I understand (more or less) that the acidification of water helps in the dissociation of water into ions. Acids easily dissociate into ions in an aqueous solution. So, accordingly, the hydrogen from the acid goes to the cathode and the hydrogen from the water makes a compound with the negative radical from the acid. In the process, oxygen is left alone and hence goes to the anode. I want to confirm whether this is the correct explanation.

I have been trying to find out why is copper sulfate acidified during its electrolytic refining. I interpret that acidifying copper sulfate makes it dissociate easily into ions but I want to know what the exact process is. It doesn't seem to me that copper sulfate may behave in the same way as water does above.

In general, I would like to understand the effect of decreasing or increasing pH on the process of electrolysis.


2 Answers 2


I'd rather not put it in such terms. Dissociation of water is determined by the constant $\ce{K_w=[H+][OH-]=10^{-14}}$, which you can't change simply by adding acid (or anything else, for that matter). Acids dissociate easily into ions, just like you said, therefore adding acid to pure water (which is a poor conductor by itself) makes the latter more conductive and thus more prone to electrolysis.

Electrolytic refining of copper is a whole different story. Indeed, once dissolved in water, $\ce{CuSO4}$ would dissociate into ions completely, no matter whether or not you acidify it. The role of acid is just to prevent hydrolysis.

In general, as you change pH, some of your ions may undergo certain transformations (like $\ce{CO3^2- -> HCO3- -> CO2\uparrow}$, or $\ce{Fe^2+ -> Fe(OH)2\downarrow}$), which in turn might or might not affect the reactions on one or both electrodes. As it happens, being too general prevents us from drawing more specific conclusions.

  • $\begingroup$ Thanks for your response. Would you please also mention how does copper sulphate hydrolyze in water. $\endgroup$ Commented Jan 12, 2017 at 10:14
  • $\begingroup$ Honestly I don't know much about hydrolysis of copper sulphate. So i want to ask whether something with happen if copper sulphate is left alone with water. Will calcium sulphate and water react to form other compounds $\endgroup$ Commented Jan 12, 2017 at 11:03
  • $\begingroup$ $\ce{CuSO4}$ by itself hydrolyzes only partially; however, with local variations of pH created by electrolysis, it may go all the way to $\ce{Cu(OH)2}$ or some basic salts which would precipitate from the solution, and you don't want that to happen. $\endgroup$ Commented Jan 12, 2017 at 11:10

You are right in that adding acid increases the concentration of dissociated ions in water, and that the hydrogen goes to the cathode but you have assumed too much with the oxygen part.

In electrolysis/electroplating, the positive ions move towards the negative cathode and are reduced, and vice-versa, the negative ions move towards the positive anode and are oxidized. For a pure acid solution (no salts), hydrogen is the only postive ion so it migrates towards the cathode, however not only do the hydroxide ions move towards the anode, but so do the conjugate bases of the acid. These conjugated bases may or may not be oxidized before the hydroxide depending on potentials and available mechanisms for reaction to occur.

For example, chlorine can oxidize before oxygen and electrolysis with chlorides forms aqueous chlorine as shown $\ce{2Cl-_{(aq)} -> 2e- + Cl2_{(aq)}}$ and sulfate can form persulfates at the anode: $\ce{2SO4^2- -> (SO3OOSO3)^2- + 2e- }$. These bases compete with the $\ce{OH-}$ in solution to be oxidized. These species also tend to be very reactive and will often oxidize the anode unless an inert material such as graphite or platinum is used (sometimes nickel will work).

In electroplating, the metal ions compete with the hydrogen ions to be reduced and indeed hydrogen gas is formed at the same time that electroplating occurs. For everr $\ce{H+}$ reduced, there is a $\ce{OH-}$ left in solution. In electrolysis, the electron to reduce the $\ce{H+}$ would come from the $\ce{OH-}$, but in electroplating it comes from the metal (i.e. $\ce{Cu -> Cu^2+ + 2e-}$). Now the electroplating process has created $\ce{Cu^2+}$ ions and left $\ce{OH-}$ ions in excess which creates $\ce{Cu(OH)2}$ which precipitates creating anode mud.

This is where the acid comes in to keep the bath clean, I'll use sulfuric acid as an example. The sulfuric acid prevents the formation of $\ce{Cu(OH)2}$ by providing $\ce{SO4^2-}$ and $\ce{H+}$ ions to react with the $\ce{Cu(OH)2}$ and keep the copper in solution $(\ce{Cu(OH)2 + H2SO4 -> CuSO4 + 2H2O})$.


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