704 



Transactions. — Chemistry . 



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one side so as to allow all the air to be driven out when 

 the cork was forced into its place. A second lOcc. of the 

 same solution were now added. The gold rod passing through 

 ., , , the cork was varnished from a point a 



little below the surface of the liquid to 

 within -Jin. of its lower end, which rested 

 on the gold plate. No. 2 test-tube was 

 filled up in exactly the same way, whilst 

 No. 3 differed only in having neither 

 gold rod nor plate. The three tubes 

 were placed in a small beaker standing 

 in water, and were then covered by in- 

 verting over them a beaker, which also 

 dipped into the water, and so protected 

 the solutions from any fumes that might 

 be in the laboratory. After standing for five days the solu- 

 tions in the upper and lower portions of the tubes were 

 removed separately and analysed in the following manner : 

 5cc. — i.e., half of the solution — from the top or bottom of each 



N 

 test-tabe was titrated with AgNOg— This gave the amount 



of free potassium cyanide remaining in the solution ; the other 

 half was evaporated to dryness with the addition of a few 

 drops of sulphuric acid, and the residue was heated first alone 

 and then with small additions of ammonium carbonate. This 

 residue, which consisted of potassium sulphate and metallic 

 gold, was weighed, and the gold determined by the usual 

 assay process. The amount of potassium sulphate obtained 

 from 5cc. of solution, both above and below the cork, was 

 thus known. From the gold found the amount of potassium 

 cyanide in the double cyanide of gold and potassium can be 

 calculated ; the amount of free potassium cyanide is also 

 known. The sum of these two values, calculated as sul- 

 phates and subtracted from the total amount of potassium 

 sulphate found, gives the amount of potassium sulphate exist- 

 ing in the solution as potash. Now, if the solution of the 

 plate be due to oxygen in the solution below the cork, potash 

 will be formed there ; therefore, in such a case, the ratio of 

 the potash found below the cork to that found above it will 

 be the ratio of the gold dissolved in these two places. If, 

 however, the solution of the plate be due to an electric 

 current generated at the surface of the solution, no potash 

 will be formed below the cork, but only at the surface of the 

 solution. Of course, there is sure to be some oxygen below 

 the cork, and this will act upon the plate independently of 

 the electrical action, and so form a small am.ount of potash 

 in the lower portion of the tube. However, to prove that a 

 current flows, and in the direction stated above, it is only 



