Question #f3a1e

1 Answer
May 12, 2016

Answer:

#"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.

https://peda.net/oppimateriaalit/e-oppi/ylakoulu/kemia1/oppikirja/kuvat/kuvagalleria-v/2mrjss/nimet%C3%B6n-cf05