# What do electrons usually do when lead acid battery is idle?

Lead acid batteries operation are governed by Redox reactions for charging and discharging. But, what do electrons usually do when lead acid battery is idle? And why?

Assumptions:

• Say more specifically when the lead acid battery is 100% charged and just before we begin to have an intentional discharge on it.
• There is no intention to look for a Quantum Mechanics answer
• Self-discharge is yet to happen, too

Update: I am not referring to the instant of time between both half reactions. I am only referring to the instant of time just before the whole redox reaction.

Efforts to answer myself but failed:

Half reaction of the redox describes what happens during discharge at the anode where the circuit is about to make most of the electrons necessary for the upcoming load but it does not show the instant of the state of electrons just before discharge:

$$\ce{Pb + SO4^2- -> PbSO4}$$

So, what may the electrons be doing at this time?

Do they just stay in their actual elements till a wire is connected so they can move out?

Or do they just hang out nearby the plates?

Note that at molecular/atomic level, there is no such a thing like being idle. If a particular system ( like the particular lead acid half-cell ) is in equilibrium, so macroscopically "nothing happens", there is ongoing dynamic equilibrium.

The rate of the oxidation reaction, providing electrons to the electrode wire, increases exponentially with raising of the electrode pontential.

The rate of the opposite, reduction reaction, absorbing electrons from the electrode wire, decreases exponentially with raising of the electrode potential.

The electrode maintains such a potential, where these rates of the opposite reactions match.

It is the same principle as for gas pressure equilibrium, temperature equilibrium, radiation equilibrium, concentration equilibrium. the both respective opposite processes are ongoing by the same same.

The NET redox reaction occurs, if the electrode has other potential than the potential of the redox equilibrium. That happens as a very temporary state just after the electrode is put into the solution, or if an external potential is forced on the electrode, e.g. by the wire. But even if the net redox reaction has zero rate, the opossite half reactions of the equilibrium are still ongoing.

Majority of the motion is due ions leaving and returning to the solid matrice of $$\ce{PB, PbO2 , PbSO4}$$ to keep the charge (dis)balance to maintain the respective electrode potential.

Both forward and reverse reactions

\begin{align} \ce{Pb + SO4^2- &-> PbSO4 v + 2 e^-\\ PbSO4 v + 2 e^- &-> Pb + SO4^2-} \end{align}

are ongoing at the same speed, effectivelly canceling each other.

The same for the other electrode:

\begin{align} \ce{PbO2 v + SO4^2- + 4 H+ + 2 e- &-> PbSO4 v + 2 H2O \\ PbSO4 v + 2 H2O &-> PbO2 v + SO4^2- + 4 H+ + 2 e- } \end{align}

• I was also referring to the instant just before the other half reaction at the cathode happens. I am not talking about the instant between the two half reactions since they happen almost simultaneously. I will update the question to indicate this clearly. But, I see your answer and will comment on in a separate comment anyways. – Gold_Sky Apr 9 at 10:59
• But they do happen simultaneously all the time. Just with different rate at different circumstances. – Poutnik Apr 9 at 11:00
• In regards your answer, you said the ions are leaving & returning the solids. This is essentially the reaction but what about just before it? if you meant they are just bouncing back and forth without doing meaningful work, is this a random motion to keep equilibrium? Or do the ions align themselves nearby the electrodes first? I am still not getting what is the "macroscopic state" of situation just before the whole redox reaction. – Gold_Sky Apr 9 at 11:08
• I thought the redox reaction will not happen till there is a wire from the anode to the cathode. What do you mean by ions leaving and entering the electrode at the same time? I just do not get this. You mean on the surface of the lead (say at the anode for example), ions of $\ce{SO4}$ enters the electrode? then what? if they leave it? for what? I just do not get it... – Gold_Sky Apr 9 at 11:33
• If you think I need to read about something before understanding your answer, please let me know because I am not familiar to equilibrium notions and a like... – Gold_Sky Apr 9 at 11:36