Question #79098

Oct 10, 2016

More reactive metals lose electrons easier and become ions in solution, whereas less reactive metals accept electrons easier in their ion form, becoming solid.

Explanation:

In displacement reactions, the solid metal is oxidised and the metal ions in solution are reduced. If these two terms a new to you, oxidation is the loss of electrons and reduction is the gain of electrons. Most people remember this with the help of the Oilrig pneumonic. Oxidation Is Loss Reduction Is Gain- OILRIG

For example, placing $Z n$ metal in a $A g N {O}_{3}$ solution,

$Z {n}_{\left(s\right)} + A g N {O}_{3 \left(a q\right)} \to Z n {\left(N {O}_{3}\right)}_{2 \left(a q\right)} + A g \left(s\right)$

$Z n$ metal displaces $A g$ ions out of solution. What is actually happening is a transfer of electrons, where $Z n$ loses electron and $A g$ recieves electrons. $Z n$ is thus said to be oxidised and $A g$ is said to be reduced. This can be represented in two half equations.

Oxidation half equation: $Z {n}_{\left(s\right)} \to Z {n}^{2 +} + 2 {e}^{-}$
Reduction half equation: $A {g}^{+} + {e}^{-} \to A {g}_{\left(s\right)}$

More reactive metals are oxidised easier than less reactive metals. Since $Z n$ is more reactive than $A g$, it can lose its electrons easier. As a result, when placed in the $A g N {O}_{3}$ solution, it tends to lose its electrons to the $A g$ ions, forming $Z n$ ions and solid $A g$.