376 



EECENT PROGRESS IN PUYSICS. 



It would have been a great support for Daniell's view if lie had 

 succeeded in isolating the hypothetical oxysulphion SO^. In order to 

 attain this, if possible, he experimented with dilute sulphuric and 

 phosphoric acids, and with a solution of sulphate of soda in an 

 U shaped decomposing cell, provided with a diaphragm below, the 

 leg containing the positive pole being cooled down to o°F, in order, 

 by this low temperature, to prevent the decomposition of SO^. But 

 these experiments did not give the desired result. 



Daniell then experimented with solutions of acid salts. He pre- 

 pared a solution of hi-sulphate of potassa. After electrolysis it was 

 found that for each equivalent of hydrogen evolved at the negative 

 pole one equivalent of oxygen had escaped at the positive pole, but 

 that the cell at the negative pole contained only y equivalent of 

 potassa, and that at the positive pole I equivalent of sulphuric acid 

 more than before, or, in other words, that for one equivalent of water 

 decomposed -} equivalent of potassa had been transferred to the nega- 

 tive, and ^ equivalent of sulphuric acid to the positive pole. 



This result appears, at the first glance, very complicated, but 

 Daniell succeeded in suggesting a simple explanation. Let us assume 

 that the fluid consists of 3 electrolytes, water, neutral sulphate of 

 potassa, and hydrate of sulphuric acid, or, according to Daniell's 

 designation, OH, SO4K and SO^H. If, now, the current is so divided 

 that one-half is conducted through HO, one-fourth through SOJ<^, 

 and one-fourth through SO4H, the direct results of the decomposition 

 will be : 



At the positive pole- 

 l equiv. 0, 

 I equiv. SO^, 

 ^ equiv. SO4, 



From OH 



From SO4K 



From SO.H 



At the negative pole. 

 ^ equiv. H, 

 ■|- equiv. K, 

 l' equiv. H. 



By the secondary effect the one-fourth equivalent of K takes up ^ 

 equivalent to form \ equivalent KO, and thereby {- equivalent of 

 H is liberated ; but at the other side SO4 separates into SO3 and O, 

 and thus we have : 



At the positive pole — 



^ equiv. 0, 

 \- equiv, SO3 -|- i equiv. 0, 

 ^ equiv. SO3 -j- |- equiv. 0, 



h equiv. SO3 -\- 1 equiv. 0, 



At the negative pole — 

 ^ equiv. H, 

 I equiv. KO -f- ^ equiv. H, 

 ^ equiv. H, 



I equiv. KO + 1 equiv. H, 



which, in fact, perfectly agrees with the results of the experiment. 



This, also, gave the key to the explanation of the decomposition of 

 diluted sulphuric acid; it is a compound electrolyte, consisting of 

 water and hydrate of sulphuric acid. The current is so divided that 



