Where negative elements are concerned, the same convention holds, but the ‹logarithmic expression for the potential of such an electrode carries a negative sign› (see footnote 1, p. 261), which must be inserted, algebraically, when the expression is used as a term in the difference under discussion. [533] If C′ > C″, the logarithm will be positive and ε_{Cu′, Cu″} will have a ‹positive› value, which means that the copper plate, Cu′, ‹which is named first in the subscript to ε›, will be charged positively, ‹when the system works›. If C′ < C″, the logarithm will be negative, which means that the first plate, Cu′, mentioned in the subscript, will receive a negative charge, ‹when the system works›. The sign is therefore intended, by the convention adopted (p. 261), to express any result for the ‹working system›, irrespective of the charge on the individual plates before they are combined. For instance, for C′ = 1 and C″ = 10^{−10}, both plates are positive, ‹before› they are connected with each other, since in each case C > K, and ε_{Cu, CuX} = (0.0575 / 2) log(C / K) = a positive value. When the plates are combined, we find from ε_{Cu′, Cu″} = (0.0575 / 2) log(C′ / C″) that the first plate, dipping in the more concentrated solution of cupric-ion, is ‹positive›, which is confirmed by experiment. [534] (1 / 10)-molar cupric sulphate, 100 c.c., containing some sodium sulphate or nitrate, to reduce the resistance, is a convenient concentration. [535] The copper plate is best freed from adhering sulphide by means of a strong cyanide solution, and re-introduced into the solution. [536] Küster, ‹Z. Elecktrochem.›, «4», 110 and 503 (1897). [537] In a solution of zinc sulphate in which [Zn^{2+}] = 0.114, the potential ε_{Zn, ZnSO_{4}}= −0.514 (the minus sign indicates that the metal named ‹first› in the subscript has a negative charge). Inserting the values for [Zn^{2+}] and ε_{Zn, ZnSO_{4}} in the general equation given on p. 261, and solving for K, we find K = 10^{17}. For [Zn^{2+}] = 0.022 and ε_{Zn, ZnSO_{4}} = −0.535, we find K = 10^{16.8}. (‹Cf.› Wilsmore's tables, ‹loc. cit.›) [538] Equilibrium will be established whenever the potential of the system is equal to 0. The potential of the system may be calculated according to the equation (see footnote 1, p. 262) ε_{Cu, Zn} = ε_{Cu, CuSO_{4}} − ε_{Zn, ZnSO_{4}} = (0.0575 / 2) (log(Cu^{2+} / K_{Cu}) − log(Zn^{2+} / K_{Zn})). The potential ε_{Cu, Zn} is 0 whenever [Cu^{2+}] / K_{Cu} = [Zn^{2+}] / K_{Zn}, ‹i.e.› when [Zn^{2+}] / [Cu^{2+}] = K_{Zn} / K_{Cu}. For ions of different ‹valence›, such as silver and cupric ions, the equilibrium equation assumes a somewhat less simple form. For Cu ↓ + 2 Ag^{+} ⇄ 2 Ag ↓ + Cu^{2+}, we have [Ag^{+}]^2 / [Cu^{2+}] = (K_{Ag})^2 / K_{Cu}.