Equation 8 is certainly not the whole story. I haven’t done the experiment
myself, but according to “common knowledge� lead sulfate is formed on the
“+� plate, more in agreement with equation 3 than equation 8.

It is a tangentially interesting question whether any of the hypothetical
reactions listed above play any role -- even as intermediates -- in the real
reaction.

That is only tangentially interesting, because whether or not those reactions
occur, we are still left with a major mystery: how and why does anything
containing the SO4 group attack the “+� electrode? Neutral H2SO4 could
reach the plate by simple diffusion, but it is present in fantastically low
concentration. The bisulfate ion is present in high concentration, but would
have to swim uphill against the electric field. The sulfate ion is present in
low concentration and would have to swim doubly hard uphill. That means that
when the cell is under heavy load, i.e. when there is a large field across the
electrolyte, the SO4-related reaction would come to a halt.

Let’s put in some numbers: The cell has an open-circuit voltage of 2.2 volts
are so. Suppose that we put it under heavy load, so that there is ? v = 0.4
volts “IR� drop across the electrolyte. As always, room temperature
corresponds to 25 meV (.025 electron-volts). Putting it all together: the
Boltzmann factor that tells you what fraction of the bisulfate ions manage to
climb the potential is exp(q ? v / kT) = exp(.4 / .025) = 9,000,000. So we
would expect the reaction to proceed millions of times slower when we need it
to proceed faster.

In the foregoing calculation, we ignored the effect of dielectric screening.
This may or may not have been the right thing. Argument pro: energy is
conserved. At the end of the day, to move a bisulfate ion up a hill 0.4 volts
high, you have to do 0.4 eV of work. Argument con: most of the height of the
hill is associated with the dipole layer at the edge of the water, at the place
where the electrolyte meets the plate; within the bulk of the electrolyte the
field is smaller. The ions can with relatively resonable probability get close
to the “+� plate, just outside the dipole layer. Unanswered question: if
they get that close, is that close enough?

**********************

Note the basis of calculation, opening sentence, paragraph four: 2.2 volts per
cell.


From

http://www.av8n.com/physics/lead-acid.htm