Question #bf3a3

1 Answer
Sep 29, 2016

Answer:

#"Zn"_ ((s)) + "Cu"("NO"_ 3)_ (2(aq)) -> "Zn"("NO"_ 3) _ (2(aq)) + "Cu"_ ((s))#

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, #"Cu"#.

#"Zn"_ ((s)) + "Cu"("NO"_ 3)_ (2(aq)) -> "Zn"("NO"_ 3) _ (2(aq)) + "Cu"_ ((s))#

The nitrate anions, #"NO"_3^(-)#, are spectator ions here, which means that you can remove them to get the net ionic equation

#"Zn"_ ((s)) + "Cu"_ ((aq))^(2+) -> "Zn"_ ((aq))^(2+) + "Cu"_ ((s))#

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, #"Cu"^(2+)#, to copper metal while being oxidized to zinc cations, #"Zn"^(2+)#, in the process.

You have

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

#"Cu"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"e"^(-)))) -> "Cu"_ ((s))#
#color(white)(aaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#

#"Zn"_ ((s)) + "Cu"_ ((aq))^(2+) -> "Zn"_ ((aq))^(2+) + "Cu"_ ((s))#