Question

Question #bf3a3

Answer 1

`"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))`