How do you determine the concentrations of reagents in a galvanic cell when the cell "dies"?

1 Answer
Apr 9, 2017

Answer:

#["Ag"^"+"] = 2.22 × 10^"-18"color(white)(l) "mol/L"; "Ni"^"2+" = "1.50 mol/L"#

Explanation:

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"#