# Question f3a1e

May 12, 2016

${\text{Cu"_ ((s)) + 2"Ag"_ ((aq))^(+) -> "Cu"_ ((aq))^(2+) + 2"Ag}}_{\left(s\right)}$

#### Explanation:

Start by assigning oxidation numbers to the chemical species involved in the reaction

${\stackrel{\textcolor{b l u e}{+ 1}}{\text{Ag"^(+))_ ((aq)) + stackrel(color(blue)(0))("Cu")_ ((s)) -> stackrel(color(blue)(+2))("Cu")""^(2+) ""_ ((aq)) + stackrel(color(blue)(0))("Ag}}}_{\left(s\right)}$

Notice that the oxidation number of silver went from $\textcolor{b l u e}{+ 1}$ on the reactants' side, to $\textcolor{b l u e}{0}$ on the products' side; this tells you that silver is being reduced, since its oxidation number is decreasing.

On the other hand, copper's oxidation number went from $\textcolor{b l u e}{0}$ on the reactants' side, to $\textcolor{b l u e}{+ 2}$ on the products' side; this tells you that copper is being oxidized, since its oxidation number is increasing.

The oxidation half-reaction will look like this

stackrel(color(blue)(0))("Cu")_ ((s)) -> stackrel(color(blue)(+2))("Cu")""^(2+) ""_ ((aq)) + 2"e"^(-)

Here each atom of copper is losing two electrons to form the copper(II) cation, ${\text{Cu}}^{2 +}$.

The reduction half-reaction will look like this

${\stackrel{\textcolor{b l u e}{+ 1}}{\text{Ag"^(+))_ ((aq)) + "e"^(-) -> stackrel(color(blue)(0))("Ag}}}_{\left(s\right)}$

Here each silver(I) cation is gaining one electron to form a silver atom.

Now, in any redox reaction, the number of electrons gained in the reduction half-reaction must be equal to the number of electrons lost in the oxidation half-reaction.

In your case, you need to multiply the reduction half-reaction by $2$ and add the two half-reactions to get

$\left\{\left(\textcolor{w h i t e}{a a a a a a a} {\stackrel{\textcolor{b l u e}{0}}{\text{Cu")_ ((s)) -> stackrel(color(blue)(+2))("Cu")""^(2+) ""_ ((aq)) + 2"e"^(-)), (stackrel(color(blue)(+1))("Ag"^(+))_ ((aq)) + "e"^(-) -> stackrel(color(blue)(0))("Ag}}}_{\left(s\right)} \textcolor{w h i t e}{a a a a a a a a a a} | \times 2\right)\right.$
color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/(color(white)(aaaaaaaaaaaaaa)#

${\text{Cu"_ ((s)) + 2"Ag"_ ((aq))^(+) + color(red)(cancel(color(black)(2"e"^(-)))) -> "Cu"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"e"^(-)))) + 2"Ag}}_{\left(s\right)}$

The balanced chemical equation will thus be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{Cu"_ ((s)) + 2"Ag"_ ((aq))^(+) -> "Cu"_ ((aq))^(2+) + 2"Ag}}_{\left(s\right)}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In this reaction, copper acts as a reducing agent because it reduces silver(I) cations to silver metal. Likewise, the silver(I) cations act as an oxidizing agent because they oxidize copper metal to copper(II) cations. 