Question

A galvanic cell is constructed with a silver-silver chloride electrode, and a nickel strip immersed in a beaker containing 5.56 x 10-2 M solution of NiC12. Determine the balanced cell reaction and calculate the potential of the cell. Enter in volts. (assume a temperature of 25°C) The answer is not .97

Answer 1

1) Ni(s) = Ni^2+(aq) + 2e^-

2) 2Ag^+(aq) + 2e^- = 2Ag(s)

E^o(cell) = E^o(cathode) - E^o(anode) = 0.80 V - (-0.23 V) = 0.80 V + 0.23 V = 1.03 V

Solve for [Ni^2+]:

Ratio of NiCl2:Ni^2+ is 1:1, so [Ni^2+] = 3.97 x 10^-2 M

Solve for [Ag^+]:

Ksp = [Ag^+] [Cl^-] ... let x = [Ag^+] = [Cl^-]

(1.77 x 10^-10) = x^2 ... solve for x

x = square root (1.77 x 10^-10) = 1.33 x 10^-5 M

Solve for Q:

Q = [Ni^2+] / [Ag^+]^2 = (3.97 x 10^-2) / (1.33 x 10^-5)^2 = 2.24 x 10^8

Solve for E(cell):

E(cell) = E^o(cell) - (0.0592 / n) log Q

E(cell) = 1.03 - (0.0592 / 2 moles of e^-) log (2.24 x 10^8)

E(cell) = 0.783 V