[544] The potential of a solution of the iron salts is given by ε = 0.058 log(10^{17} × [Fe^{3+}] / [Fe^{2+}]). In a solution of a ferric salt, if [Fe^{2+}] = 0, the potential would obviously be ∞, which could not present a condition of equilibrium. Equilibrium is established in such a solution, as will be shown further on in the text, by the liberation of chlorine and the formation of ferro-salt, according to 2 Fe^{3+} + 2 Cl^{−} ⇄ 2 Fe^{2+} + Cl_{2}, until the potential, resulting from the tendency of chlorine to form chloride-ion, just balances the tendency of the ferric-ion to form ferro-ion. But when a ferric chloride solution is used as the source of supply of positive electricity, as in the experiment described in the text, ‹both› the ferric-ion and the chlorine tend to charge the platinum electrode with positive electricity and to revert to a condition of equilibrium in reference to their individual constants. The relations are much like those between a cupric salt solution and a copper plate: if [Cu^{2+}] > K_{Cu^{2+}}, equilibrium will be established, as we have seen, by the positive charging of the plate in sufficient degree to oppose the tendency of the cupric-ion to discharge (see p. 259). But when the solution and plate are used as the source of supply for an electric current (p. 264), both the positive charge on the plate, and the tendency of the cupric-ion to discharge and acquire the concentration [Cu^{2+}] = K_{Cu^{2+}}, will supply the positive current. In calculations we ignore the positive charge already deposited on the plate and deal only with the concentration of Cu^{2+}. The chlorine, liberated in a solution of ferric chloride, plays practically the same rôle as does the copper plate in a cupric salt solution, and it can be ignored in the discussion of the combination described in the text. In a ferrous salt solution, in a similar manner, some ferric-ion must always be formed by liberation of hydrogen (see p. 282), until equilibrium is reached according to 2 Fe^{2+} + 2 H^{+} ⇄ 2 Fe^{3+} + H_{2}. Hydrogen plays here the same rôle as chlorine does in the ferric chloride solution. [545] The condition for equilibrium is [Fe^{2+}] : [Fe^{3+}] = 10^{17}, in a solution considered for itself. [546] This ratio need not be 10^{17}, since we have two solutions combined with each other and the total potential will be expressed by: ε = ε_{1} − ε_{2} = 0.058 (log(10^{17} × [Fe^{3+}]_{1} / [Fe^{2+}]_{1}) − log(10^{17} × [Fe^{3+}]_{2} / [Fe^{2+}]_{2})) = 0.058 log([Fe^{3+}]_{1} × [Fe^{2+}]_{2} / ([Fe^{2+}]_{1} × [Fe^{3+}]_{2})). Equilibrium is reached when the total potential is 0. Then [Fe^{3+}]_{1} × [Fe^{2+}]_{2} / ([Fe^{2+}]_{1} × [Fe^{3+}]_{2}) = 1 and [Fe^{2+}]_{1} / [Fe^{3+}]_{1} = [Fe^{2+}]_{2} / [Fe^{3+}]_{2}.