What reaction would take place on an electrode in a lead-acid secondary cell if it were charged and the other electrode removed?

Question

I'm reading a book that covers the basics of electrical theory and I'm at a chapter that discusses the concept of a secondary cell. The author includes the redox reactions that take place in a primary cell when it is charging/discharging.

The secondary cell is comprised of an anode (composed of PbO₂) and a cathode (composed of Pb) submerged in a solution of H₂SO₄. As it discharges both electrodes are plated with PbSO₄ until they are fully covered, at which point the reactions that produce a charge for the battery cease and the original H₂SO₄ is now diluted with H₂O.

My question focuses on the charging aspect of it not the discharging (because I have an easier time framing my question from this perspective) but I want to preface this aspect with a quote from earlier in the chapter discussing a primary cell:

"The zinc element or electrode, when immersed in the acid, starts to dissolve, leaving its electrons behind and the zinc goes into solution as Zn⁺⁺ ions. This action stops very shortly, as the zinc plate becomes negative and ceases to throw off positive ions. The amount of zinc thus dissolved is very minute. If the negative zinc plate is connected externally by a wire to the copper plate, an electrical current will flow from the negative zinc plate through the external wire to the copper plate (positive). As the negative charge of the zinc plate is thus removed, more Zn⁺⁺ ions go into the solution"

As the original secondary cell I mentioned recharges the PbSO₄ plating reverts to the composition of it's respective electrode and the diluted H₂SO₄ reverts to it's original solution. The reactions that take place when the cell is recharging as quoted from the book:

Positive Plate: PbSO₄ + 2H₂O --> PbO₂ + HSO₄^- + 3H⁺ + 2e

Negative Plate: PbSO₄ + H⁺ + 2e --> Pb + HSO₄^-

I presume that each plate is a half reaction; the negative plate is a reduction half reaction; the positive plate is an oxidation half reaction; I believe these half reactions can be combined into the full reaction: 2PbSO₄ + 2H₂O --> PbO₂ + Pb + 2H₂SO₄. Please correct me if I'm wrong about what these half reactions are or about how I combined them; I have only just recently learned about redox reactions in order to better understand this chapter.

Now for the question this has all been building up to; if the positive plate were removed during the recharge process would the half reaction that takes place on the negative plate become the half reaction: PbSO₄ + 2H₂O --> Pb + HSO₄^- + 2O⁼ + 3H⁺ for a short period of time until the solution becomes negative with O⁼ anions and ceases to throw off H⁺ cations (similar to what happens in the quote)?

If this is the case, then I don't know how the difference of charge between the reaction and product side is possible. The only thing I can imagine is that the O⁼ is just a spectator and does not participate in the actual reaction.

Answer 1

Dilute sulfuric acid forms two ions: \$H^+\$ and \$SO_4^{2-}\$.

So half reactions for Charging:

Positive Plate: \$PbSO_4 + 2H_2O \to PbO_2 + SO_4^{2-} + 4H^+ + 2e^-\$

Lead sulfate becomes Lead peroxide.

Negative Plate: \$PbSO_4 + 2e^- \to Pb + SO_4^{2-} \$

Lead sulfate becomes Lead.

Your full reaction is correct but your half reactions were wrong (ions were missing).

Source: Operation of Lead Acid Batteries

Negative plate has an excess of electrons and positive plate has an excess of holes and it is the two poles with the electrolyte that allows for the chemical reaction to build up of potential from the positive to negative poles.

If you remove the positive poles, the chemical reaction stops. Period, and literally the end of the story.

The rest is nothing good and the purely speculation.

Electrons on negative pole will recombine with \$H^+\$ ions to form hydrogen gas (I'd say a lot if battery was fully charged) and a lot of heat. Odds are Lead melts.

Holes on the positive poles attract free electrons (again a lot and more heat) as soon as the pole comes into contact with a conductive surface .

Odds are positive pole creates a spark and hydrogen gas creates an explosion with molten lead and acid burns as water vaporizes.