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My guess:
In a lead-acid battery durring "forming" of the plates the anodic reaction is:
PbO + H2O => PbO2 + 2(H+) + 2(e-)
It takes 2 moles of electons to oxidise on mole of litharge to one mole of lead dioxide. This suggests it takes at least *4* moles of electrons to oxidise solid lead to lead dioxide. I am unsure how the lead gets on the anode in the first place, that would be an "anodic reduction" reaction, can that even happen? If lead nitrate was present unionised something like this might happen requiring only two moles of electrons:
Pb(NO3)2 + 2(H2O) => PbO2 + 2(NO3-) + 4(H+) + 2(e-)
No idea how practical that is, seems to me we'd be dealing with Pb++ ions. Obviously the lead ends up being Pb++++ which is oxidation, so there is at least two moles of electrons envolved, if not more. Personally my guess is four or six Faradays for one mole of lead dioxide which says you got very close to 100% or something that deserves a nobel prize and would be the salvation of mankind.
I am probably wrong though, electrochemistry is not something I am good at.
I have just found this equation for the anode reaction?.. i still cant work out how many moles of electrons are used in this equation .. my gues is 2 moles lead? 2 of water = 4 moles?.. which is more or less the one you have explained above..
Pb2+(Ely) + 2 H2O(Ely) -> PbO2(PbO2) + 4 H+(Ely) + 2 e-