ON ELECTROLYSIS. 733 
explain every migration phenomenon, and we can see the effect of gene- 
ralising it afterwards. 
Consider, now, the mixture AC + A’C’, where A and A’ stand for anions, 
C and C’ for cations. 
Let the current be conducted by these two compounds in the ratio: 
‘X to X’, so that X + \’=1; and, in order to isolate the portion of the 
liquid near either electrode and study the changes therein occurring, we 
may picture the whole fluid as contained in two vessels united by a siphon 
tube ; and we shall call the two vessels the anode vessel and the cathode 
vessel, respectively. 
Further, we may if we like think of AC as sulphate of copper, and of 
A’'C’ as water; the products of deposition, or liberated ions, will then be 
Cand A’ respectively, and the bye-product AC’ will be free sulphuric 
acid. This example therefore belongs to case (2) above, one ion belong- 
ing to each compound ; and this will serve well enough as an example 
to work out. 
Picturing the convection of electricity by the ions through the tube, 
we see that A carries a quantity of negative electricity a, A’ a quantity 
3’ ; C carries 5 of positive electricity, and C’ a quantity $\’. So by the 
time a unit quantity of electricity has been conveyed, and an equivalent 
of anion and cation deposited on anode and cathode respectively, the follow- 
ing changes have occurred in the ingredients of the fluid in the two 
vessels. The cathode vessel has gained 4A equivalent of C from the 
anode vessel, but it has lost a whole equivalent by deposition, so its net 
loss of C, which we may write dC, is 1—3d. Of the other cation it has 
lost none, and has gained $)’ from the other vessel, so we may write 
dC’ = — 3N’. Of anodes it has lost some of both, but only by travelling 
to the other vessel, and accordingly dA =}, and dA’= 42)’. Similar 
reasoning easily applies to the anode vessel. 
Then passing from the elements only to consider the compounds, we 
recognise at once that a compound is lost by the loss of either of its 
elements, and that those elements may be either lost absolutely or may 
be found in new combinations. We thus soon perceive that the loss 
of the compound AC in the cathode vessel is equal to its loss of C, 
a.e. to its major loss; that is, d/AC) =dC=1—23). The loss of A/C’ 
in this same vessel is equal to whichever is the bigger of the two losses, 
dA’ or dC’; and as dC’ is, in the present instance, not a loss, but a gain, 
there is no doubt but that d(A’C’) = dA’ = 3N’. 
Similarly with the other vessel. 
But there remains to be considered what becomes of the unpartnered 
balance of all the various elements, for unless dC equals dA in both 
vessels, there will be in one or other an excess of either A or C—in the 
present case of A; and, again, unless dC’ equals dA’ there will be an 
excess again—in this case of C’; hence we have the new compound AC’ 
formed :—in amount, d(AC’) = dA — dC = dC’ — dA’. 
The equality of dA — dC and dC’ — dA’ is necessary, unless it is 
possible to obtain isolated ions ; and it will be found that they come out 
equal from the preceding values, and that d(AC’) = — 2’ in the cathode 
vessel, while in the anode vessel d(AC’) = — X. 
Represent all this now in a table :— 
