If I have aqueous sulfuric acid solution as electrolyte and the anode electrode is made out of copper and the cathode electrode is made out of graphite, what are the two reactions at the respective electrodes and why?

\begin{align} \text{Anode:} && \ce{Cu &-> Cu^2+ + 2e-}\\ \text{Cathode:}&& \ce{2e- + 2H+ &-> H2 (g)}\\\hline \text{Total:} && \ce{2H+ + Cu &-> H2 + Cu^2+} \end{align}

That should be the reaction, but why? Why isn't sulfuric acid participating? Why does that specific reaction happen?


1 Answer 1


To analyse this kind of problem, the first step should be identifying all the species you have in solution that could take part in any electrolytic reaction. These are:

$$\begin{array}{}\ce{SO4^2-} & \ce{H+} & \ce{H2O} & \ce{Cu}\end{array}$$

I did not add graphite to this list since it generally never participates in electrolytic processes. The next step would be to identify what can be reduced and what can be oxidised. Thankfully, all species are either in their lowest or highest common oxidation state. See the list below with oxidation states added:

$$\begin{array}{cccc}\ce{\overset{{+VI}}{S}\overset{-II}{O4}^2-} & \overset{\mathrm{+I}}{\ce{H+}} & \ce{\overset{{+I}}{H2}\overset{-II}{O}} & \overset{\pm 0}{\ce{Cu}}\end{array}$$

Obviously, the only candidates for oxidation are oxygen or copper, and the only candidates for reduction are sulfur or hydrogen. I’ll concentrate on the oxidation first, as the equations are simpler. We need to find a redox pair for each component and then decide which is more likely. For copper, the only real possibilty is generating copper(II), while for oxygen the only real possibility is generating oxygen gas. See equations $(1)$ and $(2)$:

$$\begin{align}\ce{Cu &-> Cu^2+ + 2 e-}&&E^0 = +0.337~\mathrm{V}\tag{1}\\ \ce{2H2O &-> O2 + 4 e- + 4 H+}&&E^0 = +1.229~\mathrm{V}\tag{2}\end{align}$$

I added the standard electrode potentials $E^0$, too. We can easily compare which oxidation is more likely: the smaller potential will act first, thus, copper is oxidised. Now on to the reductions. For hydrogen, it is naturally the known equation. What about sulfur in sulfate? Well, the most obvious choice is assuming $\ce{SO3^2-}$ as a reduction target. See equations $(3)$ and $(4)$:

$$\begin{align}\ce{2H+ + 2e- &-> H2 ^}&&E^0 = \pm 0~\mathrm{V}\tag{3}\\ \ce{SO4^2- + 2 e- + 2 H+ &-> SO3^2- + H2O}&&E^0 = +0.17~\mathrm{V}\tag{4}\end{align}$$

The reduction potential is actually given for the target $\ce{SO2 (aq)}$; however, that has the same oxidation state and is in equilibrium with sulfite, so they are comparable. Here, the more positive potential should actually react first. However, using the graphite electrode typically favours hydrogen evolution via overvoltage mechanisms. This is, unfortunately, something you will have to memorise.

Summing up:

  • There are only four reactions that can happen
  • of these, two can be chosen as more favourable by their respective electrode potentials
  • one choice needs to be adjusted due to overvoltage.

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