Would a drawing of the electrolysis of NaF(l) have two Pt electrodes at the anode and cathode? We used carbon in the labs.. How much of a difference does using a Pt electrode have over a carbon one?

 

Also i'm trying to figure out OVERPOTENTIAL, but i don't understand it (i dont think my book explains it very well).

 

For example, the overpotential for the discharge of H2(g) at a mercury cathode is about 1.5V, but if the reduction of Na+ is still 2.71V, why is Na(s) being produced? Or does the Eo of Na+/Na change with the addition of a mercury cathode? Where does the mercury play a part in these transformations??

Flourine gas doesn't sound too safe to me...

 

but... overpotential/voltage is usually only relevant in aqueous solutions..

I dont know. But is you used carbon rods the flourin would destroy one of them quickly. Flouring is the most reactive chemical know, so I would imagine electroysis of it would be impossible or just stupid.

It also wouldnt work too well because not too much of the HF would dissociate.

wait, you actually did this in a lab?

 

the F2 would destroy carbon, but it would be hard for it to do anything to Pt. it's what Moissan used when he first isolated F2. if you plan on trying this, just remember that making F2 is a good way to die.

Flourine gas doesn't sound too safe to me...

 

but... overpotential/voltage is usually only relevant in aqueous solutions..

Correct. Overpotential/voltage is not relevant to molten, non-aqueous electrolysis. In an aqueous environment, you wouldn't even be able to produce fluorine gas anyway. I believe you would just get electrolysis of the water and the remnants would be HF and NaOH which would get you right back to water and NaF again.

wouldn`t it form Fluorates if aqueous?

and also kick off some F2 and H2 aswell?

Fluorates do not exist. Fluorine is an element, which only exists in oxidation state 0 (as element) or in oxidation state -1 in ALL of its compounds.

 

There is "hypofluorous" acid, HOF, but this in reality contains fluorine in oxidation state -1, oxygen in oxidation state 0 and hydrogen in oxidation state +1 (as it usually is). So, HOF is totally different from the other hypohalogenites, such as the common bleach, which on very slight acidification gives HOCl.

 

Higher fluorates simply do not exist.

 

So, when NaF is electrolysed, then indeed, as Jdurg already stated, O2 and HF are formed at the anode and OH(-) and H2 are formed at the cathode:

 

anode: [ce]2H2O + 4F^{-} - 4e -> 4HF + O2[/ce]

cathode: [ce]4H2O + 4e -> 4OH^{-} + 2H2[/ce]

 

Net reaction when the solution is mixed: decomposition of water.