In short, the aim of my experiment is to determine how the concentration of sulphuric acid in solution affects the mass liberated and deposited at copper electrodes, when copper sulphate is electrolysed.

Over all data points, what I observed is that the mass deposited at the cathode was lower than the mass liberated at the anode. Theoretically, the two should be equal, and even experimentally I expected the opposite to be true, due to oxygen production taking place at the anode which could disrupt the liberation of copper. The voltage was kept constant at 3V, so I monitored the current over the duration of the experiment. I observed that the current increased, which further suggested that more ions were released at the anode than deposited on the cathode, as that represents an increase of positively charged particles in solution and thus an increase of current density.

Now, perhaps the liberation of oxygen facilitates the oxidation of the copper anode? Is this my misunderstanding?

Additionally, I was unable to find conditions which would stop the production of hydrogen at the cathode. The voltage required to electrolyse water is 1.23V, but if I set my voltage to that low, no plating took place either. Is hydrogen also expected to be produced alongside the plating process? Maybe, the hydrogen production was disrupting the plating? In that case, what is the explanation for why hydrogen was produced at the cathode in the first place? Since copper is lower in the reactivity series than hydrogen, no hydrogen should be produced.

Can someone point out the flaws in my experiment or where I've misunderstood things. I don't have much experience with electrochemistry. Thank you!

  • $\begingroup$ You better forget about the voltage, and only try to check the intensity, which depends on the surface of the electrode, and which should be around $\pu{0.02 A/cm2}$. With such a value, you should not get any Hydrogen on the cathode. The amount of sulfuric acid should not have an effect on the yield. Also try to use a high concentration of CuSO4 in the solution, around $\pu{1 M}$. $\endgroup$ – Maurice Mar 21 at 20:03
  • $\begingroup$ I was indeed using a CuSO4 concentration of 1M. What is the theoretical reason for why a lower current density would stop the hydrogen production? Also, I'm not sure what you mean by yield. Increasing the concentration of sulphuric acid increases the current density in solution, leading to a greater current and a greater yield of copper deposited. Maybe it doesn't affect the production of hydrogen. In any case, would oxygen still be produced at the anode? $\endgroup$ – PLM Mar 21 at 20:48
  • $\begingroup$ Also, consider that a) some copper may stay in solution, if you add H2SO4, and b) look for sludge at the the bottom. $\endgroup$ – DrMoishe Pippik Mar 22 at 0:36
  • $\begingroup$ @Poutnik I actually believe that hydrogen production is necessary in order for the sulphuric acid concentration to affect plating. With the reduction of hydrogen ions taking place at the cathode, a constant stream of H ions in the solution is insured, thus increasing overall current. What I observe is that if the voltage is set to too low, the current merely spikes and goes down to near 0. I assume that represents positive ions from the solution being rapidly pulled to the cathode, before the high concentration of hydrogen ions creates a layer around the cathode stopping the plating. $\endgroup$ – PLM Mar 22 at 7:52
  • $\begingroup$ Moved my comments to answer. $\endgroup$ – Poutnik Mar 22 at 8:25

$\pu{3 V}$ is too high. Kinetics of deposition of $\ce{Cu}$ is saturated, so another electroactive system kicks on, creation of hydrogen. Similarly for anode. Try to evaluate volt-ampere characteristic of the system and stay on the current plateau before gases start to be created.

Generally, is is more advantageous to use a power source keeping constant current rather than constant voltage.

If gases are being evolved, you cannot expect the dissolved and deposited mass of copper would match, as various portions of the current are spent on gases, instead of copper. The current density that Maurice suggests is a good one.

For low enough voltage, copper deposition + dissolution can absorb the current. At particular current density, one of these reactions is saturated and cannot take more current any more, being diffusion controlled. Then either hydrogen, either oxygen starts to be created, depending on the electrode where it occurs first. When it happens for the other reaction as well, the other gas gets created as well.

Each electrode has a voltampere characteristic given by I=f(E) function, that has a shape of irregular rounded step-wise slope. When a cathodic potential gets low enough value for a given compound/ion, current starts to raise. When reaction gets saturated, i becomes being controlled by ion diffusion and there is zero concentration at the electrode. Then the current reaches a plateou. The plateou ends at a potential where another redox system starts to interact. Similarly for anode.

When an external voltage is applied, electrodes gain such potentials to meet 2 conditions:

  1. The potential difference equals to the external voltage.
  2. Currents of both electrodes, according to the above function, are equal but the sign.

Lower current density means more positive cathode potential, where $\ce{Cu^2+}$ gets selectively reduced. Higher density means lower cathode potential where both $\ce{Cu^2+}$ and $\ce{H3O+}$ may get reduced. It is like if you used mildly reducing reagent, reducing just copper ions versus strongly reducing reagent, reducing also hydronium ions. Similarly for the anode.

The only necessery thing is the cathodic current matches the anodic current. Processes on both electrodes are independent. There is no "quantum entanglement" between these processes. If $\ce{Cu^2+}$ diffusion and $\ce{Cu}$ deposition can absorb all the cathodic current, there is no need of hydrogen creation.

Try to measure the steady current as a function of the applied voltage. You can use a small resistor is a serie with a cell, measuring voltage on it, which is equivalent to the passing current as $I = U_\mathrm{R} / R.$ Find the first current plateau. Or the "like to be a plateau, just if the 2 voltage steps were not so close together."

Choose the voltage from this plateau, but be aware it may change with the change of the cell geometry or electrolyte composition.

Even in cased hydronium ions do not participate on electrode reactions directly, they can participate indirectly by influencing I=f(U) characteristics of the cell via:

  • Keeping or affecting chemical state of involved ions. ( E.g. pushing equilibrium $\ce{[Cu(H2O)4]^2+ + H2O <=> [Cu(H2O)3(OH)]+ + H3O+}$ to the left.
  • Balancing the local gradient due charge disbalance
  • Increasing conductivity and therefore the solution potential difference, that subtracts from the applied voltage.
  • $\begingroup$ I really appreciate the answer. Clears things up a lot. But as I stated previously, if the hydrogen ions from the sulphuric acid do not participate in the flow of charge, then they don't have any effect on the current, right? If they were to participate in the flow of charge, then current will increase because I = Q/t. If the concentration of hydrogen ions is high, but the voltage is low, the ionization of the solute is inhibited, so the sulphuric acid concentration will only disrupt the reaction rate. $\endgroup$ – PLM Mar 22 at 8:56
  • $\begingroup$ What do you mean by "the ionization of the solute is inhibited" ? $\endgroup$ – Poutnik Mar 22 at 11:28
  • $\begingroup$ In concentrated solutions of strong electrolytes (such as sulphuric acid) there are strong forces of attraction between the ions of opposite charges — there are inter-ionic forces. Due to these inter-ionic forces the conducting ability of the ions is less in concentrated solutions. I guess I should have said that this occurs when the concentration of both hydrogen and SO4 ions is too high. This is highlighted in this thread: chemistry.stackexchange.com/questions/26781/… $\endgroup$ – PLM Mar 22 at 11:32
  • $\begingroup$ Yes that is true. Your prior formulation is confusing. But it would apply rather to unreasonably high acid concentration, and would be applicable to any soluble enough ionic compound. $\endgroup$ – Poutnik Mar 22 at 11:37

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