Question

How does zinc form an insoluble zinc hydroxide in water?

Answer 1

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.

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Consider the following Pourbaix Diagram:

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.