( 568 ) 
and riot simply y =p — ko? as was assumed in the old theory, in 
which the changes of the bordering layer were neglected. 
Had we started from the supposition, that the mercury was —, 
the solution +, a transition of 2 He + positive elections from the 
mercury into the solution — where they would have formed Han 
— would have been required, in order to render the formation of a 
new double layer possible ; and that would have caused the above 
deduction to be modified as follows. 
The part (2) would have been derived from 
; d 
But, according to (207s), this would still have yielded = — oe ds, so 
@ 
this part is not modified. The change of the (negative) electric energy 
on the other hand becomes now @ ds(V,—V49), i. e. 
—ods.hA, 
where now A is positive = ka; so finally equation (3) is yielded 
quite unmodified. 
This equation therefore is of general application, as well in the 
case, that in the solution at the bordering layer negative SO4- or Cl-ions 
++ 
occur, as in the case that there occur positive Hgz-ions. 
But — and this is a circumstance of great importance — the term 
0 e . 
os will in the two cases not have the same value for equal values 
Ww 
of w. For it would bea curious coincidence, that the change of energy 
in the bordering layer, occasioned by adding an infinitely small 
i 
quantity of Hey would te the same as that, occasioned by adding 
an equal quantity of SO, or Cl. In fact this does not happen. The 
experiments show clearly, that the curve represented by (3) is not 
symmetrical on the two sides of the point, where w is zero; and 
that the curve does not consist of one continuous parabola, but of 
two parts of quite different parabolae, which meet in the point where 
co = 0. Only one of them, namely the ascending branch (mercury +, 
solution —), presents a maximum near the point where @ = 0 (so 
not exactly at that point.) 
