Question #e2370

1 Answer
Apr 27, 2017

Answer:

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

Explanation:

The half-reactions are:

#color(white)(mmmmmmmmmmmmmmmmmmmmmmmm)E^@"/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 #"Ag"^"+""(aq)"# oxidizes the #"Cu(s)"# to #"Cu"^"2+""(aq)"# and is itself reduced to #"Ag(s)"#, which plates out.

www.fccj.us

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