Note 556 — Chapter XV · OXIDATION AND REDUCTION. II. OXIDATION BY OXYGEN, PERMANGANATES, ETC.; OXIDATION OF ORGANIC COMPOUNDS

The constants are calculated from the estimated potential of the oxygen-hydrogen cell, +1.231 volt, at 18°. (Vide G. N. Lewis, Z. phys. Chem., 55, 465 (1906); Nernst and Wartenberg, ibid., 56, 534 (1906); Brönsted, ibid., 65, 91 (1908); and a summary and discussion by Schoch, loc. cit.) At 18° oxygen, under one atmosphere pressure, gives an estimated potential εO2, HO− = +1.508 against an acid solution, in which the concentration of the hydrogen-ion [H+] = 1 (see the table at the end of this chapter). Since at 18° [H+] × [HO−] = 0.81E−14, the value for [HO−] in this acid solution is 0.81E−14. Now, for oxygen, at 18° and 760 mm., there must be some concentration of hydroxide-ion, which we will call [C], at which the tendency of the oxygen to ionize is exactly balanced by the tendency of the hydroxide-ion to form oxygen—at this point the potential is 0. For any concentration [HO−] of the hydroxide-ion, other than [C], a potential will exist εO2, HO− = 0.0575 log([C] / [HO−]). Since for [HO−] = 0.81E−14, we have a potential εO2, HO− = +1.508, these values can be introduced into the equation and the latter solved for [C]. We find thus [C] = 1.36E12, and oxygen, at 18° and 760 mm. pressure, would be directly in equilibrium with a solution in which [HO−] = 1.36E12. At 18° and 760 mm. pressure a liter of oxygen contains 1 / 23.9 mole, and thus we have for the condition of equilibrium [HO−]4 : [O2] = K : (1.36E12)4 : (1 / 23.9) = 8.2E49 = K.