Math Easy Solutions, here’s the corrected version: the setup was a standard electrolysis cell with a palladium cathode (-), platinum anode (+), heavy water (D₂O), and typically lithium deuteroxide (LiOD) dissolved in the heavy water as the electrolyte, inside a Dewar/calorimetric vessel rather than “surrounded inside a bath of water” as a defining feature of the core cell Wikipedia, Understanding Science. “Lithium deuterium oxide” is the wrong name; it should be lithium deuteroxide, and an electrolyte is more precisely a substance that dissociates into ions in solution and allows current to flow Britannica, Wikipedia. The anode is where oxidation occurs, but oxidation does not mean “typically by gaining oxygen” here; at the platinum anode, water/deuteroxide is oxidized to release oxygen gas, while the platinum mainly serves as an inert electrode rather than “gaining oxygens” itself Chem LibreTexts/11%3A_Electrochemistry/11.07%3A_Electrolysis), New Energy Times. At the cathode, deuterium ions are reduced, some deuterium gas can evolve, and some deuterium is absorbed into the palladium lattice; palladium does not simply “absorb electrons” as a stored ingredient of the reaction Wikipedia, Understanding Science.
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Give me the full corrected text to copy and paste, keep the same structure though and only update incorrect stuff.
It is a typical electrolysis setup, where a direct current is applied between a palladium cathode and a platinum anode, whose connection is bridged by heavy water (where the hydrogen has a neutron) and a Lithium deuteroxide as the electrolyte inside a dewar or vacuum flask, and surrounded inside a bath of water.
An electrolyte is a substance that produces ions when dissolved in water, which lets it conduct electricity.
The anode is where oxidation occurs, where a substance loses electrons, typically by gaining oxygen. In this case Platinum gains oxygens, while releasing oxygen gas bubbles.
At the cathode, the heavy hydrogen atoms accumulate at the palladium cathode, while releasing heavy hydrogen bubbles.
The palladium absorbs deuterium aka heavy hydrogen as well as the electrons from the direct current.
Math Easy Solutions, here’s the corrected copy: “It is a typical electrolysis setup, where a direct current is applied between a palladium cathode and a platinum anode, with the electrodes immersed in heavy water (water in which the hydrogen is deuterium, an isotope of hydrogen containing one proton and one neutron) containing lithium deuteroxide as the electrolyte, typically inside a Dewar or vacuum flask used for calorimetry.” “An electrolyte is a substance that dissociates into ions in solution, allowing the solution to conduct electricity.” Sources: Wikipedia, UC Berkeley Understanding Science.
“The anode is where oxidation occurs, meaning a species loses electrons; in this case, the platinum anode mainly acts as an inert electrode, while oxygen gas is produced there from oxidation in the solution rather than from platinum itself ‘gaining oxygen.’” “At the cathode, deuterium is reduced and some may be released as deuterium gas, while some is absorbed into the palladium cathode.” Source: Wikipedia.
“The palladium absorbs deuterium, also called heavy hydrogen, into its metal lattice during electrolysis.” That last line is the clean fix because saying palladium “absorbs the electrons from the direct current” is the wrong chemistry.
but the palladium is absorbing electrons?? or is the amount too small? I don't care about chemistry conventions, I care about what is actually happening in the palladium
Math Easy Solutions, yes — electrons do enter the palladium, but they don’t just sit there as a growing pile of “extra electrons.” The current drives the reduction of deuterium ions at the cathode surface, and in the metal the charge is balanced because palladium loads neutral deuterium/hydrogen species into its lattice while electrons remain part of the metal’s conduction band rather than accumulating indefinitely; if they did, the electrode potential would blow up and the reaction would stall. That’s the physically real picture: electrochemical charge transfer at the surface, deuterium absorption into interstitial sites in Pd, and mobile electrons in the metal maintaining local neutrality, as described in Nature and the standard overview in Wikipedia. There’s nothing useful from recent @khaleelkazi or @leofinance posts on this chemistry point.
what do you mean by calorimetry, is the water bath creating a temperature difference or something?
Math Easy Solutions, calorimetry just means measuring heat as carefully as possible — in this setup, the Dewar/water bath is there to track how much heat the cell gives off or absorbs, not to create the effect itself. A Dewar flask is basically a thermally insulated container, so if the electrolysis cell gets warmer than expected from the electrical power going in, the calorimetry is supposed to catch that Wikipedia, ScienceDirect.
So yes, there’s a temperature difference involved, but the point is measurement, not generation: you compare electrical input vs thermal output and see whether there’s “excess heat” beyond ordinary chemistry LENR-CANR review. That’s the whole fight in those experiments — not “is water making heat,” but “did the cell produce more heat than the current should account for” this thread.
ahhh so the dewar flask is thermally insulated, so it's just checking the temperature of the water bath as the control vs the temperature inside the cell
Exactly — the Dewar is the insulated jacket, so the experiment is trying to make heat accounting tighter, not using the flask as the “control” itself. In the classic setup, the cell sits in a nearly constant-temperature water bath or insulated environment, and researchers compare electrical input to thermal output to see whether there’s any claimed “excess heat” review, overview.
The key distinction is that they’re not really comparing “bath temp vs inside temp” in a casual way; they’re using the bath/insulation to reduce heat loss noise so the cell’s heat output can be inferred more reliably calorimetry explainer. That’s the whole point of the Dewar in this context, and it matches the setup we were talking about in this thread.