Post by tonyqj on Oct 24, 2016 22:01:35 GMT
Here's a fun post for everyone who thinks they know how a lead-acid battery works. You don't. Nobody does. Now that you think I'm full of shit I'll prove it.
During discharge............
The positive plate (which is actually a cathode for the purposes of the internal reaction) reduces from lead dioxide to lead sulphate. In doing so it gains two electrons from the external circuit (thus reduction) and the voltage of the reaction is 1.685 volts. This is an atomic/chemical reaction. There are no iffs or buts about it. That voltage is not negotiable, this is what it is.
The negative plate oxidises from lead to lead sulphate. In doing so it releases two electrons (thus oxidation) to the external circuit. The voltage of this reaction is 0.356 volts.
Add the two voltages (for each reaction) together and we get 2.041 volts. That is the natural voltage of that cell.
Now there either are available ions at the interface between the acid and the plates or there aren't available ions. There is no in between.
So either the plates react (thus producing those voltages) or they don't. There is no in between.
So how can the voltage ever be higher than this (which it often is)? 12.7 volts is about fully charged and rested. Where does the extra 0.65 volts come from?
Why the hell does the voltage fall as the battery discharges?
Nobody knows! No-one!
There are lots of esoteric theories about the phenomenon but there is absolutely no-one who can explain it.
I must now point out that I'm not this clever, the above was pointed out to me by Gibbo several years ago and I thought it would be fun to share it. There's actually even a bit more to add about the chemistry if anyone's interested.
Cheers,
Tony
During discharge............
The positive plate (which is actually a cathode for the purposes of the internal reaction) reduces from lead dioxide to lead sulphate. In doing so it gains two electrons from the external circuit (thus reduction) and the voltage of the reaction is 1.685 volts. This is an atomic/chemical reaction. There are no iffs or buts about it. That voltage is not negotiable, this is what it is.
The negative plate oxidises from lead to lead sulphate. In doing so it releases two electrons (thus oxidation) to the external circuit. The voltage of this reaction is 0.356 volts.
Add the two voltages (for each reaction) together and we get 2.041 volts. That is the natural voltage of that cell.
Now there either are available ions at the interface between the acid and the plates or there aren't available ions. There is no in between.
So either the plates react (thus producing those voltages) or they don't. There is no in between.
So how can the voltage ever be higher than this (which it often is)? 12.7 volts is about fully charged and rested. Where does the extra 0.65 volts come from?
Why the hell does the voltage fall as the battery discharges?
Nobody knows! No-one!
There are lots of esoteric theories about the phenomenon but there is absolutely no-one who can explain it.
I must now point out that I'm not this clever, the above was pointed out to me by Gibbo several years ago and I thought it would be fun to share it. There's actually even a bit more to add about the chemistry if anyone's interested.
Cheers,
Tony
