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Answer 1 · 2015-05-13T12:22:47

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.