Answer:
1)
In this case, the AgCl electrode is only cathode
2)
the standard EMF of this cell is 0
as Ag + Cl-(anode) + AgCl ==> Ag + Cl-(cathode) + AgCl . . .Eo cell = +0.00 V
3) 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
(0.0150 / 2.55) = +0.0657 V
4)
Cl- concentration will increase on cathode and it will decrease on the anode
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