Question #a32d3 | Socratic (2024)

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)#