I am modifying light bulbs trying to use unique bottles and jars for the envelope. I was thinking I could prevent the filament from oxidizing by removing all $O_2$ before sealing it, and an easy way to do this is just fill the bottle with water and place the negative terminal inside the jar, completely filling it with $H_2$ gas. Would the filament have any problems incandescing in pure $H_2$?
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$\begingroup$ So where does the oxygen go? And hydrogen is really good at diffusing out of sealed containers (not to mention blowing up if the glass cracks open). $\endgroup$– Jon CusterCommented Sep 18, 2020 at 20:12
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1$\begingroup$ @JonCuster generally when doing electrolysis of water, one terminal generates hydrogen gas while the other generates oxygen gas. I have no idea as to the relative purity of them though. $\endgroup$– CR DrostCommented Sep 18, 2020 at 20:24
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$\begingroup$ Well, you can't split H2O into H2 and O2 without generating the O2 - it doesn't magically disappear... $\endgroup$– Jon CusterCommented Sep 18, 2020 at 20:25
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$\begingroup$ Because glass is an amorphous material, the space between molecules will vary substantially, and even if there is one hole larger than the hydrogen molecule (and the probability of that is not small), all the hydrogen molecules will sooner or later escape. (People sometimes fill a closed space with helium to check for leaks using this very principle.) $\endgroup$– KingLogicCommented Sep 18, 2020 at 20:29
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1$\begingroup$ It's easier to generate the $H_2$ (but no $O_2$) by dissolving an electropositive metal (like $Al$, $Zn$ or $Fe$) in a strong acid like muriatic (hydrochloric) acid. Dry the hydrogen thus obtained with conc. $H_2SO_4$ or anhydrous $CaCl_2$, before using it. $\endgroup$– GertCommented Sep 18, 2020 at 20:38
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Setting aside the point on electrolysis of water to get the H2 which is not the question in the title or the final sentence with the question mark, I can see three potential issues of using hydrogen.
Reaction of the tungsten filament with hydrogen. This does not seem to be an issue, with the known tungsten hydrides (from WH to WH$_{6}$) being stable at very high pressures. OK, the filament won't become a hydride (most of which are brittle). So far so good.
Hydrogen as a fill gas in a bulb envelope. The major problem is the high diffusivity of hydrogen in almost all materials. So, you will lose your bulb atmosphere if it is hydrogen rather than the common nitrogen or argon. The bulb lifetime will be substantially lower. Losing much of the gas will also result in increased stresses on the (brittle) glass envelope so it is more likely to break suddenly.
Finally, hydrogen gas has a nearly 10x higher thermal conductivity than argon does. This means that more power will be lost as heat, so the filament bulb will be even less efficient than a regular bulb. It also means that the outside of the bulb will be substantially hotter, increasing the risk of fire.
answered Sep 18, 2020 at 21:36
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1$\begingroup$ One potential objection to point 1: the tungsten hydrides may have a lower melting point than pure tungsten, and the wire might melt if the bulb get too hot. $\endgroup$ Commented Sep 18, 2020 at 22:47
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1$\begingroup$ @KingLogic - the hydrides require GPa patois be stable, so they aren’t forming under normal bulb conditions. If filaments were titanium that would be a problem, but tungsten is just fine. $\endgroup$ Commented Sep 19, 2020 at 1:28
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$\begingroup$ What #2 seems to suggest is that over several heating and cooling cycles, the envelope will eventually have a negative pressure due to H2 diffusion outward as it pressurizes, but no subsequent infusion from the air. This causes more of the stress on the glass than I suppose any temperature gradient would, which should be small. $\endgroup$ Commented Sep 19, 2020 at 21:06