Question

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−]=1.49×10⁻² M and [Cl−]= 3.00 M , respectively.

a. Which electrode is the cathode of the cell?

b. What is the standard emf of the cell?

c. What is the cell emf for the concentrations given?

d. For the anode compartment, predict whether [Cl−] will increase, decrease, or stay the same as the cell operates.

e. For the cathode compartment, predict whether [Cl−] will increase, decrease, or stay the same as the cell operates.

Answer 1

Anode: Ag + Cl- ==> AgCl + e-
Cathode: AgCl + e- ==> Ag + Cl-
======================================...
Ag + Cl-(anode) + AgCl ==> Ag + Cl-(cathode) + AgCl . . .

a). electrode of lesser concentration of Cl^- will be cathode

b). Eo cell = +0.00 V

c) 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.
Qcell = [Cl-] cathode / [Cl-] anode
E cell = Eo cell - 0.059/n log Q = Eo cell - 0.059/1 log [Cathode]/ [anode]

= 0 - 0.059 log [0.0149]/[3.00] = -0.059 log[0.0049]]

= -0.059*-2.3 = 0.1352

How to we came to know that the 3.00 M Cl^- was in the anode and that the 0.0149 M Cl^- was in the cathode?
If they were reversed, then Q = log Q = 2.30 and the overall voltage would be negative. Voltage can not be negative for the voltaic cell.

d) At anode oxidation will occur, it means Cl^- will oxides to Cl so [Cl^-] will decrease at anode

e) For cathode reduction occur, on cathode Cl will reduce to Cl^- hence concentration of [Cl^-] will increase