Electrolytic Production of Sulphuric Acid + some questions.

Great work DeAlt! Right noe, it would be a project of only academic concern since I am still able to obtain automotive battery electrolyte at reasonable cost. But, having a workable plan as back up is getting more practical as global governance tighens its death grip on us all.

A few point and concerns if you will:

Since the method only produces 20% concentration with any respectable efficiency, boiling off to concentrate the acid is going to be a major endeavor. With battery electrolyte of 35 to 40% initial concentration, it is necessary to boil down to ONE FOURTH original volume just to obtain acid of specific gravity of 1.84! So if you want respectable amounts of 98%, then you will need MASSIVE amounts of the 20%.
Also, there is a potential for significant acid losses due to varying combinations of water and H2SO4 that boil at different temperatures. So, it might be pertanent to distill the volatiles into a receiver as apposed to just "boiling off".

There seems to be a variation to the cells resistance during startup and full run. Maybe just hooking up a high powered Variac transformer with a bridge rectifier circuit to like 3 or 4 cells in series might help this. Start with 16- 20 V per cell then as the current climbs, turn down the voltage. A 1000W dimmer might even be handy here.

Now questions:

DeAlt-do you have data on current density at your cathode and anode? What temperatures did you run at?

Secondly- I am out of time, could someone find some data on the boiling points of the various H2SO4/water combinations?

Thanks in advance

Sulphuric acid is on the DEA diversion list II and also has been noted by the UN. Quantities involved are probably large and concentration high, I don't have the details.

....some data on the boiling points of the various H2SO4/water combinations



I have a whole lot more data on H2SO4 that I gathered in this investigation - but that will have to be later...

Regards, Der Alte

Awesome work again, DerAlte! Does this work with FeSO4 or Na2SO4 too? The latter is tempting because plaster of Paris could be used as a source of SO4(2-): CaSO4 + Na2CO3 => Na2SO4 + CaCO3

Sulphuric Acid, contd.: Concentration, etc

As Chloric pointed out, a lot of effort is needed to concentrate weak sulphuric acid. The acid is somewhat unusual in that anything below around 75% is called dilute. This is about 13M. Pure acid is 18.6M.

For general reagent purposes a 10% acid (1M) often suffices. For such things as producing nitric acid, HCl, SO2, etc, around 70-95% is desirable. And for organic chemistry, in such operations as chemical dehydration or as a catalyst, the stronger the better. As a dessicant it is second to none (almost) but must be >80% for effectiveness (partial pressure of H2O is then about 10pa (n/m^2 in MKS) or 0.08mm Hg.@ 20C).

To estimate the strength nothing beats titration, but a hydrometer is very useful. If you know the strength of a sample, you can estimate what you have after concentration by comparing volumes. Sulphuric acid is essentially non-volatile, like phosphoric acid. Here are a few figures I have calculated for the relative volumes of solutions containing the same amount of acid:

2%, 90.9; 5%, 35.7; 10%, 17.2; 20%, 8.03; 50%, 2.62; 70%, 1.63; 80%, 1.33; 95%, 1.05; 100%, 1.00

You cannot concentrate beyond about 97% by heating. At that temp. it forms a sort of constant BP mixture which is actually a decomposition product into H2O and SO3. This will recombine at a lower temp, so in a sense, H2SO4 can be ‘distilled’ to get about 97.5%, but the pure acid cannot be gained by heat alone. You have to add SO3 to partially hydrated acid to get 100% or oleum (see thread on this; UTSE.)

H2SO4 forms distinct hydrates. You can actually concentrate weak acid (<37%) by freezing. I have tried it. It is not a practical method because the ice forms as a gelatinous slurry, almost impossible to separate. I saw a patent for doing this, by centrifugation. Also, the monohydrate H2SO4.H2O (84.4% by weight), freezes out at 8.6C, a tetrahydrate at about -33C. See the freezing curve:



Heating in a microwave oven works well, up to the point where the vapor pressure of H2SO4 becomes high enough to cause corrosion. Partial pressure of the acid (not the water component) at the boiling temperatures is approximately as follows;

80% BP ~205C PP of acid ~ 0.3mm Hg.
74% BP ~180C PP of acid ~ 6*10^-2 mm Hg.
67% BP ~160C PP of acid ~ 4*10^-3 mm Hg.
62% BP ~140C PP of acid ~ 2*10^- 4 mm Hg.

So anything below 70% should not produce noticeable fumes (only steam).

One problem with the MW oven is the very high dielectric const of H2SO4 (~100) plus the very high conductivity of conc. acid. This tends to put a high stress on the magnetron (due to mismatch) which is likely to run hot and may trip the thermal breaker which some have – or blow the magnetron/and or PS in a cheapie.

Traditional boiling in an open vessel works well. I have had little bumping until you get to around 70%. Use a bit of unglazed porcelain or even silica sand to aid stable boiling. And be very careful of the hot acid (over 300C) when it gets near maximum concentration of 95+%, where the vapor pressure is over 75 mm Hg, producing copious noxious acid fumes. I saw a picture once of what hot conc. acid does to a hand – not pretty.

A convection oven works slowly and is less efficient, but the movement of vapor off the acid by the fan helps evaporation. There is the same problem of possible corrosion, so 70% odd is about the limit.

Finally, the amount of heat required to evaporate off water is rather high. I calculate, crudely, that to evaporate 1g acid in a 20% to a 95% solution takes about 12KJ. This includes the heat required to bring to temp., enthalpy of evaporation of water and the enthalpy of dilution which has to be returned on concentration. Thus, although the electrolysis is not very efficient beyond 20%, the price on concentrating is such that one might as well go to 30% or even higher and concentrate as far as possible that way, by electrolyzing off the water! Pity one can’t use the H2 and O2 in some way, it seems very wasteful to just let them dissipate.

A lot of work for half a liter of acid! Enough said,

Regards, Der Alte

Der Alte - greetings!

I came across the following electrolytic method for producing sulphuric acid. What is interesting is that it claims to produce concentrations as high as 95%. The procedure is from "The Manufacture of Chemicals by Electrolysis" a treatise which appeared in 1919, and describes methods in use at the turn of the century. The internet copy is confusing to read, has many mistakes and the diagrams are missing, but someone may be interested in trying this out.

"ELECTROLYTIC SULPHURIC ACID is produced and a con-
centration as high as 95 per cent, obtained by oxidising
sulphurous acid in a diaphragm cell with a cylindrical
nickel cathode and an anode of platinum gauze. A porous
cup or cell which acts as cathode is filled with sulphuric
acid or sodium sulphite, and the outer anode compartment
contains a solution of sulphur dioxide which is kept
saturated during the process by passing in the gas
continuously. The current density used is about 1 amp. per
dm. 2"

I guess SO2 could be produced from a simple sulphur burner.
Use a platinum coated Ti mesh electrode.

May not be practical, but it seems to be an interesting amateur exercise!

Further ideas to increase current efficiency (CE)

If we waive the condition of very little sodium in the acid produced, we can start with NaHSO4 in both anode and cathode cells. Then initially, on electrolysis, the ions passing through the diaphragm will be chiefly Na+ from anode side to cathode, and HSO4- in the reverse direction (ignoring the much smaller SO4-- ion concentration). Simultaneous enrichment of HSO4- and depletion of Na+ will occur at the anode cell. 1 Faraday will move ~1 mol Na+ (leaving behind 1 mol HSO4-, and will also move ~1 mol HSO4- into the anode cell, for a total of 2 mol HSO4- per Faraday not neutralized by Na+. Initial CE, g/AH of increase in unneutralized SO4 concentration, twice the previous amount or about 7.3g/Ah.

The general condition while both cells are acidic can be shown as follows, (brackets = relative amount for pH~1 in both cells):
Anode cell to cathode cell: <-- Na+ (major); <-- H+ (minor) …. (1).
Cathode cell to anode cell: --> OH-(very small); -->SO4-(minor); -->HSO4- (major) ..(2)
At cathode: 2H+ + 2e- --> H2(g)
At anode: 2HSO4- --> 2HSO4* + 2e-; HSO4* + H2O --> H+ + SO4- + ½ O2(g)
* = transient radical.

As the anode cell increases its acid content its pH falls; further, as Na+ ions are abstracted [Na+] decreases. Hence the efficiency of the Na+ transfer decreases as [H+] increases. Simultaneously, the [HSO4-] drops in the cathode cell and [SO4--] increases, and the transfer of SO4 moiety decreases for a given current.

Once the cathode cell approaches pH 7, the efficiency of SO4 transfer has dropped to 50% and further increases in pH cause it to drop much lower, due to increasing [OH-]. At the anode pH decreases further, [H+] increases and the transfer rate of Na+ to the cathode cell diminishes. It is thus theoretically impossible to deplete the anode cell entirely of Na+ ions.

I have attempted some rough calculations to see what this variation of efficiency might be theoreticlly. We are using very concentrated solutions so the accuracy is likely to be very approximate, due to variations in ionic activity, but such calculations show a trend. I used a pH calculator to estimate the ionic activities, but although there is some correction, the concentrations are way outside the range of accurate results. I will not go into detail because the calculation proved long and tedious. It does not help at all that sulphuric acid is diprotic!

The results can be applied to the original case discussed previously, where there are no Na+ ions in the anode cell (well, very few.)

The best presentation seems to be graphical. The key parameter is the ratio of [Na+] to [total SO4], i.e. or alternatively [H+] or pH. The dependent quantity is the efficiency as compared to one mol HSO- per Faraday = 3.66g H2SO4/Ah. A separate graph of pH vs. [SO4--] +[HSO4-] is also shown.




The efficiency of Na+ transfer from anode cell (A) to cathode cell (C) is
100*[Na+]/{[Na+]+[H+]}

and of SO4 ions C --> A is
100*{[HSO4-] + [SO4--]}/{[HSO4-] + 2[SO4--] + [OH-]}

The pH refers to solutions in the region of 2-3 M where ion activity tends to be variable, typically. Accuracy not to be expected except near the critical neutrality point. If it looks like a traditional titration curve, it is – the only difference is that only concentrations vary instead of concentration and volume.

This graphs indicate that in the case where cathode cell C contains NaHSO4 initially (50% Na to SO4 ion ratio) the expected efficiency is about 88% of 3.66 = 3.22 g/Ah. (The commercial NaHSO4 usually contain 8% Na2SO4 and this drops the efficiency to about 3.1g/Ah). This accords closely with what I measured initially earlier. When the C contents become Na2SO4 (neutral, at 66.7% Na) this efficiency drops to 50% exactly = 1.83gAh. Further electrolysis gives diminishing returns; at 90% Na it is only 0.4g/Ah. If the aim were to produce pure NaOH in C, it would take forever. Hence an indicator or pH meter is a help if you want to keep efficiency high. The ‘efficient’ region is narrow, between 50 to 66.7% Na/SO4 ratio. This means only ½ of the sulphate in NaHSO4 can be transferred efficiently from C to A.

If both A and C are started with NaHSO4, as much Na is transferred to C as SO4 to A initially. This results in about twice the above rate of enrichment of H2SO4 in A, i.e. about 6.4g/Ah. As C becomes less acidic, the rate drops as before. When C becomes neutral, this rate is 50% of initial Na to C from A, 1.83g/Ah. The C to A transfer rate of SO4 moiety depends on the relative C concentrations, and the relative size of A and C. It helps to have a large cathode cell C to avoid this dropping rapidly – a coaxial cell usually guarantees the volume of C > volume of A.

I have tried this method. The expected efficiencies are indeed much higher, but of course you still have some sodium in the acid produced.

In the case of only acid in both cells (concentration mode, as suggested by chloric above), relative efficiencies of about 98% ( ~3.5 g/Ah) would be expected (no Na+). This method can also be used to get rid of residual sodium efficiently. However, the overall process efficiency drops to about the 1.8 g/Ah for both processes in sequence because the electrolytic process is used twice.

There is no bar to producing concentrated acid, at least up to 45%, I have found in my latest experiments.

One other experiment is worth mentioning, even though it proved unproductive.

Consider a 3 part coaxial cell. The anode cell contains the PbO2 anode; the center cell saturated NaHSO4; cathode could be Fe or carbon, Ni, etc. Anode cell would be primed with H2SO4, very dilute; cathode with NaOH, very dilute, the purpose being to increase initial conductivity. The starting NaHSO4 concentration is about 4.8+ M in the center cell. On electrolysis the following should occur (arrows show flow direction.:

Cathode cell from center cell: <--Na+ (mainly); <-- H+ (minor)
Cathode cell to center cell: --> OH-
Anode cell from center cell: --> HSO4- (mainly); --> SO4-- (minor)
Anode cell to center cell: <-- H+
In center cell, H+ + OH- --> H2O (controlled by pH level)
At cathode: 2H+ + 2e- <-- H2(g)
At anode: 2HSO4- --> 2HSO4* + 2e-; HSO4* + H2O --> H+ + SO4- + ½ O2(g)
* = transient radical.
At the center to cathode barrier: OH- + H+ --> H2O

In words, the NaHSO4 in the center cell is depleted equally of Na+ and HSO4- ions while at the same time water is generated in the center. This produces, ideally, a solution of H2SO4 at the anode and one of NaOH at the cathode, pure, simultaneously.

The expected maximum CE for this process is ~1 mol/F of HSO4- or Na+ transferred, i.e. the ~3.2 g/Ah level of H2SO4. This rate will not decrease much until the pH goes above about 2, when SO4- - and H+ ions begin to predominate in the center cell. This can be postponed by adding crystalline NaHSO4 to the center cell, which is slowly diluted by water @ ~1g/Ah. This CE would be 7 times greater than the average 0.4g/Ah got in the previous post.
I tried this. It works but the overall resistance of the cell was surprisingly high, 0.5 amp for 16V impressed, and that for over 400 cm^2 of center pot surface. This was traced to a large drop across the center/cathode barrier. My guess is that the water produced by the last reaction above is creating a layer in the 5mm thick terracotta pot that is essentially pure and non conductive. The current efficiency was about as expected but the power efficiency abysmal. At the current obtainable it takes forever to get any acid and alkali. So another good idea bit the dust! If my theory of water production is correct, a much thinner diaphragm should help to reduce this dreadful power inefficiency.

A Few Experimental Observations

The NaHSO4/H2SO4 method, outlined previously, was repeated. Essentially the same results were obtained as above, except the electrolysis was stopped once the cathode cell became alkaline. The anode contents were then used to start off a new electrolysis, the NaHSO4 cathode level being renewed. 43.5% acid was obtained in this way at fair efficiency, around 2g/Ah.

The two cell NaOH/NaHSO4/H2SO4 version was run for some time at the rather low current level of ~0.5 amp. Titration of the NaOH against roughly standardized HCl showed that the resultant cathode solution was about 0.8N, the anode content being about 2M H2SO4. The efficiency was > 1.5g/Ah, but difficult to estimate due to the changing nature of the current under severe temperature fluctuations. (Run outside on one of our rare cold nights.)

The NaHSO4/NaHSO4 2-cell method produced acid at relatively high efficiency, difficult to estimate because of remaining Na+ content. One interesting point was that persulphate and/or H2O2 was also produced. The acid bleached the indicator I used (phenol red). The resultant anode liquor produced bubbles on standing, suspected of being oxygen but not proven. This acid could, but has not yet been, purified of sodium and further concentrated by electrolysis in a H2SO4/H2SO4 cell.

I now have over a kg of Na2SO4. It crystallizes in huge crystals as Na2SO4.10H2O if done slowly – some I got were about 10 cm by 1.5cm, prismatic. The solution supersaturates with great ease. If rapidly cooled, a single crystal added will cause the entire precipitation in a few seconds. You can see the crystals growing as long needles. It is not possible to keep these crystals unless the relative humidity is > ~ 50% - they effloresce. And at 32C they melt, or dissolve in their water of crystallization. Since NaHSO4 is far more soluble, the two salts can be separated this way and the bisulphate returned for electrolysis.

NaHSO4 also forms large crystals as monohydrate, but these are quite different in shape, being compact and not needles. I was not able to get a perfect crystal, but they are monoclinic or rhomboidal, not far from cubic, as far as I could judge.

@chief
WRT Sulphites, etc. :- I think sulphites or metabisulpites (like Na2S2O5) would work in a divided cell, but poorly. H2SO3 is a weak acid (Ka1 = 1.89 , Ka2 = 7.21). In the strong acid environment of the anode cell, the SO3- - ions entering would tend to become un-ionized H2SO3 and their concentration would be low compared with the HSO4-. SO2 might be evolved. Anodic oxidation only takes place on the ions, so efficiency would be poor. In an undivided cell sulphate would result. However, SO2 dissolved in an undivided cell, especially if bubbled over the anode, should work. Thiosulphates, and the various thionates and thionites also would do something, but possibly also produce elemental sulphur. NaHSO4, being an industrial byproduct, is very cheap; I’m sure it must be cheaper than Na2S2O5.
Regards, Der Alte