CV is usually used to monitor corrosion (from what I have learned) because, if it is a reversible reaction, we can see the cathodic and anodic peak in the voltammogram. I know that the scan rate influences the intensity of the current peak and I also know why that happens. However, I just can't seem to find the reason why different solutions result in different current peaks in the voltammogram. Especially if the solutions get more acidic the current peak changes.

So my question is why do the peaks in the voltammogram change when we change the solution?


Recording a CV assess both so-called half cell reactions. If balanced individually, some of them only need only the transfer of an electron, e.g. for

$$\ce{Fe^{2+} -> Fe^{3+} + e-}$$

Other half-reactions include either $\ce{H+}$ or $\ce{OH-}$, and their consumption / generation alters the pH value of the solution, e.g.

$$\ce{MnO4^- + H2O + 3 e- -> MnO2 + 4 OH-}$$ or $$\ce{MnO4^- + 4 H3O+ + 2 e- -> Mn^{2+} + 6 H2O}$$

Since each of them are characterized by a standard electrode potential, altering the pH value of the solution thus may alter the reaction pathway accessible; the reaction of permanganate to manganese dioxide is typical for a reaction in a basic aqueous reaction, the one to yield $\ce{Mn^{2+}}$ typical for an acidic one. Indeed pH electrode is an application of this pH dependence which may be quantified by the Nernst equation.

While the examples shown deal with small inorganic ions, the same principle is valid for recording a CV for organic molecules, too. For the couple quinone / hydroquinone, the variation of the pH value thus may yield a plot like the following:

enter image description here


The changes for the redox couple, recorded in different solutions of different pH value equally were plot with the cell potential in function of said variation of the pH value of the analyte:

enter image description here

Note: compare the half reactions with each other. If both the number of electrons, as well as the number of protons exchanged are equal, the potential of the cell will not be affected by a change of the pH value.

reference: pH Dependent Redox Couple: An Illustration of the Nernst Equation, Walczak et al., J. Chem. Educ. 1997, 74, 1195-1197; doi 10.1021/ed074p1195

  • $\begingroup$ Thanks for the answer, but I still don't really understand it. In the figure the peaks barely change, it's just that the potential has to change to get the reaction going. I have found other figures in papers where the peaks change due to a different solution. I don't understand why that happens. $\endgroup$ – Paperreader Aug 8 '20 at 13:36
  • 1
    $\begingroup$ The peak position, e.g. about 0.38 V (pH = 1.6), 0.27 V (pH = 3.4), ..., 0.80 V (pH = 6.3) changes significantly because these are recordings in different, independently prepared samples. It is not that there was one beaker with the hydroquinone / quinone couple at pH 1.6 you record one CV which after this cycle happens to have a pH = 3.4 because of the previous recording of the CV. The buffer acetate/phosphate concentration was about 0.1 M, hydroquinone only 1 mM. CV does not convert synthetically useful amounts of matter, as electrosynthesis (youtube.com/watch?v=8UPQLiR4Fsk) $\endgroup$ – Buttonwood Aug 8 '20 at 15:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.