Question #a32d3

Zinc metal will indeed react with vinegar, which is a dilute solution of acetic acid ,`"CH"_3"COOH"`, to form zinc acetate, `("CH"_3"COO")_2"Zn"` and hydrogen gas, `"H"_2`.

Zinc will be oxidized to zinc cations, `"Zn"^(2+)`. At the same time, hydrogen will be reduced to hydrogen gas, `"H"_2`.

The balanced chemical equation for this reaction looks like this

`"Zn"_text((s]) + 2"CH"_3"COOH"_text((aq]) -> ("CH"_3"COO")_2"Zn"_text((aq]) + "H"_text(2(g]) uarr`

The acetate anion, `"CH"_3"COO"^(-)`, is a spectator ion, which means that it does not influence the reaction. The net ionic equation for this reaction will look like this

`"Zn"_text((s]) + 2"H"_text((aq])^(+) -> "Zn"_text((aq])^(2+) + "H"_text(2(g]) uarr`

Zinc loses its two electrons to form the zinc cation. In the process, its oxidation state goes from `color(blue)(0)` to `color(blue)(+2)`.

`stackrel(color(blue)(0))("Zn") -> stackrel(color(blue)(+2))"Zn" + 2"e"^(-) ->` the oxidation half-reaction

Hydrogen will pick up these electrons to form hydrogen gas. In the process, its oxidation state will go from `color(blue)(+1)` to `color(blue)(0)`

`2stackrel(color(blue)(+1))("H"^(+)) + 2"e"^(-) -> stackrel(color(blue)(0))"H"_2 ->` the reduction half-reaction

Put the two half-reactions together to get

`stackrel(color(blue)(0))("Zn") + 2stackrel(color(blue)(+1))("H"^(+)) + color(red)(cancel(color(black)(2"e"^(-))))-> stackrel(color(blue)(+2))"Zn" + color(red)(cancel(color(black)(2"e"^(-)))) + "H"_2`

Which is of course the net ionic equation

`color(green)("Zn"_text((s]) + 2"H"_text((aq])^(+) -> "Zn"_text((aq])^(2+) + "H"_text(2(g]) uarr)`