Two questions about lead-acid chemistry

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Ed Ferris - 30 Jun 2009 03:15 GMT

Here's a question I've never seen mentioned, much less answered, about the
lead-acid cell: Why don't we get lead hydrogen sulfate instead of lead
sulfate crystals after a partial discharge?

Has anybody tried phosphoric acid instead of sulfuric as the electrolyte?
Lead phosphate is even less soluble than lead sulfate and might prevent
sulfation (or "phosphation") on standing discharged. Would we get the
dihydrogen phosphate first, and is it soluble?

I still think a dissolved oxidant (such as CrO3) and a porous membrane
makes more sense. True, it leaks into the anode half-cell eventually, but
in the end we are all dead, including our lead-acid cells.

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Poutnik - 02 Jul 2009 21:28 GMT

Ed Ferris's previous post was like this :

> Here's a question I've never seen mentioned, much less answered, about the
> lead-acid cell: Why don't we get lead hydrogen sulfate instead of lead
> sulfate crystals after a partial discharge?

Supposing Pb2+ + 2 HSO4- -> PbSO4(prec) + 2 H+
is releasing Gibbs energy,bpushing it to right side.

> Has anybody tried phosphoric acid instead of sulfuric as the electrolyte?
> Lead phosphate is even less soluble than lead sulfate and might prevent
> sulfation (or "phosphation") on standing discharged. Would we get the
> dihydrogen phosphate first, and is it soluble?

Less solubility means also less reactivity, questionable if reversible,
also phosphoric acid has less conductivity and different behavior.
It would be totally different system.

> I still think a dissolved oxidant (such as CrO3) and a porous membrane
> makes more sense. True, it leaks into the anode half-cell eventually, but
> in the end we are all dead, including our lead-acid cells.
>

Try it and tell us :-)

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Poutnik
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Poutnik - 02 Jul 2009 21:31 GMT

Poutnik's previous post was like this :

> Supposing Pb2+ + 2 HSO4- -> PbSO4(prec) + 2 H+
> is releasing Gibbs energy,bpushing it to right side.

Pb2+ + HSO4- -> PbSO4(prec) + H+ , for sure.

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Poutnik
The best depends on how the best is defined.

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Ed Ferris - 03 Jul 2009 01:53 GMT

> Poutnik's previous post was like this :
>>
>> Supposing Pb2+ + 2 HSO4- -> PbSO4(prec) + 2 H+
>> is releasing Gibbs energy,bpushing it to right side.
>> >
> Pb2+ + HSO4- -> PbSO4(prec) + H+ , for sure.

Rather,

Pb(HSO4)2 = PbSO4 + H2SO4

The only way this reaction goes to the right is if the sulfate precipitates
(from battery acid) and the hydrogen sulfate is soluble. Do we know this
is true? The CRC Handbook says both are slightly soluble in H2SO4.

Substitute any other cation for Pb++ and the reaction goes to the left.

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Poutnik - 03 Jul 2009 05:54 GMT

Ed Ferris's previous post was like this :

> Rather,
>
[quoted text clipped - 5 lines]
>
> Substitute any other cation for Pb++ and the reaction goes to the left.

Your reaction in fact never happens.
In aquatic solution of acids and salts,
the reactions occurs with ions.

In solution these salt compounds do not exist, there exist ions.
( Well, in some degree there exist non charged complexes,
but they are dissociated to high degree)
Once there is HSO4-, it does not care,
if it belongs to Pb2+ or to H+.

Reactions that runs all the time in both directions are
H2SO4 <-> H+ + HSO4-
HSO4- <-> H+ + SO4(2-)
and possibly
M(2+) + SO4(2-) <-> MSO4(prec)
M(2+) + 2 HSO4- <-> M(HSO4)2 (Prec)

To form solid salt, the multiplication of concentration of involved
ions need to cross a given value.
In acid lead accumulator precipitation of PbSO4
keeps Pb2+ concentration at so low level
( about 10^-10 mol/l or so),
that formation of precipitated Pb(HSO4)2 cannot occur.

In general all depends on solubility ratio of both salts and on pH.
Well, more precisely also on level of ion complex creation.
If one of salts is much less soluble than the other,
you cannot ( at least easily ) get the other.

Both KHSO4 and K2SO4 are comparably soluble,
so what you get depends on pH.
But you do not get from solutions.
Pb(HSO4)2, Cu(HSO4)2, Zb(HSO4)2, Fe(HSO4)2......
because MSO4. nH2O precipitates sooner.
You only get concurrent presence of ions.

The fact is, that PbSO4 similar as BaSO4
is partly soluble in concentrated H2SO4,
where is minimal concentration of SO4(2-)
to keep them precipitated.
As far as there is mainly HSO4- you can say
you have solution of Pb(HSO4)2.
But you will not get it.

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Poutnik
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Ed Ferris - 03 Jul 2009 16:44 GMT

> As far as there is mainly HSO4- you can say
> you have solution of Pb(HSO4)2.
> But you will not get it.

So if you take an aqueous solution of one mole of FeSO4 and add one mole of
H2SO4 to it, you have one mole of Fe++ and two moles of HSO4- in solution,
but if you evaporate the solution you get FeSO4 and H2SO4 again, not
crystalline Fe(HSO4)2. Similarly, the lead-acid cathode reaction could
form Pb(HSO4)2 as an intermediate but would precipitate PbSO4 and re-form
H2SO4 from it. This would explain why you don't get a stepwise reduction
in potential as the acid is consumed and the electrolyte density approaches
1. This would also imply that you don't need to worry about the solubility
of Pb(H2PO3)2 when trying phosphoric acid as the electrolyte, since Pb2
(PO3)2 will precipitate anyway.

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Ed Ferris - 03 Jul 2009 16:58 GMT

> This would also imply that you
> don't need to worry about the solubility of Pb(H2PO3)2 when trying
> phosphoric acid as the electrolyte, since Pb2 (PO3)2 will precipitate
> anyway.

Make that PO4 instead of PO3. Three minus charges, four oxygens.

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Poutnik - 03 Jul 2009 19:58 GMT

Ed Ferris's previous post was like this :

>So if you take an aqueous solution of one mole of FeSO4
>and add one mole of H2SO4 to it,
>you have one mole of Fe++
>and two moles of HSO4- in solution,

Better said summary amount
of H2SO4, HSO4- AND SO4(-2) is 2 mols.

>Similarly, the lead-acid cathode reaction
>could form Pb(HSO4)2 as an intermediate
>but would precipitate PbSO4 and re-form H2SO4 from it.

In fact Pb2+ precipitates with existing SO4(2-),
that is corrupting acidobasic equilibrium,
and HSO4- dissociates to H+ and SO4(2-)
to keep c(H) . c(SO4) / c(HSO4) = K2
( dissociation constant of HSO4- )
Well, activity to be more precise.

Unless you have Pb(HSO4)2 in hand,
you can forget considering it as compound or salt,
but think about it as ion coexistence in solution.

If you dissolve 2 mols of NaCl and 1 mol of K2SO4,
or 2 mols of KCl and 1 mol of Na2SO4,
these salts does not exist in solution,
neither react each other.
Each kation belongs
to each anion in the same degree.
There are just K+, Na+, Cl- and SO4(2-) ions.

>This would explain why you don't get
>a stepwise reduction in potential
>as the acid is consumed
>and the electrolyte density approaches 1.

There is always excessive amount of H2SO4,
There is roughly 20-40% H2SO4, depending on
charge degree.

>This would also imply that you don't need
>to worry about the solubility of Pb(H2PO4)2
>when trying phosphoric acid as the electrolyte,
>since Pb3(PO4)2 will precipitate anyway.

Sure, but anyway, nothing from sulphuric system
can be applied to phosphoric system
without proper analysis. It is totally different system,
not just one acid replaced by another.

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Natural Science Forum / Chemistry / Electrochemistry / July 2009

Two questions about lead-acid chemistry