I'm a Chemistry teacher and have students who are performing an experiment to determine the effect of copper ion concentration on the rate of solid copper formation at the cathode in an electrolytic cell.

I've had students do this experiment in the past and they invariably find that an increase in concentration results in an increase in the rate of copper formation. At first glance, I assumed the theory underpinning this experiment could be explained quite simply using collision theory...

Put simply, an increase in copper ion concentration means that there is an increased number of copper ions at the surface of the cathode. This, in turn, results in copper ions being reduced to solid copper at an increased rate due to a larger number of successful collisions between copper ions and electrons per unit time.

After doing some reading online, however, I can't seem to find a clear-cut description of why increasing electrolyte concentration increases the rate of electrolysis. Is the relationship not as straightforward as I think it is? If not, how could I explain it without overwhelming my students (remembering that they're school students, not PhD students).


3 Answers 3


There are electrode reactions controlled by electron transfer(slow ones) or by diffusion(fast ones).

Depending on choice of forced electrode potentials, electrolyzer geometry and ion concentration, many reactions can be arranged to be electron-transfer limited or diffusion limited.

If the cathode potential is decreased below its equilibrium potential, the cathode current density starts to exponentially grow (the start of the polarographic step).

With the further decreasing of the cathode potential, the current density does not grow infinitely, but starts to be limited by cation electro-migration, with cations starting to be depleted at the electrode surface.(the end of the polarographic step).

Finally, further decrease of potential leads to the electrode current plateau unless there is yet other electrochemical system able to be electro-reduced.

Electro-deposition of copper is charge-tranfer controlled at low current densities and diffusion controlled at high current densities.

There is direct relation between ion concentration and the current density limit, caused by ion diffusion. In the limiting state, all ions arriving at the cathode due electro-migration, are immediately consumed by the reaction.

There is electrode adjacent electrolyte transition layer where cation concentration of bulk electrolyte starts to drop toward zero when approaching the electrode.

At equilibrium, the cathode reduction and oxidation currents cancel each other to the zero net current.

Note that in context of electrode processes, more important than absolute currents are current densities.

When the electrode potential lower than the equilibrium one is externally applied, the cathode increases reduction current and decreases oxidation current. This decreases the inner cathode potential until the potential difference across the wire causes the current equal to the net redox electrode current.

On the electrolyte side, the net redox current causes cation (or generally reduced object ) depletion, until the extra electrostatic gradient caused by ion displacement forms electromigration current equal to the cathode net redox current.

If the forced potential is too low and the redox net current too high, even almost total depletion is not able to cause high enough electro-migration current. As consequence, the initial net redox current is not sustainable and decreases to match the maximum migration current the electrolyte is able to provide. This is concentration dependent. The wire current adapts to the sustainable net redox current by the inner electrode potential shift toward the external forced potential.

  • $\begingroup$ Thanks for your answer. Out of interest, if a reaction is controlled by electron transfer, what does that mean for the effect of concentration on the rate? $\endgroup$
    – Vandelay
    Jan 21, 2022 at 14:47
  • $\begingroup$ Assuming anode and cathode are Cu, the number of Cu atoms oxidized equates the number of Cu ions reduced. Because as such the solution remains neutral. But the more charge carriers in the solution, the higher the rate until either rate of oxidation or reduction limit the overall rate (depends on the electrodes' surface, the applied electrical potential, etc.). $\endgroup$
    – Buttonwood
    Jan 21, 2022 at 14:54
  • $\begingroup$ Effect of concentration on electron transfer controlled reaction is small. It is like 2 reactions in a serie: diffusion + electron transfer. When one is much slower than the other, it does not matter much how fast is the fast one. $\endgroup$
    – Poutnik
    Jan 21, 2022 at 14:55
  • $\begingroup$ Right. So I'm still a little bit confused here. Can the increase in deposition rate with increasing concentration (for copper electrolysis) be attributed to there being more copper ions at the electrode (i.e., higher frequency of collisions)? Is an increased rate of diffusion to the cathode (from the bulk solution) also responsible? I'm assuming here that limiting current density hasn't been reached. $\endgroup$
    – Vandelay
    Jan 21, 2022 at 15:06
  • $\begingroup$ Surface concentration of copper ions is diffusion controlled. Electrode reaction cannot go faster than is the ion arrival rate. you have also to distinguish the momentary rate and a steady rate. The former can be much faster than the latter. $\endgroup$
    – Poutnik
    Jan 21, 2022 at 15:08

The rate of electrolysis is the amount of copper $n\ce{(Cu})$ deposited at the cathode per second (in mol/s). This amount $n\ce{(Cu})$ is proportional to the time $t$ and to the current $I$ (in Amperes) according to the Faraday's law : $n = It/zF$. Now the current $I$ depends on the voltage $U$ according to Ohm's law : $I = U/R $ where $R$ is the resistance of the solution. This resistance $R$ is such that its inverse $1/R$, called conductance $G$, is equal to the conductivity $\sigma$ times the cell constant $K$ , so that $1/R = \sigma K$, where $K$ is an experimental geometrical constant depending on the distance between the electrodes. Then the conductivity $\sigma$ is equal to the concentration $c\times \Lambda $, where $\Lambda$ is a (nearly) constant for all $\ce{CuSO4}$ solution : $\Lambda$ ($\ce{CuSO4})$ = $55+80 = 135$$ ~ \pu{S cm^2 mol^{-1}}$. Combining all these formula leads to $$n_{Cu}(t) = \frac{It}{zF} = \frac{U}{R}\frac{t}{zF} = U\times \sigma K~ \times \frac{t}{zF} = U\times c\times \Lambda \times K~ \times \frac{t}{zF} = \frac{135 U K t}{zF}\times c$$ or : $$I = 135 ~UKc$$ As a consequence, if the voltage $U$ is maintained constant in a given cell, the current $I$ is proportional to the concentration $c$ of the solution.

  • $\begingroup$ It seems to me that you assume I is proportional to U and $\sigma$ what is obviously false. At least, and just as the first approximation, there must be subtracted voltage extrapolated for zero current. $\endgroup$
    – Poutnik
    Jan 24, 2022 at 9:02
  • $\begingroup$ @Poutnik. What do you have against Ohm's law: U = RI ? $\endgroup$
    – Maurice
    Jan 24, 2022 at 15:31
  • $\begingroup$ Applying it on electrolytic cell is kind of stretch. Better approximation is I = ( U - U0)/R, and still just a course approximation. If a cell is to be approximated by an electronic system, it would be quite a complex system, not reducible to a single resistor. U=RI is applicable for conductivity measurement by AC voltage of small amplitude. $\endgroup$
    – Poutnik
    Jan 24, 2022 at 15:36
  • $\begingroup$ @Poutnik What is the value of $\pu{U_o}$ in an electrolysis of a $\ce{CuSO4}$ solution between two copper electrodes ? $\endgroup$
    – Maurice
    Jan 24, 2022 at 16:18
  • $\begingroup$ Make the graph I=f(U) and U_0 will be the intersection of the graph tangent and axis X. Even if it was initially zero for shared electrolyte, it would soon becomes nonzero due migration gradients. Now, take the case of charging of the Li-on cell. $\endgroup$
    – Poutnik
    Jan 24, 2022 at 16:24

Copper ion concentration is a variable, of course. However, the rate of copper deposition is DIRECTLY related to the current being passed. Well, except for the possible complexities addressed below.

If increasing the copper ion concentration increases the rate of copper deposition, it is likely because the power supply is not regulated. The resistance of the solution drops, the voltage does not, more current flows, and the result is that more copper deposits.

In any experiment, all the variables ought to be held constant except the one (or two) that you wish to vary. In this experiment, copper ion concentration is one variable, temperature another, applied voltage, applied current, stirring, electrode area, and on and on. Many of these variables cause effects which are better explained by reference to more fundamental variables. The one variable most fundamental in explaining the rate of copper deposition is the rate of electron flow, which is measured by the current.

There is a situation where I believe this anomalous result could occur even if you hold the current constant. If the anode were an inert material, like a carbon rod or platinum, if you have some, then, over some time period, in the dilute copper sulfate solution (assuming sulfate), the initial current would be carried to the cathode solely by copper ions, because there are essentially no other cations (pH near neutral). However, after enough current has passed and deposited copper, the copper ion concentration drops and is replaced by H$^+$ ions. When these carry the current, they evolve H2 gas, and the rate of copper deposition decreases. Aha! But you claim that you see no (or almost no) gas bubbles at the cathode. OK, perhaps you have set the current so low that the H2 dissolves before bubbles can form and rise. H2 could then evolve from the surface slowly and escape without a trace.

There's another effect that can reduce the possibility of H2 evolution: suppose you have a magnetic stirrer in the vessel (not absolutely necessary, but it helps). The O2 evolved at the anode might react with H2 in the solution, or at the electrodes (especially platinum, if you used that). This causes a reduction of evidence at the electrodes that electrolysis is occurring - an example of Faradaic (or coulombic) inefficiency. This would be totally unexpected (unless you need it to explain the results).

Now if you do the exact same experiment with a more concentrated copper sulfate solution, the generation of H$^+$ ions will be slower, so the deposition of copper will not be reduced so much during the same time of observation. Strictly speaking, it is not the higher concentration that has caused the change, but rather the extra amount of copper ion that supplies the extra copper metal. The exact same effect could be caused by incereasing the amount of dilute solution so that the copper ion is not depleted as rapidly as in the smaller amount of solution.

  • $\begingroup$ Let's imagine that all variables are held constant except for the copper ion concentation and the current increases (which often happens). Isn't the increase in current DUE to the reaction occurring at a faster rate, and not the other way around? $\endgroup$
    – Vandelay
    Jan 24, 2022 at 3:07
  • $\begingroup$ "In an electrolytic cell a current is passed through the cell by an external voltage, causing an otherwise nonspontaneous chemical reaction to proceed. In a galvanic cell the progress of a spontaneous chemical reaction causes an electric current to flow." from Wikipedia. Vandelay specified an electrolytic cell. An applied voltage is forcing the reaction. A decrease in solution resistance is what allows (or forces) more current if the voltage is held constant. The increase in current IS the reaction being forced (or allowed) to occur at a faster rate. $\endgroup$ Jan 24, 2022 at 16:46
  • $\begingroup$ Thanks for your response. So the current is essentially just a manifestation/measure of the reaction rate? (That gels with my understanding, by the way, based on Faraday's law, but I've read some annoyingly contradictory things online.) $\endgroup$
    – Vandelay
    Jan 25, 2022 at 10:15
  • $\begingroup$ @Vandelay: I think you've got it. It can get complicated because you have cations going one way, anions going the other, electrons going in and coming out - but not individually crossing the electrolyte as in a wire. Not to mention the complexities of the electrode processes and the change in concentrations and double layers and diffusion, etc, etc. These can all be addressed, but the simplest measurements are voltage, current, pH, possible gas evolution, Cu deposit weight, and concentration. Measuring these should enable a good explanation of your results. $\endgroup$ Jan 25, 2022 at 14:32

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