I am planning to try copper plating a piece of metal by performing electrolysis on an aqueous solution of copper sulfate. I plan run an electrical current with the metal I want to plate as the cathode and a platinum anode. I understand that the copper ions will become reduced when they plate the metal, but I am curious what will happen to the sulfate ions. Will it oxidize to form sulfur trioxide with the produced oxygen from the electrolysis?

  • $\begingroup$ Related post: check the standard reduction potentials. $\endgroup$
    – ringo
    May 16, 2015 at 20:38

1 Answer 1


A typical solution for copper plating contains about $0.5\ \mathrm{mol\ l^{-1}}$ $\ce{CuSO4}$ and $0.5\ \mathrm{mol\ l^{-1}}$ $\ce{H2SO4}$:

$$\begin{align} c(\ce{Cu^2+})&=0.5\ \mathrm{mol\ l^{-1}}\\ c(\ce{H+})&=1\ \mathrm{mol\ l^{-1}}\\ c(\ce{SO4^2-})&=1\ \mathrm{mol\ l^{-1}}\\ \end{align}$$

The positive ions $\ce{Cu^2+}$ and $\ce{H3O+}$ are attracted to the cathode. The redox potentials $E$ for $\mathrm{pH} = 0$ show that cathodic reduction of $\ce{Cu^2+}$ to $\ce{Cu}$ is preferred over reduction of $\ce{H+}$ to $\ce{H2}$:

$$\begin{alignat}{2} \ce{Cu^2+ + 2e- \;&<=> Cu}\quad\quad &&E^\circ = +0.340\ \mathrm{V}\quad\quad &&&E=E^\circ+\frac{0.059\ \mathrm{V}}{2}\log 0.5 = 0.331\ \mathrm{V}\\ \ce{2H+ + 2e- \;&<=> H2}\quad\quad &&E^\circ = +0.000 \end{alignat}$$

At the anode, the redox potentials $E$ imply that oxidation of $\ce{H2O}$ to $\ce{O2}$ is preferred over oxidation of $\ce{SO4^2-}$:

$$\begin{alignat}{3} \ce{O2 + 4H+ + 4e- \;&<=> 2H2O}\quad\quad &&E^\circ = +1.229\ \mathrm{V}\\ \ce{S2O8^2- + 2e- \;&<=> 2SO4^2-}\quad\quad &&E^\circ = +2.01\ \mathrm{V} \end{alignat}$$

Nevertheless, $\ce{SO4^2-}$ actually is oxidised at the anode: $$\ce{SO4^2- -> SO4- + e-}$$ The reason for this reaction is that negative ions are attracted to the anode. Under neutral conditions, $\ce{OH-}$ ions would migrate to the anode where they would be oxidised to $\ce{O2}$. However, under the given acidic conditions, the concentration of $\ce{OH-}$ is very low. The predominant negative ions are $\ce{SO4^2-}$, which migrate to the anode.
Since the produced concentration of $\ce{SO4-}$ is low, the $\ce{SO4-}$ ions do not combine to $\ce{S2O8^2-}$. Instead $\ce{SO4-}$ reacts with water: $$\ce{4SO4- + 2H2O -> 4SO4^2- + O2 + 4H+}$$

Therefore, the net reaction is the above-mentioned oxidation of $\ce{H2O}$ to $\ce{O2}$. The $\ce{SO4-}$ ions act as catalyst:

$$\begin{align} \ce{4SO4^2- &-> 4SO4- + 4e-}\\ \ce{4SO4- + 2H2O &-> 4SO4^2- + O2 + 4H+}\\ \hline\ce{2H2O &-> O2 + 4H+ + 4e-} \end{align} $$

  • $\begingroup$ Hmmm... I thought the hydrogen sulfate ion is not a strong acid and will not ionise completely. So is it safe to assume c(H+) = 1 M? $\endgroup$ Jan 3, 2018 at 12:14
  • $\begingroup$ I have never seen this mechanism proposed for the catalysis of the oxidation of water using sulfate as the catalyst. Any references to support? $\endgroup$ Jan 3, 2018 at 12:16

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