Capt. NealŪ wrote:
"Ronald Raygun" wrote
Therefore charge does travel through the battery. We have a complete
circuit, with external flow of electrons and internal flow of H+ ions.
There's where I disagree. Charge does not 'travel through' the inside
of a battery.
The battery is empty when all the sulphuric acid is used up and has been
turned into lead sulphate on both plates. Let's look at the battery just
before it's completely empty, when the last ten dissociated H2SO4 molecules
are still swimming around, and let's consider what happens from then until
they're all gone.
Your diagram shows the dissociation of H2SO4 as partial into H+ and HSO4-,
but in some ways it's easier to think in terms of full dissociation into
2H+ and SO4--. It makes no difference to the bottom line in terms of
charge accounting, it's just a little more confusing to work with the
partially dissociated model.
Think of these 10 molecules as spread out uniformly through the remaining
electrolyte, and let's have the battery oriented with its - terminal (Pb
plate) in the West and the + terminal (PbO2 plate) in the East, and let's
imagine the electrolyte divided into five vertical slices perpendicular
to the East-West line, and let there be two molecules (i.e. two SO4-- ions
and four H+ ions) in each slice. Call the slices A, B, C, D, E, with A
adjacent to the West plate and E to the East plate.
On average, one of the SO4-- from each slice will end up on each of the
plates, and all four H+ will take part only in the Eastern reaction.
The plate reactions a
West: Pb + SO4-- + 2H+ -- PbSO4 + 2H+ + 2e-
East: PbO2 + SO4-- + 4H+ + 2e- -- PbSO4 + 2H2O
Pragmatically, though, both SO4-- ions from both A and B will go West,
while both from both D and E will go East. From C, one will go West and
one East. But all H+ from all cells will go East.
We must have a neutral charge change within each slice, but as you'll see,
there will be charge flowing across all slice boundaries.
At the boundary between the West plate and slice A, A's own two SO4-- ions
will travel West, as will the three which came passing through from slices
B and C. Net flow across this boundary: -10 West.
At the AB boundary, we have A's 4 H+ going East, and B's and C's three
SO4-- coming West. So as far as slice A goes, we have -10 going out on
the Western front, -6 coming in from the East (subtotal -4 out), and A's
four H+ going out East. Final charge total is neutral *for the slice*,
but of course at the *boudaries* we have a net flow -10W in the W and
-6W and +4E in the East, which is equivalent to -10W.
Slice B: Its two SO4-- go West, one SO4-- passes through EW from C, all
its 4 H+ go East and A's 4 H+ pass through WE. Since all the "passers
through" cancel out, the net change to B is -4W and +4E which is neutral,
but the BC boundary has one SO4-- going West (-2W) and eight H+ going
east (+8E) which again is equivalent to -10W.
Slice C: One of its SO4-- goes West, one East. Net -4 out so far. All
its 4H+ go East. Net neutral. 8H+ passing through West to East.
CD boundary: one SO4-- East (-2E) 12H+ East (+12E). Net +10E which
is equivalent to -10W.
I'll leave what happens in slices D and E and at the D/E and E/Eplate
boundaries as an exercise.
The crux is that that no matter where you "slice" the electrolyte,
there will be charge flowing across that slice boundary, equal to -10
going West inside the battery for every -10 going East outside the
battery.
There is no conductor
The electrolyte is sulphuric acid. I think you'll find that's a conductor.
It's a water based solutions with ions, i.e. charge carriers, swimming
around in it. If more charge carriers swim one way than the other,
then you have a net flow of charge, i.e. a current, and that makes the
medium a conductor.
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