2
$\begingroup$

I'm currently studying in the International Baccalaureate model. This high school model asks their students to do an individual work called the Internal Assessment. One of the subjects I study is chemistry, and my Chemistry Internal Assessment is based on the production of hydrogen (a source of energy) in the water electrolysis as a way to save electrical energy. I was going to do an experiment to measure the production of hydrogen in the electrolysis of water with various concentrations of four different solutes (NaCl, NaOH, sulphuric acid and sodium bicarbonate, but any electrolyte will do). However, with the Covid-19 outbreak, I cannot go to the laboratory, so in order to get the data I need, I searched on google databases with the information I wanted, but I found barely nothing. These links (https://www.scirp.org/pdf/NR_2013032814044640.pdf) (https://pubs.rsc.org/en/content/articlehtml/2017/se/c7se00334j) are the only useful pdfs I found, and I wonder if anyone could help me giving me some databases with the data I need.

$\endgroup$
7
  • 1
    $\begingroup$ I am afraid the amount of H2 produced by electrolysis does not depend on the concentration or of the formula of the chosen salt or hydroxyde, provided it is a soluble alcaline compound. It only depends on the intensity of the currant. $\endgroup$
    – Maurice
    Commented Mar 27, 2020 at 20:32
  • $\begingroup$ it should, as it enhances electrical conductivity. moreover, there is a point where the solute concentration is so high that the electrolysis of water will stop, and the electrolysis of solute begins. that is the point of my investigation. $\endgroup$ Commented Mar 27, 2020 at 20:38
  • $\begingroup$ The problem of performing an electrolysis in a solution containing nearly no water is a tricky problem. it may happen with concentrated sulfuric acid and concentrated solutions of NaOH. By memory, it seems that electrolysis of sulfuric acid produces peroxydisulfuric acid. I am not a specialist in this domain, but it deserved to be studied. $\endgroup$
    – Maurice
    Commented Mar 28, 2020 at 10:56
  • $\begingroup$ Of course is a problem, I want to see what is the optimal concentration of solute (the maximum velocity at the production of hydrogen and oxygen), where the solute does not react with either the anode and cathode. In the laboratory I was only going to use sodium bicarbonate for safety reasons, but now that I redirected my study and I use databases, I don't have to worry about that, that is why now I use more solutes. $\endgroup$ Commented Mar 28, 2020 at 11:07
  • $\begingroup$ Sodium bicarbonate is the least candidate to start such a study. it is not enough soluble in water. $\endgroup$
    – Maurice
    Commented Mar 28, 2020 at 11:12

3 Answers 3

1
$\begingroup$

You need to look for the data in a different way. You can use electrical conductivities to get the current flow the solution. From the imagined current flow you calculate how much hydrogen would be generated.

$\endgroup$
0
$\begingroup$

Alejandro, you are right, the electrolytes will make a difference. The difference is not in terms of hydrogen production per se, as Maurice mentions in the comments, it will be in terms of energy saving. Electrolytic reactions are "exact" in the sense that if 100 electrons reach the electrode from a battery, 100 hydrogen ions will be reduced. In other words,

1 mole of electrons will reduce exactly 1 mole of a monovalent ion, 1/2 mole of a divalent, 1/3 mole of a trivalent ion and so on.

You would probably like to have an electrolyte which has the least electrical resistance. Now you may recall that the proton in water (= hydrogen ion) has the least electrical resistance known today. What comes next? Hydroxide ions.

All you have to search and study is the conductivity of electrolytes at various concentrations.

$\endgroup$
4
  • $\begingroup$ that is very helpful. Thank you, but even though hydrogen has the least electrical resistance, oxygen has not. This is very important, because we could storage hydrogen and oxygen produced at the electrolysis to then, in a galvanic cell, combine these two elements to produce energy. If the oxygen does not electrolyse, we would lose an important element. $\endgroup$ Commented Mar 28, 2020 at 10:44
  • $\begingroup$ @Alejandro. What do you want to say, when stating that "oxygen does not electrolyse". As written, this sentence does not seem to have any meaning $\endgroup$
    – Maurice
    Commented Mar 28, 2020 at 10:49
  • $\begingroup$ Sorry, I wanted to say oxidize $\endgroup$ Commented Mar 28, 2020 at 10:51
  • $\begingroup$ I mean hydrogen ion has the highest conductivity not hydrogen. Be careful in terminology. Whenever there is a reduction, there is an oxidation. $\endgroup$
    – ACR
    Commented Mar 28, 2020 at 14:41
0
$\begingroup$

I did find a work, 'Electrolyte Engineering toward Efficient Hydrogen Production Electrocatalysis with Oxygen-crossover Regulation under Densely Buffered Near-neutral pH Conditions', with comments of interest.

I start with some background from the introduction:

In recent decades, drastic progress in solar fuel production has occurred: photovoltaic cells can generate electricity from the sunlight,[1−4] whereas photoelectrochemical [5] and photocatalytic [6−11] water splitting can directly produce hydrogen and oxygen by harvesting the sunlight. An electrolyzer can electrochemically split water molecules,[12,13] which can be conjugated with the photovoltaic cells.[14,15] Importantly, during the photoelectrochemical and photocatalytic water splitting, what takes place on the surface is electrocatalysis, in association with the photonic processes at the bulk of photon absorber and the interface between the photon absorber and electrocatalyst.[11] Electrocatalysis will thus certainly play a crucial role in the solar fuel production process in the future.

And further concentration-related statement taken from the abstract, where HER refers to the hydrogen evolution reaction:

The choice of electrolyte in terms of its identity and activity drastically altered the HER performance. Electrolyte properties (activity coefficient, kinematic viscosity and diffusion coefficient) accurately described the mass-transport contribution, which was easily isolated when a highly active Pt catalyst was used. The HER rate on the Pt was maximized by tuning the solute concentration (typically 1.5 – 2.0 M). Moreover, the kinematic viscosity and oxygen solubility under such densely buffered conditions governed the oxygen mass-transport flux in the electrolyte, which in turn tuned the cross-over flux. At near-neutral pH, as high as 90% selectivity toward the HER was achieved even under an oxygen saturated condition, where only a 40 mV overpotential was needed to achieve 10 mA cm−2 for the HER. This information can be regarded as an important milestone for achieving a highly efficient water splitting system at near-neutral pH.

Of potential interest are associated graphs, starting with Figure 1, on Page 31, to Figure 9.

$\endgroup$
0

Not the answer you're looking for? Browse other questions tagged or ask your own question.