# Does a battery exert energy keeping a capacitor charged?

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CWatters
Homework Helper
Gold Member
He produced 1 liter of hydrogen gas, which is 0.044643 moles, because 1 mol H2 = 22.4L
I thought it produce a 1 liter mixture of hydrogen and oxygen? That would be 0.66L (0.03 mole) Hydrogen and 0.33L Oxygen. It could even be less because the mixture was bubbled through water at 91 degrees so the mix is probably not at STP.

I agree with your figure for the input power.

Revising the figures for the output gives..

The output energy (from combusting the hydrogen is potentially) =
(0.03 moles H2) * (237.5 kJ/mol) = 7125J

with the efficiency working out at about

(7125/15163) * 100 = 47%

I thought it produce a 1 liter mixture of hydrogen and oxygen? That would be 0.66L (0.03 mole) Hydrogen and 0.33L Oxygen. It could even be less because the mixture was bubbled through water at 91 degrees so the mix is probably not at STP.

I agree with your figure for the input power.

Revising the figures for the output gives..

(0.03 moles H2) * (237.5 kJ/mol) = 7125J

with the efficiency working out at about

(7125/15163) * 100 = 47%
Hmm I think you're right, I thought the hydrogen was being separated from the oxygen for some reason.

Either way though, don't you need oxygen for the combustion reaction i.e. 2 H2(g) + O2(g) ->2 H2O(g) + 475 kJ ?

I could be wrong, but that's what I thought

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Nevermind I think you're right that it's 47% efficient

Where does all that energy go i'm wondering?

And btw i think the hydrogen combustion energy output is 286 kJ/mol, so the efficiency would be

0.03 * 286 = 8580 kJ

-> 8580/15163 * 100 = 56.59%

CWatters
Homework Helper
Gold Member
I think most of the loss goes into heating the water.

In theory you only need something like 1.2-1.4V to split water so running at 12-14V is wasteful..

https://en.wikipedia.org/wiki/Electrolysis_of_water

Electrolysis of water
is the decomposition of water (H2O) into oxygen (O2) and hydrogen gas (H2) due to an electric current being passed through the water. The reaction has a standard potential of −1.23 V, meaning it ideally requires a potential difference of 1.23 volts to split water.
Overpotential
Real water electrolysers require higher voltages for the reaction to proceed. The part that exceeds 1.23 V[24] is called overpotential or overvoltage, and represents any kind of loss and nonideality in the electrochemical process.

For a well designed cell the largest overpotential is the reaction overpotential for the four-electron oxidation of water to oxygen at the anode; electrocatalysts can facilitate this reaction, and platinum alloys are the state of the art for this oxidation. Developing a cheap, effective electrocatalyst for this reaction would be a great advance, and is a topic of current research; there are many approaches, among them a 30-year-old recipe for molybdenum sulfide,[25] graphene quantum dots,[26] carbon nanotubes,[15] perovskite,[27] and nickel/nickel-oxide.[28][29] The simpler two-electron reaction to produce hydrogen at the cathode can be electrocatalyzed with almost no overpotential by platinum, or in theory a hydrogenase enzyme. If other, less effective, materials are used for the cathode (e.g. graphite), large overpotentials will appear.
Not sure if this has changed but most Hydrogen used to be made from fossil fuel because its cheaper than using electricity. ..

https://en.wikipedia.org/wiki/Hydrogen_production

As of 1999, the majority of hydrogen (∼95%) is produced from fossil fuels by steam reforming or partial oxidation of methane and coal gasification with only a small quantity by other routes such as biomass gasification or electrolysis of water.[/quote]

I know the energy on the plates is U = CV^2/2 where U is the energy, C is the capacitance, and V is the voltage.

In this case I'm assuming that the plate configuration creates a parallel plate capacitor

You're saying that some of this energy is heating the water, right? That makes sense. Along with what berkeman mentioned about leakage, I guess that's two sources right there the system is losing energy.. interesting

CWatters
Homework Helper
Gold Member
I don't think there is much to be gained by thinking of the cell as a capacitor. There will be some capacitance but it's not really relevant to the operation of the cell. Once the capacitance is charged (occurs in fractions of a second) no further energy is needed to charge that capacitance.

No I'm not saying that some of the energy used to charge the capacitor is wasted.

The leakage current in a capacitor is normally designed to be as small as possible. That's not the case with an electrolysis cell. You actually want current to flow between the plates of an electrolysis cell.

berkeman
Mentor
I know the energy on the plates is U = CV^2/2 where U is the energy, C is the capacitance, and V is the voltage.

In this case I'm assuming that the plate configuration creates a parallel plate capacitor

You're saying that some of this energy is heating the water, right? That makes sense. Along with what berkeman mentioned about leakage, I guess that's two sources right there the system is losing energy.. interesting
I agree with CWatters. The electrolysis cell is not modeled as a capacitor, and there are no significant reverse leakage currents.

It is a load that you drive with a voltage. There is a forward current that flows. That voltage and current do work -- some of that work splits the water into H2 and O2, and some of it does work on the water (by driving ions through the water), which is lost as heat.

Maybe not in this case there isn't, but in the case of my undergrad project it is relevant. And to my understanding current does not flow between the capacitor plates in pure water situations, which is why people use electrolyte because

Electrolysis of Water: Pauling, Linus (1970) General Chemistry, Section 15-2. San Francisco.

"If a water-soluble electrolyte is added, the conductivity of the water rises considerably. The electrolyte disassociates into cations and anions; the anions rush towards the anode and neutralize the buildup of positively charged H+ there; similarly, the cations rush towards the cathode and neutralize the buildup of negatively charged OH− there."

But for the most part capacitors produce electric fields; not current.

the reason water is being split is because of the opposite charges on the capacitor plates. one plate's charge is positive, and the other plate's charge is negative, so the negative oxygen atoms are attracted to the positively charged plate, and the positive hydrogen ions are attracted to the negatively charged plate.

it is not beneficial if there is current running between the plates, because in the case there is current: There would be no difference in having a capacitor setup like we've been discussing, and a straight up wire in water.

As mentioned above, electrolyte is added, but not to explicitly allow current to flow between the plates, but instead to neutralize the build up of H+ ions and OH- ions on either plate, so that the process of generating electric field, and likewise creating an anode and a cathode can continue

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berkeman
Mentor
Capacitors don't pass DC currents...

Capacitors don't pass currents at all

berkeman
Mentor
Sure they do, AC currents are conducted by the changing electric field.

Sure they do, AC currents are conducted by the changing electric field.
The current doesn't pass between the plates though, if that's what your saying

berkeman
Mentor
IIRC Lithium batteries leak less than Aluminum electrolytics, but other than that, batteries self discharge more than capacitors leak.
After 30 years of designing battery chargers and power supplies for many companies, and reading the customer like I was playing poker, it seemed that a circuit that lost less than a micro Amp when not in use was ok. Designing a circuit that leaks less than a uA but can stand up to lightning, takes some work.

what material do you use for dielectric in the capacitor?

berkeman
Mentor
what material do you use for dielectric in the capacitor?
Who are you asking, and what do you mean?

There are a number of common dielectrics used. Is this a separate topic you want to ask about, or related to your electrolysis questions somehow?

Who are you asking, and what do you mean?

There are a number of common dielectrics used. Is this a separate topic you want to ask about, or related to your electrolysis questions somehow?
i'm asking Clark Magnuson, because he brought up designing circuits, so i assume he has experience building capacitors, and yes the dielectric is relevant to electrolysis, because water in a capacitor acts as a dielectric with coefficient value ~80.4, and I am wondering if there is anything he knows that he can tell me about the properties of water, and/ or dielectrics in general in capacitors.

berkeman
Mentor
Ah, it helps to remove ambiguity when you want to ask someone if you quote part of their reply. That makes it clear who you are asking. You can do that by clicking "Reply" in the lower right of a post, or just click-drag to select part of a post and click Reply to quote just that part of the post.

I'll tag @Clark Magnuson in case he doesn't have auto-notification of replies enabled in his account.

I never designed capacitors other than ones I did not want. That was the 10pF /foot of parallel wires or traces. Multiply by 2pi for coax shield.

I have anecdotes from designing with commercial capacitors. Some of the things we look at are:
1) ESL effective series inductance. The leads or wires to the capacitor are 1uH/ meter that adds to the cap's own problems.
2) ESR effective series resistance. For high frequency this is more important than the value of the capacitor in Farads.
3) Maximum RMS current. Must get the heat out of the capacitor.

If there is an electrolyte, it can leak and cause end of life. I test the capacitor with an oscillator, hooked up to an amplifier, hooked up to transformer, hooked up to a DC bias, hooked up to the capacitor under test in an oven. I weigh the capacitor before and after testing. I pick up the capacitor with tweezers, as my finger prints weigh more than the amount of Hydrogen outgassing I am trying to measure.

Some of the best capacitors I have used are multilayer ceramic. They are available in a range of dielectric materials.