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A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is based on the following half-reaction:
AgCl(s)+e−→Ag(s)+Cl−(aq).
The two cell compartments have [Cl−]= 1.60×10−2M and [Cl−]= 3.00M , respectively.
a) What is the standard emf of the cell?
b) What is the cell emf for the concentration given?
Can someone explain to me how to do this
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Answers (1)
The overall cell reaction is
Anode: Ag + Cl- ==> AgCl + e- . . .Eo = -0.22 V
Cathode: AgCl + e- ==> Ag + Cl- . .Eo = +0.22 V
=========================================================
Ag + Cl-(anode) + AgCl ==> Ag + Cl-(cathode) + AgCl . . .Eo cell = +0.00 V
A small amount of voltage is produced because the Cl- concentrations are different in the two cells; this voltage can be calculated using the Nernst equation.
E cell = Eo cell - 0.059/n log Q = 0.00 V - 0.059/1 log ([Cl- cathode] / [Cl- anode])
= 0.00 V - 0.0295 log (3.00 / 0.0160) = 0.00 V - (0.0295) log (0.0160 / 3.00) = +0.067 V
How did I know that the 3.00 M Cl- was in the anode, and that the 0.0160 M Cl- was in the cathode?
If they were reversed, then Q = log Q = 2.27 and the overall voltage would be negative. The problem stated that the cell was voltaic (produces a positive voltage).