# Why does magnesium deposit during electrolysis of a molten mixture of MgI2 and AlBr3?

During the electrolysis of mixed salts, I read, and was taught, that the cation with a higher ionization energy would deposit at the cathode. I was also told that you cannot use the standard electrode potential to determine which cation would be reduced.

Is this ionization energy referring to the first ionization energy? If it is, why doesn't the second and third ionization energy factor into this problem seeing that magnesium would have a 2+ charge and aluminum would have a 3+ charge?

Yes, it is true that you cannot use the standard electrode potential to determine which cation would be reduced and vise versa during an electrolysis of mixed molten salts, in your case, $$\ce{AlBr3}$$ and $$\ce{MgI2}$$. That's because most tabulated values of standard electrode potentials are determined for aqueous solutions. Therefore, they can be used only as approximate guidance. Instead, electronegativity values (EN) can be used to estimate the stronger oxidizing and reducing agents. For example, EN values of $$\ce{Br}$$ and $$\ce{I}$$ are 2.8 and 2.5, respectively. Therefore, we can conclude that $$\ce{I-}$$ is the stronger reducing agent between two elements (because $$\ce{I}$$ is less EN than $$\ce{Br}$$). Similarly, between $$\ce{Al}$$ and $$\ce{Mg}$$, $$\ce{Al^3+}$$ is the stronger oxidizing agent because $$\ce{Al}$$ is more EN than $$\ce{Mg}$$ (1.5 vs 1,2, respectively), so $$\ce{Al^3+}$$ gains electrons easier.
Therefore, possible cathode half-reaction (reduction): $$\ce{Al^3+(l) + 3e- -> Al(l)}$$
And Possible anode half-reactions (oxidation): $$\ce{2I-(l) -> I2(l) + 2e-}$$
• I suppose a contributing factor can be the tendency of aluminium to form halogen anions, like $\ce{AlF6^3- AlCl4^-}$. I see as an option to exist $\ce{MgX+ + AlX4-}$ or $\ce{Mg^2+ + 2 AlX4-}$ – Poutnik Apr 27 '19 at 8:22
• @ Poutnik: Thank you for your input. Yet, I didn't see the existence of your version in molten mixture. So for I found is simple molten $\ce{AlCl3}$ giving only $\ce{Al^3+}$ and $\ce{Cl-}$ ions, which converted to $\ce{Al(l)}$ and $\ce{Cl2(g)}$ only. So, I believe $\ce{Al^3+}$ the one reducing at cathode. Also, I'd like to see where is OP get " magnesium deposit during electrolysis of a molten mixture.." part, which is not clear in the text. – Mathew Mahindaratne Apr 27 '19 at 21:43