[OTI]

So I have sodium NaI and Fe(NO₃)₂ in aqueous solution and I'm performing electrolysis. Find the half reaction at the anode and the cathode.



So I know the ions I have to work with (not given): Na⁺, I^-, Fe⁺², NO₃^-, H⁺, and OH^-

I'm not exactly clear on which half reactions will run though.

I'm guessing the half reaction 2I^-    I₂ + 2e^- happens at the anode because it has the lowest potential (my teacher told me that the lowest potential stuff is the one that always get's oxidized).

I seriously have no idea for the cathode but my gut says it's probably Fe⁺² because my teacher said that it's really hard to get sodium through electrolysis when with other stuffs (for a lack of a better term).

Yup. Please tell me if I'm on the right track or if I'm way off (has happened numerous times). Thanks.

[Borek]

Fe²⁺ and Na⁺ are not the only cations in the solution.

And when it comes to the reduction the situation is similar to the oxidation, just reversed - now the first thing to get reduced is the substance with the highest redox potential.

[Enthalpy]

How alkaline would an NaI solution be? My simplistic thought is that pure water conducts electricity little (or extremely little as compared with any solution) so if H⁺ is even more scarce than in pure water, its contribution must be negligible. Current must result from Fe²⁺ and Na⁺ essentialy, but at least Na⁺ won't evolve as it reacts with water. So H₂ may evolve, but not as a consequence of H⁺ movement.

Or did I miss something?

[Borek]

if H⁺ is even more scarce than in pure water

That means alkaline solution, doesn't it?

Or did I miss something?

Fe(OH)₂ pK_sp=15.1

[OTI]

Okay, that makes sense. So the lowest redox potential thing is oxidized and the highest potential one is reduced. Thanks!