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Reading up on the theory of cyclic voltammetry, I note that the current at the inversion potential (that is, the point at which the applied potential goes negative) in every reversible CV trace I have seen is nonzero. My intuition says that this is related to the finite speed of enlargement of the depletion zone around the working electrode and suspect that given sufficient time this current will drop away to zero as the bulk solution homogenises. Is this correct? If not, what accounts for the current?

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  • $\begingroup$ Do you mean the point at which the $\Delta$ of the applied potential goes negative? Or do you mean the point at which $V_{\text{app}}$ goes back to $V_0$? I think you should add a graph to your question! $\endgroup$ – CHM Sep 21 '12 at 23:45
  • $\begingroup$ I'm re-reading Skoog and Holler, ch. 25. Extremely interesting stuff. From what I remember pre-reviewing, CV studies diffusion controlled redox processes, so yes, the current should tend to a minimum, but definitely not 0. There are basically two currents, faradaic and non-faradaic (or capacitive, because of the capacitor that forms at the electrode boundary), competing, the latter being impossible to eliminate. I can't offer a better answer, but you got me interested in Voltammetry again! $\endgroup$ – CHM Sep 21 '12 at 23:59

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