\mathrm{d}t} = C\frac{\mathrm{d}V}{\mathrm{d}t}$. Now when you put a sawtooth wave across the capacitor, $\frac{\mathrm{d}V}{\mathrm{d}t}$ flips back and forth between a positive and a negative value, hence so does $I$, giving the characteristic square shape.