# Question e2370

Apr 27, 2017

Once all the $\text{Ag"^"+""(aq)}$ has been plated out as $\text{Ag(s)}$, there is nothing left to react with the $\text{Cu(s)}$.

#### Explanation:

The half-reactions are:

$\textcolor{w h i t e}{m m m m m m m m m m m m m m m m m m m m m m m m} {E}^{\circ} \text{/V}$
2×["Ag"^"+""(aq)" + "e"^"–" → "Ag(s)"]; color(white)(mmmmmmmml)"+0.7994"
1×["Cu(s)" → "Cu"^"2+""(aq)" + "2e"^"-"]; color(white)(mmmmmmmm)"-0.337"
stackrel(————————————————————)(color(white)(mmll)"2Ag"^"+""(aq)" + "Cu(s)" → "2Ag"^"+""(s)" + "Cu"^"2+""(aq)"); stackrel(———)(0.4624)#

The $\text{Ag"^"+""(aq)}$ oxidizes the $\text{Cu(s)}$ to $\text{Cu"^"2+""(aq)}$ and is itself reduced to $\text{Ag(s)}$, which plates out.

Once all the $\text{Ag"^"+""(aq)}$ has been plated out as $\text{Ag(s)}$, there is nothing left to react with the $\text{Cu(s)}$, so the reaction stops.