Question

Question #9390e

Answer 1

You would see an oxidation reduction reaction occur in the beaker.

The net ionic form of this reaction would be:
`Zn`(s)`+Cu^(2+)`(aq)`->Zn^(2+)`(aq)`+Cu`(s)

Let's break this equation into half-reactions to explain what is happening.

The zinc is being oxidized and replaces the `Cu^(2+)`(aq) in solution:
`Zn`(s)`->Zn^(2+)`(aq)`+2e^-`

The copper is reduced and replaces the `Zn`(s) on the zinc rod:
`Cu^(2+)`(aq)`+2e^(-) ->Cu`(s)

Notice : The 2 electrons lost in the oxidation of the zinc metal allow for the copper to be reduced.