# Question a32d3

Feb 18, 2016

Zinc will be oxidized to zinc cations.

#### Explanation:

Zinc metal will indeed react with vinegar, which is a dilute solution of acetic acid ,$\text{CH"_3"COOH}$, to form zinc acetate, ("CH"_3"COO")_2"Zn" and hydrogen gas, ${\text{H}}_{2}$.

Zinc will be oxidized to zinc cations, ${\text{Zn}}^{2 +}$. At the same time, hydrogen will be reduced to hydrogen gas, ${\text{H}}_{2}$.

The balanced chemical equation for this reaction looks like this

${\text{Zn"_text((s]) + 2"CH"_3"COOH"_text((aq]) -> ("CH"_3"COO")_2"Zn"_text((aq]) + "H}}_{\textrm{2 \left(g\right]}} \uparrow$

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

${\text{Zn"_text((s]) + 2"H"_text((aq])^(+) -> "Zn"_text((aq])^(2+) + "H}}_{\textrm{2 \left(g\right]}} \uparrow$

Zinc loses its two electrons to form the zinc cation. In the process, its oxidation state goes from $\textcolor{b l u e}{0}$ to $\textcolor{b l u e}{+ 2}$.

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 $\textcolor{b l u e}{+ 1}$ to $\textcolor{b l u e}{0}$

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

$\textcolor{g r e e n}{{\text{Zn"_text((s]) + 2"H"_text((aq])^(+) -> "Zn"_text((aq])^(2+) + "H}}_{\textrm{2 \left(g\right]}} \uparrow}$