[bikesandcars]

I've searched the internet and this forum but unfortunately can't find the chemistry behind what I'm seeing in my "lab" (garage).

I started with a sulfated lead acid battery.  It measured 12 volts after charging but would not produce current.  It is a 1 year old battery with no other issues.  I let it discharge and stay discharged and lead sulfate (pbS04) turned chrystaline on the plates

I added Magnesium sulfate and distilled water (epsom salts) as outlined in several online posts.  The solution I made was almost fully saturated (heated h20 to boiling, added MgSO4 until saturation).

I removed all existing water (weak acid) from the battery.

I added the MgS04 solution (I assume since it dissolved it is MG 2+, H2O, H2SO4, is this a wrong assumption?)

I charged the battery (2 amps on a 6 AH battery)

The battery now works fine.

I assume you could model the battery like this
   -                                                                                    +
------------------------------------------------------------------------
| pb(s),  pbSo4(s) | H2O(aq), H2S04(aq), Mg2+, H+  | PbSO4, PbO|
________________________________________________________

My question [edited] 

How (chemically) does MgSO4 de-sulfate the battery?

How would I model this reaction?

What I know from experimentation is that charging seems to be required to make this work (the battery will not self-charge after adding MgSO4). 

A healthy battery (charged) has minimal if any PbSO4 in it, and no crystaline PbSO4 (sulfation).  Without PbSO4 there isnt a need for Mg.  I'm assuming once the battery plates are de-sulfated I should drain the acidic MgSO4 solution (possibly some MgSO4 precipitate), rinse the battery, and fill with pure electrolyte (H2SO4).

I could be completely wrong, but I'd really like to understand how this chemistry works so I can get the proper way to maintain my batteries.

[Arkcon]

OK.  This is a lot for our chemists here to work with.  I'm going to try to help as much as I can.  Mostly, I'm going to try to simplify the question, removing a block of, "and then I, and then I, and then ..."

I've searched the internet and this forum but unfortunately can't find the chemistry behind what I'm seeing in my "lab" (garage).

I started with a sulfated lead acid battery.  It measured 12 volts after charging but would not produce current.  It is a 1 year old battery with no other issues.  I let it discharge and stay discharged and lead sulfate (pbS04) turned chrystaline on the plates

OK.  So far, this is consistent with the definition of a discharged lead-acid storage battery.  So far, so good.

I added Magnesium sulfate and distilled water (epsom salts) as outlined in several online posts.  The solution I made was almost fully saturated (heated h20 to boiling, added MgSO4 until saturation).

Kay, see, I hadn't heard of using magnesium sulfate to "fix" lead-acid storage batteries.  So I'm a little out of my depth here.  So I'm gonna take it at face value.  If I had boiling dH2O, added more magnesium sulfate than would dissolve, and cooled it, I would call that completely saturated, or possibly supersaturated, which would soon be completely saturated with a mass of excess magnesium sulfate at the bottom.  So, you're using terminology I don't understand, that chemists don't use.  Just letting you know, if the boards you've been using so far say one thing, and we say another, you can't be upset if this thread goes nowhere.

I removed all existing water (weak acid) from the battery.

OK, that's clear.  Replace electrolyte (mostly water) with saturated magnesium sulfate solution.

I added the MgS04 solution (I assume since it dissolved it is MG 2+, H2O, H2SO4, is this a wrong assumption?)

Kinda, yeah.  I don't expect there is any H₂SO₄ in a saturated solution of magnesium sulfate.  You have water, and Mg²⁺ and SO₄^2- ions, yes.

I charged the battery (2 amps on a 6 AH battery)

The battery now works fine.

Who-hoo.  I was worried it wouldn't work.

I assume you could model the battery like this
   -                                                                                    +
------------------------------------------------------------------------
| pb(s),  pbSo4(s) | H2O(aq), H2S04(aq), Mg2+, H+  | PbSO4, PbO|
________________________________________________________

Yes, I think you can now model the battery like that.  After charging, the sulfuric acid electrolyte would reform in solution from the battery plates.  And the magnesium ions are still there.  The sulfate ions too, because this is a mix of sulfuric acid and water

My question:  by charging, are the H+ ions created during "electrolysis" converting the MGSO4 to H2SO4, thereby freeing the MG+ ion to grab the crystalized PbS04  and replace the Pb with MG creating MgSO4 which in turn reverts to H2SO4 in a loop.  Thus creating an acidic mixture high in MgSO4 concentration?

OR: 

Is it the Mg 2+ Ion replacing the Pb in PbSO4 just out of chemistry (without electric current)?

How would I model this reaction?

What I know from experimentation is that charging seems to be required to make this work (the battery will not self-charge after adding MgSO4). 

A healthy battery (charged) as minimal if any PbSO4 in it, and no crystaline PbSO4 (sulfation).  Without PbSO4 there isnt a need for Mg.  I'm assuming once the battery plates are de-sulfated I should drain the acidic MgSO4 solution (possibly some MgSO4 precipitate), rinse the battery, and fill with pure electrolyte (H2SO4).

I could be completely wrong, but I'd really like to understand how this chemistry works so I can get the property way to maintain my batteries.

OK, you're asking a lot here.  And some of it has forgotten some of the previous steps, and some is definitely wrong by chemical concepts.  But briefly, yes, now that the battery is fully charged.  The magnesium sulfate containing electrolyte should be dumped and replaced with fresh, plain old sulfuric acid electrolyte.  The magnesium sulfate may have helped you charge the plates, but it isn't a good electrolyte for a storage cell.  And I don't know why.  However, I base my suggestion on that if it was a good idea, it would always be there.

[bikesandcars]

Thanks   I'm obviously not a chemist (an engineer and a mechanical one at that).  Thanks for the corrections and sympathetic understanding of what I'm trying to do.

I support the theory that the Mg would be there if it was beneficial, one would think capitalism would support that anyway.  I'm not a big conspiracy guy, but you could also say that it's not there because its good for business.  At this point I don't have a good understanding as to why it shouldn't be there.  Maybe it used to be or still is for some batteries.

I'm still a little unsure of how to model this chemically as it's been a long time since I played with equilibrium equations.

Mg + SO4 <> MgSO4
2H + O <> H2O
Pb + SO4 <> PbSO4
PbO + SO4 <> O + PbSO4

I would love a little more education if someone feels inclined

[Sylphy]

Hi Bikesandcars,
I know that it is 2 years later since your last post, but over the last year I have been fascinated by, and experimenting with MgSO4 and other sulphate electrolytes in Lead-acid batteries for solar off-grid energy storage.
I wanted to ask if you have done any more experimenting since your last post?, or if you got further info on the actual chemistry?,
I've chatted to some good friends of mine who are chemists, and attempted to discover some possible equations for the process, and my current thoughts are this: (Note, I am not a chemist by any stretch of the imagination , but I do learn fast, and have been doing a lot of research, hopefully other people with greater understanding in this area can enlighten me)

So, the standard equation for Lead-acid chemistry (Discharge) (both half equations together), according to wikipedia, is:
Pb(s) + 2H2SO4 (aq) + PbO2(s)  <=> 2PbSO4(s) + 2H2O @ 2.0v

Half equations for Discharge are:
@ Neg plate (anode)
Pb(s) + HSO4- => PbSO4(s) + H+ + 2e-  @ 0.356v

@ Pos plate (cathode)
PbO2 (s) + 3H+ + HSO4- + 2e- => PbSO4 (s) + 2H2O  @ 1.685v

Since Pb compounds are mostly insoluble, there is never any significant amount of lead in the solution, so the electrolyte is mostly made up of H2O + H2SO4.


Now, if we look at adding MgSO4 to the battery, instead of H2SO4 (fully replacing the electrolyte, not just adding to it), I wonder if we get the following balanced equations:

Pb(s) + PbO2(s) + 2MgSO4(aq) + 2H2O => 2PbSO4(s) + 2Mg(OH)2

The Half equations being:

@ Neg Plate
Pb(s) + SO4^2- + 2e^- + Mg²⁺ + 2H⁺ => PbSO4(s) + Mg(OH)₂

@ Pos plate
PbO2(s) + SO4^2- + Mg²⁺ + OH^- => PbSO4(s) + 2e^- + Mg(OH)₂

An interesting thing here is that if these equations are correct, then the presence of MgSO4 instead of H2SO4, causes a Mg(OH)2 solution, which is an Alkaline, along with removing any H2SO4 or HSO4^- ion from the equations, would turn this battery into a lead-alkaline battery instead yeah?.
A fully charged battery of this type would have PbSO4(s) in precipitate, with Mg(OH)2 in solution, and upon discharge, would have MgSO4 in solution, and Pb back on the plates.
Would it, therefore, theoretically be more tolerant of deeper-discharges (say, instead of a 30% DOD, a 70% DOD), without causing epic damage to the plates / electrolyte?, since the plates aren't being subjected to the H2SO4. I assume though, that Alkaline solutions would still have an effect on the Pb (MgSO4 seems to have a pH of between 9 to 11, depending on concentration, which makes it a pretty effective alkaline).

I understand how tetravalent Pb4+ (at the Pos plate, from PbO2) becomes reduced to form divalent Pb²⁺ liberating + charge to the + plate, and that (at the Neg plate, from Pb(s) ), Pb is oxidised to form divalent Pb²⁺ liberating - charge to the - plate,
but I have wondered what processes the Mg is undergoing during the charge/discharge process, and if it would return so cleanly into MgSO4 under discharge, or if, instead, it would remain as Mg(OH)2, and the SO4 ion from the PbSO4 would react with the water, to form HSO4^- ions?

--------
Looking at the Metal Activity Series Table:
Pb²⁺ is way at the bottom (as in, it needs a strong oxidising acid to react with), and it is less reactive than Mg²⁺ (which reacts with a far greater range of acids)

So, my thoughts suggest that having the Pb + Mg together in the same electrolyte means that the Mg²⁺ (as an ion in solution) is going to have a greater displacing effect on the Pb²⁺, and that the Pb²⁺ will be more submissive in this arrangement.

Maybe this accounts for how the Mg²⁺ ion will remove the sulphation on the Pb plates, due to it's greater reactivity (its greater ability to lose e^- more readily to form + ions, corrode more easily and become a stronger reducing agent (electron donor)).

Perhaps it is, literally, displacing the Pb²⁺ of the PbSO4 at the surface area of where the Mg²⁺ in electrolyte meets the PbSO4 on plates, like the Pb moves out of the way, and lets the Mg take it's partner, which, being so easily water soluble, has no trouble dissolving into solution, thus removing the SO4 ion from the Pb plate.

I have wondered, if that is the case, why the Mg²⁺ as an ion in solution, would form Mg(OH)₂, rather than remain as MgSO4(aq)?
Also, I wonder if this means that the Mg would NOT displace onto the Pb plates themselves, unlike a less reactive metal such as Cu, which seems to be displaced onto the Pb plates, when the Pb comes into solution when using CuSO4.

Hmmm..

I have read online about other people having big success of lead-acid battery rejuvenation using Alum (K₂Al₂(SO4)₄)
Both K and Al are way higher on the reactivity table of metals than Pb, and so I wonder if it is having a similar effect to the Mg, but with an even greater effect due to K and Al being even greater in reactivity than that of Mg as well?

MgSO4 is highly water soluble compared to PbSO4, which tends to remain more as a precipitate.

Lastly, in my experiments, I have different batteries of differing molarity going on, to test the results.
Just for reference, in case it's helpful to anyone else, molar mass of MgSO4.7H20 (epsom salts) = 246.47g/mol

This means that:
0.3mol solution of MgSO4 = 75g per 1 Litre of H2O
1mol solution of MgSO4 = 246.47g per 1 Litre of H2O
5mol solution of MgSO4 = 1.2Kg per 1 Litre of H2O

In my experiment with the 0.3mol solution, I found that the battery (it was a 100A 12v deep cycle) did only present a small spark when shorted out across the terminals, and when shorted out through a shunt, it initially sparked very little, and had a small Amp reading across the shunt, approx. 3amp, however this began to rise, and over a course of 5 minutes, the shunt was reading approx. 50amps, and the 4Gauge wire connecting the terminals was feeling quite warm.
The battery itself felt warm, and the hottest point was in the cell that was the most sulphated of them all when I did the electrolyte conversion, from its previous lead-acid life.

I have wondered if the solubility of the PbSO4(s) back into solution, and then for the Pb to return to the plate is playing a role in throttling the "speed" of the reaction, and therefore the maximum short-circuit amp bandwidth, so to speak.

I have not yet run the same experiment on the 1mol and 5mol solution setups to see if the same thing occurs.
-------

Can anyone add to / enlighten / correct my equations and calculations for all of this?,
I too am interested in alternative battery electrolytes, and with all the armchair scientists on internet forums out there naysaying to the possibility of using MgSO4 as an additive, let alone a stand-alone electrolyte, it would be great to refine and come up with an understanding of what processes and equations are actually going on.

Anyone got any funky ideas?


- Sylph H

[Arkcon]

There was another thread on this topic here: http://www.chemicalforums.com/index.php?topic=68057.msg245422#msg245422