What is the state of aggregation of iodine formed during electrolysis of molten nickel(II) iodide?
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.