How does zinc form an insoluble zinc hydroxide in water?

1 Answer
Jul 10, 2017

It does so in moderately basic solution, modeled by the reaction seen here. Note that in overly basic solution, #"Zn"("OH")_2(s)# will dissolve and #"Zn"("OH")_4^(2-)(aq)# will form instead.


Consider the following Pourbaix Diagram:

https://www.researchgate.net/

This graphs the EMF of the redox process against the #"pH"# of the solution.

The higher up you go in the diagram, the more oxidized the species are, and the farther right you go, the more base-ic the species become.

You can see #"Zn"("OH")_2(s)# forms in #"pH"# #9-11# solutions. In acidic conditions, the redox half-reactions for this process would be:

#"Zn"(s) -> "Zn"^(2+)(aq) + cancel(2e^(-))#
#2"H"^(+)(aq) + cancel(2e^(-)) -> "H"_2(g)#
#"--------------------------------------------------"#
#"Zn"(s) + 2"H"^(+)(aq) -> "Zn"^(2+)(aq) + "H"_2(g)#

In basic conditions, we just have to add #2 xx "OH"^(-)# to both sides to represent such conditions by neutralizing all the strong acid.

#"Zn"(s) + 2"H"^(+)(aq) + 2"OH"^(-)(aq) -> "Zn"^(2+)(aq) + 2"OH"^(-)(aq) + "H"_2(g)#

This gives:

#color(blue)("Zn"(s) + 2"H"_2"O"(l) -> "Zn"("OH")_2(s) + "H"_2(g))#

which is the reaction you asked about here.