4
$\begingroup$

Aluminium is coated with a thin layer of Aluminium Oxide. Why then do we write the reaction as

$$\ce{2Al + 3HgCl2 → 2AlCl3 + 3Hg}$$

Shouldn't we first write an equation showing how $\ce{HgCl2}$ reacts with $\ce{Al2O3}$ ?

$\endgroup$
3
  • $\begingroup$ I think the reaction takes place in an inert atmosphere. The reaction is simply a reduction of $\ce{Hg^2+}$ to $\ce{Hg}$ $\endgroup$
    – Aditya Dev
    Commented Feb 25, 2016 at 8:37
  • 5
    $\begingroup$ The reaction that happens is the one described. But the reason it happens is because the reaction with mercury salts disrupts the surface layer of alumina that otherwise protects the aluminium from routine oxidation (as many other substances do). Once the reaction has started no further protective layer forms so even minor defects in the protective alumina layer lead to a runaway disruption that allows further reaction. Small physical effects in the alumina are usually fixed in air, but won't "heal" when other reactions are occurring. $\endgroup$
    – matt_black
    Commented Feb 25, 2016 at 11:35
  • $\begingroup$ Socratic Q&A $\endgroup$
    – ananta
    Commented Jun 19, 2023 at 17:39

1 Answer 1

-1
$\begingroup$

The equation is correct assuming the aluminium is pure. In reality, $\ce{HgCl2}$ won't readily react with aluminium sheet since both are solids and $\ce{HgCl2}$ is a weakly-dissociated salt meaning that even in aqueous environment there won't be much dissociation into $\ce{Hg^2+}$ and $\ce{Cl^-}$. The molecule would simply go into solution as is.

Maybe there will be reaction if you mix aluminium and mercury(II) chloride powders and heat the mixture - still there will be formation of mercury(II) oxide rather than metallic mercury:

$$\ce{Al2O3 + 3 HgCl2 -> 2 AlCl3 + 3 HgO}$$

I think the reaction really goes off in aqueous solution or at least moist environment, where some mercury(II) chloride gives acidic reaction and we get some hydrogen chloride that will readily dissolve the oxides, allowing for the displacement reaction.

Note that aluminium amalgam is commonly prepared by first scratching the metal surface under water or under mercury (as otherwise it would immediately re-oxidize to alumina), grinding the two metals together or first etching the metal with dilute $\ce{HCl}$ or $\ce{NaOH}$. Under water, the aluminium is protected from air and will form hydroxide rather than oxide, which reacts much faster with aqueous $\ce{HgCl2}$.

Indeed aluminium with thick oxide layer reacts very slowly with $\ce{HgCl2}$. It eats away the oxide layer but at the cost of some dissolved mercury(II) turning into insoluble $\ce{HgO}$. There is a neutralization going on rather than displacement plus other side reactions (like formation of insoluble calomel - $\ce{Hg2Cl2}$, formation of oxychlorides etc.).

The typical procedure for preparing aluminium amalgam is therefore properly degreasing and etching aluminium with solvents and dilute base first, then washing off excess solvent, hydroxides and base (the metal is still wet) and then getting it into contact with dilute mercury(II) solution (which can now be very dilute as we don't waste $\ce{HgCl2}$ for the etching).

$\endgroup$
4
  • $\begingroup$ Interesting.... I am Libor too, also from Brno. $\endgroup$
    – Poutnik
    Commented Jun 19, 2023 at 13:55
  • 1
    $\begingroup$ Chem+Math Expression formatting reference: MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial // MathJax is preferred not to be used in CH SE Q titles. $\endgroup$
    – Poutnik
    Commented Jun 19, 2023 at 13:56
  • $\begingroup$ HgO is decomposed by mild heat. Al2O3 is one of the most exothermic compounds. $\endgroup$
    – jimchmst
    Commented Jun 19, 2023 at 15:56
  • $\begingroup$ What is the meaning of the sentence : "I am Libor too" , sent by Poutnik 2 hours ago ? $\endgroup$
    – Maurice
    Commented Jun 19, 2023 at 16:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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