# Question #9390e

May 13, 2015

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

The net ionic form of this reaction would be:
$Z n$(s)$+ C {u}^{2 +}$(aq)$\to Z {n}^{2 +}$(aq)$+ C u$(s)

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

The zinc is being oxidized and replaces the $C {u}^{2 +}$(aq) in solution:
$Z n$(s)$\to Z {n}^{2 +}$(aq)$+ 2 {e}^{-}$

The copper is reduced and replaces the $Z n$(s) on the zinc rod:
$C {u}^{2 +}$(aq)$+ 2 {e}^{-} \to C u$(s)

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