1487 
3. Explanation of different electromotive processes by the appli- 
cation of the idea: 
“Solubility product of a metal”. 
a. Dissolving of metals by halogen. Suppose a metal to be 
immerged in a vessel of pure water, and then a halogen e.g. chlo- 
rine, to be added, it will quite depend on the situation of the two 
following equilibria 
MSM OR ea. (da) 
ande Dera). ACL EELDE ine x, (6) 
what will bappen. 
If the concentration of the electrons in the ionisation equilibrium 
(la) of the metal is greater than that in the equilibrium (6), the 
chlorine will take away the electrons, which causes the homo- 
geneous equilibrium (la) to shift to the righthand side, and the 
heterogeneous equilibrium between metal and electrolyte is broken. 
The metal will then send metal atoms, metal ions, and electrons 
into solution, and so the metal ean entirely go into solution on 
sufficient addition of chlorine. 
It now follows from what precedes that it will depend on the 
concentration of the electrons emitted by the metal whether with 
the prevailing chlorine concentration a shifting of (la) from left to 
right is possible. 
The electron-concentration mentioned here will depend on the 
value of the solubility product of the metal, and it may, therefore, 
be predicted that metals with a relatively large solubility product 
will dissolve in chlorine water, whereas metals with a small solu- 
bility product will not be attacked. It is, however, the question 
whether there is a metal so noble that the concentration of the 
electrons which ensues through the ionisation of the metal is smaller 
than corresponds with equilibrium (6). 
lt follows already from this that the base metals possess a rela- 
tively great, and the noble metals a relatively small solubility product. 
b. Dissolving of metals in pure water. Now we might also 
answer the question why one metal dissolves in water and why 
another does not. 
For this purpose we have to consider the equilibria: 
MEM HO +. Rn ta) 
and ORO OH JT en) 
If we now have a metal with a great solubility product, then 
(4) is comparatively great in (la), and the reaction (7) can proceed 
