Answer:
2Ag(s) + Cu^2+(aq) ----------> 2Ag^+(aq) + Cu(s)
Explanation:
Ag(s)/Ag^+ (aq) is the anode as shown while Cu^2+(aq)/Cu^2(s) is the cathode.
E°cell= E°cathode -E°anode= 0.34 -0.80= -0.5V
The cell is not spontaneous as written because E°cell is negative. This implies that the electrodes of the cell must be interchanged to make the cell spontaneous.
-log (1×10^-12) is how you calculate the pOH which in this case is 12
