256 OXIDATION AND REDUCTION ELEMENTS, chap.
decomposition of the water work is done. In other words, the water spontaneously decomposes until this concentration is reached ; the liquid therefore contains OT X 10"*' gram- molecules of hydrogen, and half as much oxygen per litre at 20°. In this calculation Smale's result, E = 1*062, and Bunsen's absor^ition coefficients of the two gases in water have been used.
If the concentration of the oxygen, by being in equilibrium with the oxygen of the air, is kept constant (25 X 10"* gram-molecules per litre at 20°, according to Bunsen), the quantity of hydrogen is also determined (0*1 x 10~* gram- molecule per litre), for the product of the concentrations must, at any given temperature, be constant (see p. 85).
Strength of the Polarisation Current — Suppose we work with an electromotive force, E^ which is not sufficient to produce an evident separation of gas. Further, suppose that the quantity of dissolved oxygen in the water near the anode is kept constant by being in equilibrium with the oxygen of the air. The electromotive force E then increases propor- tionally to the logarithm of the concentration of the hydrogen (/) near the cathode, so that —
E=A + Tx 10-Mog/,
where -4 is a constant.
A polarisation current is produced by the diffusion of the dissolved hydrogen from the cathode into the water, which, according to our assumption, contains less hydrogen. The quantity of hydrogen which diffuses in a second must, ceteris 2mribic8, be proportional to the excess pressure of the hydrogen at the cathode over that in the liquid (see p. 153). This latter is so small that it may be entirely neglected. The quantity of hydrogen which diffuses is replaced by that separated by the polarisation current in one second, and this is proportional to the current strength i of the polarisation current. We therefore obtain —
i = const. /, and E = Ai + T x 10"^ log i,
where -4i is a new constant.
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