Notation of the reduction and oxidation of copper typically includes the aqueous state. Illustrated below: $$\ce{Cu(s) -> Cu^2+(aq) + 2e-}$$ This event usually occurs at 0.34 V. What if the copper was in an organic environment instead? I ask because I am doing cyclic voltammetry of a capacitor with copper as the current collector and need to measure the non-faradaic capacitance. Our current setup is using $\ce{KOH}$ as the electrolyte and we are basically just seeing a non-reversible/quasi-reversible CV like below:

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It is certainly possible for metals, including copper, to corrode in non-aqueous environments. The details depend on the precise nature of the solvent, electrolytes, and other potentially complexing species present. For example, in polar protic solvents (like water) that don't have any complexing ligands for copper, the equilibrium potential for copper oxidation will likely be the same as in water. But oxidation potentials will likely shift in solvents that are non-polar, in ways that depend strongly on the composition of the electrolyte.

However, if you are using potassium hydroxide as the electrolyte, your conditions might be more aqueous than you think. You can think of "hydroxide" as a type of water that likes to exist at high pHs. And hydroxide itself generates high pHs.

Elecrochemical water oxidation occurs via this reaction:

$$\ce{2 H2O -> O2 + 4 H+ + 4e-}$$

The pH-dependent equilibrium potential is $1.23~\rm{V} - 0.059\frac{\rm V}{pH}\times pH $, which at a pH of 14 comes out to 0.4 volts. So you might also possibly be irreversibly oxidizing hydroxide to oxygen during your electrolysis.

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    $\begingroup$ I would agree that oxygen is probably being produced under these conditions, especially since the forward sweep extends so far. One should also note that potassium hydroxide is not particularly soluble in many organic solvents (apart from short-chain alcohols) and that things like lithium perchlorate or tetra-alkyl ammonium salts are usually used as electrolytes for non-aqueous electrochemistry. $\endgroup$ – Michael DM Dryden Jul 10 '15 at 21:39
  • $\begingroup$ Out of curiosity, how can you infer that oxygen is being produced from the forward sweep? I'm having trouble understanding how to interpret our CVs and could use any advice. $\endgroup$ – John Snow Jul 11 '15 at 19:55
  • $\begingroup$ Hi curt, I wanted to check out the pH dependency you were talking about. But I looked at the equation you gave and it doesn't make sense the way it's written. The pH cancels out and doesn't have an effect. Is this a typo? $\endgroup$ – John Snow Jul 14 '15 at 20:53
  • $\begingroup$ Sorry, I'm meaning a bit confusing with my nomenclature. The "V/pH" bit is a unit, volts per pH unit. The "pH" by itself means the actual pH. So at a pH of 0 (standard state), the potential is 1.23 V, but for each change of 1 pH unit, the voltage goes down by 59 mV. $\endgroup$ – Curt F. Jul 15 '15 at 3:17

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