Question

Hayden McNeil, LLC d. Consider the Nernst equation (or equilibrium concepts) and predict whether the cell potential would be higher or lower than the standard cell potential for (i) a cell with 0.10 M Agt and 5.0 M Ni2+ (ii) a cell with 5.0 M Ag and 0.10 M Ni2+ e. If this cell operates for one lab period (170 minutes) with a current of 0.250 A, calcu- late the mass of silver that will react. Show your work.

Answer 1

Assuming
d) i) We know, E = E°(cathode) - E°(anode) and from the reactivity series: E°(Ni2+/Ni) = -0.25 V and E°(Ag+/Ag) = 0.8V Half cell with higher E' should be taken as the cathode. So here, cathode = Ag+/Ag and anode is Ni2+/Ni Ece = 0.8-(-0.25) = 1.05 V Overall cell reaction is: Ni + 2Ag+ ----> 2Ag + Ni 2+ hence number of electrons involved are n= 2 Nernst equation: Ecell = E°(cell) - 0.059/2 log([Ni2+]/[Ag +)^2) = 1.05 - (0.059/2)*log(5/(0.1^2)) = 0.97 V So the cell potential is 0.97 V. ii) in this case [Ag+1=5M and [Ni2+] = 0.1M Ecell = E°(cell) - 0.059/2 log([Ni2+] / [Ag+1^2) = 1.05 - (0.059/2)*log(0.1/(5^2)) = 1.12 V So the cell potential is 1.12 V.

e) 1 = 0.250A means 0.250 C/s total charge involved Q = 0.250C/s*(170*60s) = 2550 C Number of moles of Ag involved = Q/nF n = 2, F = 96485 mol-1 Number of moles of Ag involved = 2550 / (2*96485) = 0.01322 moles Atomic mass of Ag = 107.87 g/ mol Mass of silver involved = 0.01322*107.87 = 1.43 g