Question #e2370

1 Answer
Ernest Z.
Apr 27, 2017

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.

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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.

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