2009/10/30 Horace Heffner <[email protected]>: ... >> No it isn't. A hot cathode means the global reaction at the cathode is >> exothermic, nothing else. Palladium hydride formation is exothermic >> (it releases around 11 kJ/mol heat for a loading factor of 0.8), so a >> hot cathode is to be expected, correct me someone if I am wrong. >> >> Michel > > I was under the impression the palladium was mostly plated out before SPAWAR > jacked up the voltage and current to get the live results.
I answered that (Abd made the same objection), don't you agree that steady hydrogen absorption spots in steady state should be hotter due to steady hydride formation, and conversely desorption spots should be cooler? I visualize heat flowing through the lattice in the opposite direction to the steady state deuterium flows, but this may be just my imagination! > In any case much > of the i*W power is dissipated as heat at the cathode and anode surfaces, in > the two molecule interface layer, at least i*0.6 V worth on both the cathode > and anode where these potential drops occur. Not so Horace, I made the same flawed reasoning the other day (and later acknowledged it was flawed), a total power of i*1.54V is indeed consumed at the interfaces, but it is not dissipated there, it is converted to chemical energy, which can later be retrieved by recombining the deuterium and oxygen. Same thing for the electrical energy required to pull the Pd out of solution, it shouldn't be dissipated as heat. Energy released by PdD formation is another story, that I expect to be released as heat. Michel
