1
$\begingroup$

I'm trying to find an answer as to how lowering the pH of the CuSO4 solution during electroplating of copper affects the amount of copper deposited onto the cathode.

My very basic understanding is that if the concentration of hydrogen is too high, then there is greater competition and hydrogen will be reduced instead of the copper. (But why?)

Also the reason why CuSO4 is acidified is to prevent copper from hydrolysing from the solution as it prevents the oxidised Cu2+ from reacting with the OH- ions to form Cu(OH)2 precipitate.

However, I was wondering if this could be related to electrode potentials. That the increase of H+ concentration in the electrolyte affects the electrode potential, which is why increasing H+ increases the competition and makes it more likely for H+ to be reduced.

Im trying to explain why there is an optimal pH for the electroplating of copper and I need background information as to why too low pH or too high pH decreases the amount of copper deposition on the cathode.

$\endgroup$
1
$\begingroup$

In electrolysis, $Cu^{2+}$ is reduced in two steps, first to $Cu^+$ and later on to $ Cu$. Of course $Cu^+$ is automatically disproportioned into $Cu$ and $Cu^{2+}$ in acidic conditions. But in neutral solutions, $Cu^+$ may react with $H_2O$, forming an unwanted red precipitate of $Cu_2O$. This will affect electroplating of copper.

$\endgroup$
0
$\begingroup$

If the available Cu ions in the electrolyte are decreased (ex. CuSO4 electrolyte isn’t replenished using an inert anode OR a copper anode passivates due to polarization) the excess H+ combines with the remaining Sulfate ions to produce H2SO4 which will react with any Cu2O and form more CuSO4.

During the temporary decrease of available Cu ions, the conductivity of the electrolyte drops, and in the event that the voltage is set higher than what it would take to electrolyze water (1.23v), Hydrogen will start bubbling at the cathode, thereby decreasing the amount of copper deposited on the cathode. This also can happen if the anode/cathode ratio is too low which will cause a higher current density at the anode causing more oxidation than can be reduced by the surface area of the cathode.

If I set my power supply to 31v/5A, if there are sufficient Cu ions available between the electrolyte and the anode, the voltage will never go above 0.6v, but if I have contamination in my electrolyte (nickel/iron sulfate, or excess chloride ions) or the anode/cathode ratio falls too much due to the anode being consumed, the voltage will spike and I’ll get uneven plating and Cu metal accumulation on the bottom of my cell. I’ve had this happen when a bubble in the middle of my anode opened up exposing a large surface area of un-oxidized copper causing all the current to flow quickly through that area releasing a ton of oxygen bubbles at once and my electrolyte to cloud completely black temporarily.

I think to answer your question, the optimal pH for electroplating is more in relation to the availability of Sulfate ions than the excess of H+. This can be adjusted by adding additional H2SO4 to the electrolyte which will be an additional source of sulfate ions and also lower the pH.

$\endgroup$

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.