# Question 60418

Nov 3, 2015

$Z {n}^{\circ}$ + $2 N a O H$ + $2 {H}_{2} O$ $\rightarrow$ $N {a}_{2} Z n {\left(O H\right)}_{4}$ + ${H}_{2}$ (redox reaction)

#### Explanation:

Based on the metal activity series, zinc is not active enough to replace the $N a$ atom in the solution. So there should be no visible reaction.

$Z {n}^{\circ}$ + $2 N a O H$ $\ne$ $Z n {\left(O H\right)}_{2}$ + $2 N a$

But laboratory results tells us otherwise because zinc (along with aluminum, copper, tin, lead, and beryllium) is considered as amphoteric, a molecule which reacts to bases and acids alike.

Also, the reaction is not an acid-base, as one might expect but more of an oxidation-reduction reaction with the following half-equations:

Oxidation: $Z n$ ($s$) $\rightarrow$ $Z {n}^{\text{2+}}$ ($a q$) + 2e^–#

Reduction: $Z n {\left(O H\right)}_{4}^{\text{2-}}$ ($a q$) + $2 {e}^{-}$$\rightarrow$ $Z n$ ($s$) + $4 O {H}^{-}$($a q$)