The reactions are:
Anode: #color(white)(m)"1×"["Ni(s)" → "Ni"^"2+""(aq) (1 mol/L") + "2e"^"-"]#
Cathode: #"2×"["Ag"^"+""(aq) (1 mol/L") + "e"^"-" → "Ag(s)"]#
Cell: #"Ni(s)" + "2Ag"^"+""(aq)" ("1 mol/L") → "Ni"^"2+" "(aq)" ("1 mol/L") + "2Ag(s)"#
#E_text(cell)^@ = "0.25 V + 0.7994 V" = "1.049 V"#
As the cell runs down, #["Ni"^"2+"]# increases and #["Ag"^"+"]# decreases.
This continues until
#E_text(cell) = E_text(cell)^° - (RT)/(nF)lnQ = 1.049 V - (RT)/(nF)lnQ = 0#
#(RT)/(nF)lnQ = "1.049 V"#
#(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 298 color(red)(cancel(color(black)("K"))))/(2 × "96 485" color(red)(cancel(color(black)("J·V"^"-1"))))lnQ = 1.049 color(red)(cancel(color(black)("V")))#
#"0.012 84"lnQ = 1.049#
#lnQ = 1.049/"0.012 84" = 81.70#
#Q = e^"81.70" = 3.04 × 10^"35"#
Thus, when the cell is dead,
#Q =(["Ni"^"2+"])/(["Ag"^"+"]^2) = 3.04 × 10^35 #
Almost all the #"Ag"^"+"# ions (1 mol/L) will have disappeared.
Since the molar ratio of #"Ag"^"+":"Ni"^"2+" = 2:1#, the loss of 1 mol/L #"Ag"^"+"# will cause the formation of 0.5 mol/L #"Ni"^"2+"#.
# ["Ni"^"2+"]# will have increased to its maximum value of 1.5 mol/L.
∴ #1.5/( ["Ag"^"+"]^2) = 3.04 × 10^35#
#["Ag"^"+"]^2 = 1.5/(3.04 × 10^35) = 4.93 × 10^"-36"#
#["Ag"^"+"] = 2.22 × 10^"-18"color(white)(l) "mol/L"#
