During my water electrolysis with NaOH, I measured the voltage and current, and the volume of gases (hydrogen and oxygen) displaced at every 30-second interval. 10g NaOH in 2L water for 8 minutes.

Then, I was able to calculate the rate of reaction of the gases in terms of volume per unit time. Dividing the rate of reaction by power gives the efficiency of the trial (how much oxygen or hydrogen are produced per unit energy). As I vary the voltage from 5V to 10V and then to 15V, there is a decline in efficiency (less gases produced for each joule).

What is the reasoning for this? My hypothesis is that while the rate of production is directly proportional to current, as I increase the voltage, both current and voltage increases which causes the power to increase at a quicker rate. Ex. $P=IV$, then we increase $V$ to 2$V$, $I$ therefore increases to 2$I$ due to $V=IR$. Therefore $4P=2I \cdot 2V$ where the rate of reaction only increases by a factor of 2. Or is energy lost to heat? But this will not fully account for the large decline as shown in the following picture.

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2 Answers 2


One easier way to measure the amount of products formed (albeit not 100% correct*) would be to indeed measure the charge that passed through the circuit as Poutnik mentioned.

In an electrochemical system there are different processes which all lead to losses. Not all of them follow Ohm's law $IR=V$. First there's a minimal potential required by the Nernst equation (approximately 1.23V at STP), the equilibrium potential.

However there are also kinetic losses, at some small finite potential over the equilibrium potential one will find that the current is still approximately 0. This is because the rate of charge transfer (kinetics at the electrode) at the electrode-electrolyte interface is determined by the applied potential as well. The dependence of this rate on potential follows an approximately exponential course: $i = i_0e^{\eta_{kinetic}/b}$ with $\eta_{kinetic}$ called the 'kinetic overpotential' which is the applied potential minus the equilibrium potential (and the potential of the solution $\eta_{kinetic}=U_{applied} - U_{equilibrium} - U_{electrolyte}$).

The next loss is the conduction of the electrolyte, which follows ohmic behavior. So: $\eta_{ohmic} = I*R_{electrolyte}$, which is the "Ohmic overpotential".

Finally, we need the hydroxide/water to get to the electrode and O2 and H2 out the solution. This is mass transport and is governed by the Nernst Planck equation and there is not a simple expression for this loss, but only starts being significant when you go to very high currents. At some point you will find that no matter how much potential you apply no increase in current happens, this is due to mass transport losses (as water, H2 and O2 transport is independent of the potential). One gives this also an overpotential: $\eta_{transport}$.

The total current-potential behavior is thus not straight forward, these simple 3 phenomena already are quite complex and I have not described each contribution in enough detail. For more information I would definitely start looking at electrochemistry textbooks. In general the current is some complex function of the potential, and thus the efficiency at different rates (i.e. currents) is different: $I=f(U_{applied})$ and $efficiency=\frac{U_{applied}}{U_{equilibrium}}$.

Finally, all losses end up as heat.

*(Current going to side reactions or double layer charging of your electrodes)

  • $\begingroup$ Thanks, but is there any theoretical data (methods to find the supposed rate of reaction to power ratio) to compare with my experimental results? As you mentioned, this is extremely complex, so is calculating the theoretical ratio feasible? $\endgroup$
    – Dian Sheng
    Jan 13, 2023 at 7:40
  • $\begingroup$ You could start with measuring the resistance due to the electrolyte, you can measure the conductivity and then also the distance between the two electrodes. This would give you the resistance due to the electrolyte which will probably be a large part of your losses. If you know the surface area of your electrodes and the material you could also find kinetic parameters for the kinetic overpotential. $\endgroup$
    – Noah
    Jan 13, 2023 at 12:27

The amount of products is determined by passed charge $Q = \int_t{I \cdot \mathrm{d}t}$, not by spent energy $E = \int_t{U \cdot I \cdot \mathrm{d}t}$.

The relation of the charge $q$ and amount of substance $n$ is $q = znF$, where F the the Faraday constant $F = e \cdot N_\mathrm{A} \approx \pu{94685 C mol-1}$ and $z$ is the number of electron per particle.

Majority of energy even for $\pu{5 V}$ is wasted. Even worse for $\pu{10 V}$ and $\pu{15 V}$. You need not a source of high voltage. You need a low voltage source with low internal resicence, able to provide high current.

Theoretical voltage needed to electrolyze water is $U_\mathrm{theor} = \pu{1.23 V}$. Practical needed voltage lays in range $\pu{1.5 - 2.0 V}$, due kinetic reasons, depending on used electrodes, solution, geometry of the cell and required production rate.

The power being lost (as heat) $$P_\mathrm{lost} = ( U - U_\mathrm{theor})I \tag{1}$$

and the efficiency $$\eta = \frac{U_\mathrm{theor}}{U} \cdot 100\ \% \tag{2}$$

As first (linearized) approximation:

$$\eta = \frac{U_\mathrm{theor}}{U_\mathrm{theor} + RI} \cdot 100\ \% \tag{3}$$

where $\mathrm{R}$ is the Ohmic resistance of the electrolytic cell. Note that $\mathrm{R}$ is not constant, but depends on current, temperature, mixing, geometry, solution and other factors.

There is trade off between the electrolysis efficiency and the production rate. The maximum efficiency is near zero production rate and decreases with the rate.

  • 1
    $\begingroup$ You are right. But you should not use informations which are difficult to understand like "thrown out of a window" or "hard source" or "soft source", or "absolute loses" or "relative loses", or "minimal loses of both kinds means also the minimal rate". $\endgroup$
    – Maurice
    Jan 12, 2023 at 9:36
  • $\begingroup$ What is difficult to understand? Those are frequently used terms. But OK, I have rephrased the considered parts. $\endgroup$
    – Poutnik
    Jan 12, 2023 at 10:09

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