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