0
$\begingroup$

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

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

whereas the global reaction will be

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

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

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

the same as for

$$\ce{I2(s) + 2 e^- -> 2 I^-(\color{red}{l}) }?$$

$\endgroup$
6
  • 4
    $\begingroup$ ?!? $\ce{NiI2}$ melts at 780 °C. How are you going to have liquid $\ce{NiI2}$ in a cell at 25 °C? $\endgroup$
    – MaxW
    Commented May 24, 2020 at 21:11
  • $\begingroup$ The cell is not at 25 deg C. The room in which the cell is is at 25 C. $\endgroup$
    – user256872
    Commented May 24, 2020 at 21:12
  • 3
    $\begingroup$ Well iodine boils at 184 °C. What do you think will happen to it? $\endgroup$
    – MaxW
    Commented May 24, 2020 at 21:20
  • $\begingroup$ My guess is that the iodine will be a gas. $\endgroup$
    – user256872
    Commented May 24, 2020 at 21:23
  • 3
    $\begingroup$ I took a liberty to correct formatting, rewrite the title and add a brief detail that your cell has a heating element to avoid confusion with the RT and m.p. of the salt (you got me here, too). Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. $\endgroup$
    – andselisk
    Commented May 24, 2020 at 21:37

1 Answer 1

3
$\begingroup$

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: $$\ce{NiI2 (l) <=> Ni^2+ (l) + 2I- (l)}$$
  • 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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.