Question

Question #f3a1e

Answer 1

`"Cu"_ ((s)) + 2"Ag"_ ((aq))^(+) -> "Cu"_ ((aq))^(2+) + 2"Ag"_((s))`

Explanation:

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

`stackrel(color(blue)(+1))("Ag"^(+))_ ((aq)) + stackrel(color(blue)(0))("Cu")_ ((s)) -> stackrel(color(blue)(+2))("Cu")""^(2+) ""_ ((aq)) + stackrel(color(blue)(0))("Ag")_ ((s))`

Notice that the oxidation number of silver went from `color(blue)(+1)` on the reactants' side, to `color(blue)(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 `color(blue)(0)` on the reactants' side, to `color(blue)(+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, `"Cu"^(2+)`.

The reduction half-reaction will look like this

`stackrel(color(blue)(+1))("Ag"^(+))_ ((aq)) + "e"^(-) -> stackrel(color(blue)(0))("Ag")_ ((s))`

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

`{(color(white)(aaaaaaa)stackrel(color(blue)(0))("Cu")_ ((s)) -> stackrel(color(blue)(+2))("Cu")""^(2+) ""_ ((aq)) + 2"e"^(-)), (stackrel(color(blue)(+1))("Ag"^(+))_ ((aq)) + "e"^(-) -> stackrel(color(blue)(0))("Ag")_ ((s)) color(white)(aaaaaaaaaa)| xx 2) :}`
`color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/(color(white)(aaaaaaaaaaaaaa)`

`"Cu"_ ((s)) + 2"Ag"_ ((aq))^(+) + color(red)(cancel(color(black)(2"e"^(-)))) -> "Cu"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"e"^(-)))) + 2"Ag"_((s))`

The balanced chemical equation will thus be

`color(green)(|bar(ul(color(white)(a/a)color(black)("Cu"_ ((s)) + 2"Ag"_ ((aq))^(+) -> "Cu"_ ((aq))^(2+) + 2"Ag"_((s)))color(white)(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.