Question #60418

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Question #60418

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Answer 1 · 2015-11-03T12:07:15

$Z n^{\circ}$ $Zn^@$ + $2 N a O H$ $2NaOH$ + $2 H_{2} O$ $2H_2O$ $\to$ $rarr$ $N a_{2} Z n {(O H)}_{4}$ $Na_2Zn(OH)_4$ + $H_{2}$ $H_2$ (redox reaction)

Explanation:

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

$Z n^{\circ}$ $Zn^@$ + $2 N a O H$ $2NaOH$ $\neq$ $!=$ $Z n {(O H)}_{2}$ $Zn(OH)_2$ + $2 N a$ $2Na$
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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: $Z n$ $Zn$ ( $s$ $s$ ) $\to$ $rarr$ $Z n^{2+}$ $Zn^"2+"$ ( $a q$ $aq$ ) + $2e^–$

Reduction: $Zn(OH)_4^"2-"$ ( $aq$ ) + $2e^-$ $rarr$ $Zn$ ( $s$ ) + $4OH^-$ ( $aq$ )

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Question #60418

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