In regard to the first of these experiments Traube[20] has carefully investigated what takes place when palladium hydrogen is allowed to remain in contact with water and oxygen. Hoppe-Seyler had noticed that under these conditions small quantities of hydrogen dioxide were formed, but he attributed this to the union of the active oxygen with the water. Traube, on the other hand, finds that in the formation of hydrogen dioxide under these circumstances there is nothing formed which has oxidizing properties, not even indigo sulphuric acid is oxidized. He shows by direct experiments that nascent hydrogen can not by its action on oxygen produce active oxygen or ozone. He finds the action of palladium hydrogen to be analogous to that of zinc and other metals, which when allowed to oxidize slowly in contact with air and water give rise to the formation of hydrogen dioxide. The process is to be regarded rather as a reduction of molecules of oxygen than as an oxidation of water. Traube represents the action by the following equations:--
+--------------------+--------+ OH | OH|H O| / | Zn | | = Zn + H_{ 2}O_{ 2} | Zn + OH|H + O| \ +--------------------+--------+ OH
+--------------------+----------+ | Pd_{ 2}H OH|H | | + | + O_{ 2} | = 4 Pd + 2 H_{ 2}O + H_{ 2}O_{ 2} | Pd_{ 2}H OH|H | +--------------------+----------+
He proves by direct experiments that no active oxygen is formed during this process, and points out that the oxidations observed by Hoppe-Seyler and Baumann must have been brought about by the hydrogen dioxide. But this would not account for the oxidation of carbon monoxide, for it has been previously shown by Remsen[21] that hydrogen dioxide cannot oxidize carbon monoxide, not even when it is heated to its point of decomposition. Traube[22] therefore repeats Baumann’s experiment, he finds that palladium hydrogen in the presence of water and oxygen does oxidize carbon monoxide; but as he had shown that no active oxygen was formed during the process, and as the hydrogen dioxide could not cause the oxidation, he concluded that the palladium itself must play an important role in the reaction. By further experiments he soon became convinced that there are two stages in the process. 1st the palladium hydrogen acting on water and oxygen forms hydrogen dioxide, 2nd the hydrogen dioxide in the presence of palladium oxidizes the carbon monoxide. Traube[23] introduced into a glass flask containing carbon monoxide a dilute solution of hydrogen dioxide and a small piece of palladium foil, previously ignited, the action was allowed to continue for 22 hours after which the CO was replaced by air free from carbon dioxide. After leaving the flask the gas passed through a solution of barium hydroxide; an abundant precipitate was formed, showing that in this case the quantity of carbon dioxide formed was greater than in the first experiment, in which he used palladium hydrogen, water, oxygen and carbon monoxide.