Copyright © 2004 jsd
The factory makes each cell in the battery as follows: Connected to the “-” terminal is a thick, porous plate of metallic lead. Connected to the “+” terminal is a plate consisting mostly of porous lead dioxide paste, supported on a thin metal grid. In between the plates is fairly concentrated sulfuric acid (about 4M).
By way of preliminaries, we note that sulfuric acid is a strong acid with respect to its first proton, so even before the acid has been added to the battery, the following reaction has gone to completion:
This reaction is associated with the bulk electrolyte, independent of the plates of the battery. It is not an electrochemical reaction and is independent of the charging and discharging of the battery. We do not expect sulfuric acid to release its its second proton with any great likelihood – only a fraction of a percent – since Ka2 is only 0.012.
As the battery discharges, the following half-cell reaction takes place at the “-” plate:
This reaction makes a lot of sense. As discussed in reference 1, when the cell is under load, there is an electric field in the electrolyte that causes negative ions (in this case bisulfate) to drift toward the “-” plate. The negative ion is consumed by reacting with the plate. The reaction also produces a positive ion (proton) which drifts away under the influence of the aformentioned field. Two electrons are left behind in the plate, to be delivered to the terminal. There’s nothing surprising about any of that.
As written, equation 2 represents the discharge operation. To represent the charge (aka recharge) operation, just reverse the direction of the arrow. The result makes sense, too.
Meanwhile, the other half of the discharge reaction takes place at the “+” plate. This can be written in various ways, of which the following is conventional:
Alas, equation 3 is not very easy to understand, as will be discussed in section 2. But first, let’s discuss a few things that do make sense:
At each plate, the lead compounds are insoluble and stay attached to the plate; there is never any significant amount of lead in solution. The other reaction products (water and sulfuric acid) are completely soluble. These properties are an important part of why the cell is rechargeable. (In contrast, in non-rechargeable batteries, there are typically reaction products that become unavailable for re-use, departing as a gas or as an insoluble precipitate.)
If we add the two half-cell reactions together, we get the full-cell discharge reaction:
During the discharge operation, acid is consumed and water is produced. During the charge operation, water is consumed and acid is produced. Because sulfuric acid is much denser than water, a widely-used technique for checking the state-of-charge of a battery is to measure the specific gravity of the electrolyte. (Non-experts sometimes guess that the change in density is due to the presence of heavy lead compounds in solution, as if some sort of lead-plating reaction were involved, but this is not the case.)
During discharge, at the “-” plate, the lead is oxidized from metallic Pb to divalent Pb(II), while at the “+” plate, the lead is reduced from tetravalent Pb(IV) to divalent Pb(II). Of course during recharge, the opposite redox reactions occur.
Beware of some tricky terminology:
From this we can conclude that for a rechargeable battery, you should never mark the words “cathode” or “anode” on the terminals. See reference 2 for more about this, including a proper definition of “anode” and “cathode”.
It is not easy to understand what sort of microscopic processes are associated with equation 3. It is distressingly hard to look up information about this. A standard book on lead-acid batteries is reference 3, but it doesn’t delve into this issue. Some tantalizing web pages are reference 4 and reference 5. If anybody knows any good references on this subject, please let me know.
In particular, in equation 3 it’s not at all clear how or why the bisulfate ion moves uphill (against the gradient of the electrochemical potential) to approach the “+” plate.
There are two ways of looking at this:
One way to look at this starts by observing that the electrochemical potential is a potential i.e. the energy of the bisulfate ion is a function of position, independent of how the ion got to that position.
The height of this potential can be considered a “barrier” to the reaction.
In equilibrium, i.e. when the cell is open-circuited, the electrochemical potential is everywhere an equipotential, i.e. the height is independent of position. The mystery arises when I place the cell under load. This manifestly shifts the potential. The terminal marked "+" will become less positive, i.e. more negative, becoming unattractive to negatively-charged ions. We can calculate the density of ions using the Boltzmann factor, ρ ∝ exp(−E / kT).
This can be called the "static" or "integral" view of the situation.
In equilibrium, i.e. when the cell is open-circuited, diffusion is the only game in town. Then, when I place the cell under load, there will be a contest between diffusion and drift. Non-experts may be tempted to imagine a situation where diffusion continues to dominate, but this is not possible, because of the Einstein relation:
|D = kT b (5)|
where D is the diffusion constant for a particular species and b is the mobility for that species.
When I place the cell under load, the electric field is sufficiently large that it causes negatively-charged ions to drift away from the “+” electrode. Diffusion in the opposite direction will be of only minor significance, leading to depletion of bisulfate ions at the "+" plate, which in turn makes equation 3 appear inconsistent with the commonly-observed properties of the cell.
The Einstein relation guarantees that the “integral viewpoint” and the “differential viewpoint” make exactly the same predictions.
We now try to shed some additional light on the mystery, by contrasting equation 3 with various hypothetical reactions. Among other things, we need to see whether we can wriggle out of the difficulty by finding an acceptable alternative to equation 3 (but alas we will not really find one).
It would be much easier to understand the microscopics of such reactions. For equation 8 in particular, we can re-use the logic that we used for equation 2: A positive ion (proton) drifts in under the influence of the electric field in the electrolyte, and is consumed in a reaction. The reaction produces a negative ion (hydroxyl) which drifts away under the influence of the aforementioned field. Two electrons are in effect transferred from the plate into the electrolyte, so that two units of positive charge are delivered to the terminal. The reverse of equation 8 probably looks more like:
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 Δ 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 Δ 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?
In any case, there is some remaining mystery even if you completely buy into the screening argument. By the usual Clausius-Mossotti argument (reference 6), the field inside a spherical hole in a dielectric is 1/3rd of the way between the fully-screened value and the unscreened value. So if we put in the numbers, we get the cube root of the previous number, i.e. a factor of 200. That’s a lot less than 9 million, but it’s still a big enough factor to predict a disastrous effect on the battery under load.
In addition to that major mystery (concerning the bisulfate ion), we have a related minor mystery concerning the protons: Is the reaction really third order in proton concentration, as suggested by equation 3? If the reaction proceeds via some number of intermediate steps, perhaps along the lines of the hypothetical reactions given in this section, there would be a lower-order dependence on H+ activity.
Copyright © 2004 jsd