# Question bf3a3

Sep 29, 2016

${\text{Zn"_ ((s)) + "Cu"("NO"_ 3)_ (2(aq)) -> "Zn"("NO"_ 3) _ (2(aq)) + "Cu}}_{\left(s\right)}$

#### Explanation:

When you add a strip of zinc metal to a solution of copper(II) nitrate, a single replacement reaction takes place.

Zinc will displace copper from the solution. This will result in the formation of aqueous zinc nitrate, "Zn"("NO"_3)_2, and copper metal, $\text{Cu}$.

${\text{Zn"_ ((s)) + "Cu"("NO"_ 3)_ (2(aq)) -> "Zn"("NO"_ 3) _ (2(aq)) + "Cu}}_{\left(s\right)}$

The nitrate anions, ${\text{NO}}_{3}^{-}$, are spectator ions here, which means that you can remove them to get the net ionic equation

${\text{Zn"_ ((s)) + "Cu"_ ((aq))^(2+) -> "Zn"_ ((aq))^(2+) + "Cu}}_{\left(s\right)}$

Another way to think about this reaction involves recognizing the fact that electrons are being transferred here, which implies that this is also a redox reaction.

Zinc metal reduces copper cations, ${\text{Cu}}^{2 +}$, to copper metal while being oxidized to zinc cations, ${\text{Zn}}^{2 +}$, in the process.

You have

"Zn"_ ((s)) -> "Zn"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"e"^(-))))#

${\text{Cu"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"e"^(-)))) -> "Cu}}_{\left(s\right)}$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

${\text{Zn"_ ((s)) + "Cu"_ ((aq))^(2+) -> "Zn"_ ((aq))^(2+) + "Cu}}_{\left(s\right)}$