Metallic iron, manganese, cobalt, and nickel at a red heat remove oxygen from water with liberation of hydrogen: 3Fe + 4H2O = Fe3O4 + 3H2; 2C0 + 2H2O = CoO + O2. Conversely, a current of hydrogen passed over these oxides at a red heat will combine with their oxygen, reducing them to metal. This is an instance of mass-action. From the equations given above, it is seen that hydrogen is formed; it does not remain in the tube to re-form water; if it did, there would be a state of balance or equilibrium, all four substances remaining together in proportions depending on the temperature and on their nature; in the current of steam, however, the hydrogen is carried on, and is no longer present to act on the oxide of the metal. And in the converse action the hydrogen conveys the steam away, so that it can no longer be deprived of oxygen by the metal.

As already remarked, carbon monoxide has a similar reducing action on the oxides of the more easily reducible elements. The product in this case is the dioxide, CO2, for example, Fe2O3+ 3CO = 2Fe + 3CO2. This action requires a red heat. Another reducing agent, applied by fusing the oxide with it, is potassium cyanide, KCN; it is converted into the cyanate, KCNO. The metal thallium may be prepared by its help, T12O + KCN = 2TI + KCNO. As the cyanide is somewhat expensive, it is used only in special cases.

An instance has already been given of the mutual reduction of two compounds in the case of nitrogen. Similar instances are known with lead and with sulphur. The chief ore of lead is the sulphide, a natural product termed galena. It is roasted, i.e. heated in contact with air to a red heat. After a portion has been oxidised to sulphate, PbS 4-2O2 = PbSO4, the temperature is raised, when the sulphide and the sulphate mutually reduce each other: PbS + PbSO4 = 2Pb4-2SO2. With sulphur the partial burning of sulphuretted hydrogen may be explained in a similar manner; the reaction, 2H2S + O2 = 2H2O + S2, may be represented as the formation of water and sulphur dioxide by the complete combustion of one-half of the hydrogen sulphide, and its reaction with the remaining sulphide, thus : 2H2S + SO2 = 2H2O + 3S. And, as a matter of fact, that reaction does take place on mixing the two gases in the required proportion of two volumes of hydrogen sulphide with one of sulphur dioxide.