SRL Archive - Science

So, I've done (at school) this little experiment with a 'Hofmann voltameter'. It splits water into hydrogen and oxygen.. However I'm not entirely sure which half reactions occur at the cathode…

Either (electrode potential of 0,00):
or (electrode potential of -0,83)

(The one at the anode is (electrode potential of +1,23): )

The problem is that I'm not entirely sure if the hydron (H+) is able to go from the anode to the cathode, thus not sure if the top or the bottom reaction occurs..

However, the experiment showed us we needed around 1,40 Volt to get the device running.. Thus meaning the electrode potential must a bit lower than that.. If we then look back at the half-reactions it must be the upper one that occurs since this is one has the least electrode potential and making the theoretical voltage you need 1,23 Volt..

Wikipedia says it's the lower.

From: nielsie95
Wikipedia says it's the lower.

Yes, however that is not what I've found out :redface:.. The German wiki says it's the upper..

The German wiki says you get both :p

From: nielsie95
The German wiki says you get both :p

There doesn't arise any hydroxide though ;)

anode: 2 H2O –> 02 + 4e- + 4 H+
catode: 4 H2O + 4e- –> 2 H2 + 4 OH-

Copied that from my chemisty book.

EDIT: Assuming you are using sodiumsulphate or something else that is not as eager to oxidize nor simplify as a electrolyte. (dunno if those are the correct terms, soz!)

Actually it's neither of the two… in reality it are way more complex molecules that react. However if you look closely, with both reactions, the only thing that actually changes it's oxidation level are protons. So you would have 2H+ + 2e- —> H2. The way these protons are attached to other molecules in the environment is hardly important. Do remember your total equation has to be balanced and as your anode reaction gives you H+…..

Trust the German wiki, Germans are smart!