What is the state of aggregation of iodine formed during electrolysis of molten nickel(II) iodide? Assume the electrolysis of molten $\ce{NiI2(l)}$ (inert electrodes) in a cell. The cell with a heating element is in the room; the room is at $\pu{25 °C}$ and $\pu{101.3 kPa}.$

The half-reaction for oxidation at the anode would be

$$\ce{2 I^-(l) -&gt; I2(\text{state}) + 2 e^-},$$

whereas the global reaction will be

$$\ce{NiI2(l) -&gt; Ni(s) + I2(\text{state})}.$$

Given that iodine is solid at room temperature, should it be marked as a solid in the equations? However, given that the $\ce{NiI2}$ is liquid, I'd think that the newly synthesized iodine would be a gas for a while?

Also, what observations could one make during the electrolysis? I.e. will the iodine accumulate on the positive electrode as a solid, or will it exit the cell in the form of a gas? 



Also, is the reduction potential for

$$\ce{I2(s) + 2 e^- -&gt; 2 I^-(\color{red}{aq})}$$

the same as for

$$\ce{I2(s) + 2 e^- -&gt; 2 I^-(\color{red}{l}) }?$$
Basically, during the electrolysis process of molten nickel(II) iodide ($\ce{NiI2}$), $\ce{I– (l)}$ ions would be oxidized to $\ce{I2 (g)}$ at the anode, and $\ce{Ni^2+ (l)}$ ions would be reduced to $\ce{Ni (s)}$ at the cathode. My state assignments were based on following facts assuming the temperature of electrolytic cell is kept below $\pu{900 ^\circ C}$:


Melting point of $\ce{NiI2}$ is about $\pu{797 ^\circ C}$. Therefore, electrolytic cell must be kept above $\pu{800 ^\circ C}$. Thus following equlibrium will be maintained during electrolysis: $$\ce{NiI2 (l) &lt;=&gt; Ni^2+ (l) + 2I- (l)}$$
Boiling point of $\ce{I2}$ is about $\pu{134 ^\circ C}$. Hence, $\ce{I2}$ would be released as a gas at anode.
Melting point of $\ce{Ni^\circ}$ metal is about $\pu{1455 ^\circ C}$. Hence, $\ce{Ni}$ would be deposited as a solid at cathode. If you want to keep nickel metal in liquid form, your electrolytic cell must be kept above $\pu{1500 ^\circ C}$.


I think I have given enough information you to figure out the rest.