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Apparently Hydrogen/Oxygen are liberated when a Lead-acid battery is charged.

If true, how does one calculate the expected volume & rate at which each gas is liberated when a battery is charged?

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Hello Everyone, It goes a bit deeper into Chemistry for the exact calculation. But anyways, As I've already mentioned, It's a variable based on how much % the battery has charged... –  Waffle's Crazy Peanut Oct 11 '12 at 17:21
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Apparently Hydrogen/Oxygen are liberated when a Lead-acid battery is charged

NOT Always. Just explaining how? Lead-Acid Battery comes under Secondary cells. An LA battery usually has plates of lead & lead oxide (when fully charged) or lead sulfate (when fully discharged) in an electrolyte of 35% sulfuric acid and 65% water solution. Indeed, Over-charging could lead to evolution of hydrogen and oxygen due to electrolysis of water. Actually it's a reversible reaction.

An understandable (overall) reaction while charging,

$${PbO_2+Pb+2H_2SO_4 \to 2PbSO_4+2H_2O}$$

Actual Chemical reaction:

During charging, $$Pb+SO_4^{2-}\to PbSO_4+2e^-(anode)$$ $$PbO_2+SO_4^{2-}+4H^++2e^-\to PbSO_4+2H_2O (cathode)$$

LAB

As a lead-acid battery charge nears completion, $H_2$ gas is liberated at the negative plate, and $O_2$ gas is liberated at the positive plate. This action occurs 'cause the current required to charge the battery exceeds the current necessary to reduce the remaining amount of lead sulfate on the plates. The excess current ionizes the water in the electrolyte. $H_2$ being less dense than air, leaks out of the battery. The hydrogen gas that batteries make when charging is very explosive. Thus, $H_2$ is emitted only when the charging current exceeds the current causing electrolysis. BUT, Caution is must while doing home-experiments here... Also If temperature exceeds, $H_2$ and $O_2$ gases expand and BOOM. Maybe that's why a battery shouldn't near 50 C.

how does one calculate the expected volume of each gas liberated when a battery is charged?

Over-charging

It could be seen from the graph that $H_2$ is emitted anyways. But, The rate of emission depends on how much percent the battery has been charged... It's better to charge the battery between 40% - 70% (safety limit) for a longer lifetime which is ideal in everyday life.

Regarding the calculation, It goes somewhat deeper in Chemistry I think so, which I can't spit a word from my mouth. But, I'm quite sure that, it varies over the time of charging. BTW, Lower Flammability limits (a critical limit mentioned in percent-volume, above which mixture of air and the gas would possibly ignite) are used for calculating the explosion of batteries. For hydrogen, It's 4.00 %-volume.

More than just current, number of electrons & voltage are required for estimating that. This paper would be useful to burst your brains out...! What I'd suggest - Do some titration...

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Interesting, this makes calculating the freed gas much more difficult. The paper you linked to does not exactly contain much information on the expected amount of hydrogen freed. Do you have any further information on that? –  Claudius Oct 11 '12 at 14:55
    
Nice paper on venting hydrogen. –  Bobbi Bennett Oct 11 '12 at 20:01
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This is really more of a question about chemistry rather than physics, although ‘expected volume’ might give it a hint of physics…

Anyway, you can get the volume of a given gas with a specific pressure (usually 1013 hPa) and temperature (probably about 300 K) from the number of molecules in that gas using either the Ideal Gas Law or the Van der Waals equation. However, for both hydrogen and oxygen, the Ideal Gas Law is probably a sufficient approximation, especially given the error in temperature and pressure to be expected here.

Next, you will need the number of hydrogen/oxygen molecules (note that both hydrogen and oxygen form $H_2$ and $O_2$ molecules!) to plug into these equations. For this, you will need the reaction equation of your battery which should give you a relation between the number of electrons put into it and the number of gas molecules released.

To then get the number of electrons put into the battery (i.e., how often the reaction in your equation takes place), you need to compute the charge of the battery – this can be done in many different ways, but if you know the current output $I$ of the battery and the time $t$ it can deliver this current, the total charge $Q$ is simply

$$ Q = I \cdot t \quad .$$

EDIT: Please see Crazy Buddy’s answer for an explanation why the above text is mostly moot. It still applies if you have anything else producing gas from a chemical reaction connected to an electrical current, but not really to the question.

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+1: Yeah, I agree with that... But, I think the answer expected by OP is some kinda random values not some historic definition. Anyways, It goes deeper into Chemistry I think so... –  Waffle's Crazy Peanut Oct 11 '12 at 16:06
    
@CrazyBuddy: Random values? –  Everyone Oct 11 '12 at 16:39
    
@Everyone: No, I just meant that you'd be expecting with some examples. Don't take it seriously... :) –  Waffle's Crazy Peanut Oct 11 '12 at 16:45
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