What happens when tellurium(II) chloride reacts with water?

Dec 28, 2015

Tellurium(II) chloride is hydrolyzed by water.

Explanation:

Tellurium(II) chloride, ${\text{TeCl}}_{2}$, will be hydrolyzed by water to form tellurium metal, tellourous acid, ${\text{H"_2"TeO}}_{3}$, and hydrochloric acid, $\text{HCl}$.

The balanced chemical equation for this reaction looks like this

$2 {\text{TeCl"_text(2(s]) + 3"H"_2"O"_text((l]) -> "Te"_text((s]) + "H"_2"TeO"_text(3(aq]) + 4"HCl}}_{\textrm{\left(a q\right]}}$

An interesting thing to notice here is that tellurium, which exists in its $+ 2$ oxidation state in tellurium(II) chloride, will be reduced to tellurium metal and oxidized to its $+ 4$ oxidation state in tellorous acid.

This implies that you're dealing with a disproportionation reaction, which is the name given to a chemical reaction in which the same chemical species undergoes both oxidation and reduction.

I think that the same reaction pattern can be expected for tellurium(II) bromide, ${\text{TeBr}}_{2}$. The hydrolysis of this compound would produce hydrobromic acid, $\text{HBr}$, instead of hydrochloric acid

$2 {\text{TeBr"_text(2(s]) + 3"H"_2"O"_text((l]) -> "Te"_text((s]) + "H"_2"TeO"_text(3(aq]) + 4"HBr}}_{\textrm{\left(a q\right]}}$