The following are the answers for the given problem
Explanation:
a) we have the half reaction:
AgCl (s) + e- = Ag(s) + Cl-(aq)
Ag(s) + Cl-(aq) = AgCl (s) + e-
According to several sources, and textbooks, the less concentrated ion will be the in the cathode cell to promove the reduction reaction. In this case, we can say that Cl- = 0.0143 M is the electrode in the cathode cell,because is reducting from zero to 1, so, it's the cathode.
b) As we are having a two silver-silver electrode, both reactions have the same emf in each reaction so:
d) As oxidation is ocurring in the anode according to the half reaction above, we can say that Cl- is consumed, therefore it will decrease as the cell operates.
e) As reduction is ocurring in the cathode, according to the half reaction above, we can say that Cl- is produced, therefore it will increase as the cell operates.