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"
