# Role of the counter/auxiliary electrode in cyclic voltammetry

I am doing a school project on cyclic voltammetry and am currently studying the topic. I have learnt that the reaction mostly takes place around the working electrode where the current is measured together with the counter/auxiliary electrode. My question is, how does the counter/auxiliary electrode pass current without oxidising or reducing the analyte?

For example, I were to increase the potential in a solution of $\ce{Fe^3+}$ ions. The working electrode would be the cathode and would reduce the ions, giving them electrons and forming $\ce{Fe^2+}$ ions. In this case, what would happen around the counter electrode? How would the electrode be able to pass current to balance it? Would it oxidise the newly formed $\ce{Fe^2+}$ ions? Wouldn't this result in a never ending process? If someone has already asked this question, I apologize. Please link the page, thank you.

• In order for the solution to remain electrically neutral something has to be oxidized and something else has to be reduced. – MaxW Jan 21 '16 at 16:24

You are right that the counter electrode must balance the redox reactions happening at the working electrode. For every electron that goes to reduce an $\ce{Fe^3+}$ ion at the working electrode, an electron must be collected from an oxidation reaction at the counter electrode.
What that reaction actually is depends on a lot of things. First, we must consider what is actually in the solution to oxidize at the counter electrode. In an $\ce{Fe^3+}$ solution, unless the working electrode is very close to the counter electrode, the newly formed $\ce{Fe^2+}$ ions will not diffuse efficiently to the counter electrode, so something else must balance the bulk of the current. In many aqueous solutions, this might be the water itself being oxidized, unless there is a more easily oxidized substance in the solution (a supporting electrolyte, for instance). If the counter electrode material is chosen poorly, it may even be the counter electrode itself oxidizing.