A voltaic electrochemical cell consists of a silver electrode in contact with 200 mL of 0.1 M AgNO3 and a magnesium electrode in contact with 300 mL of a 0.1 M Mg(NO3)2 solution. We have the following data: Ag+ + e- --> Ag (E° = 0.8 V) Mg2+ + 2e- --> Mg (E° = -2.372 v) a) Calculate the value of E for the cell at 25°C b) A current is drawn from the cell until 1.2 g of silver has been deposited at the silver electrode. Calculate the value of E for the cell at this stage. c) Determine E for the cell once the cell has reached equilibrium. a a) 3.142 V; b) 2.958 V; c) zero b a) 3.172 V; b) 3.122 V; c) 3.122 V c a) 3.172 V; b) 3.122 V; c) 3.172 V d a) 3.142 V; b) 2.958 V; c) 2.958 V Attempt at solution: Overall reaction: 2Ag+ + Mg(s) -> 2 Ag(s) + Mg2+ E0cell = 3.172 Q = [Mg2+]/[Ag+]^2 solid phase components not included For part a) Nerst equation with E0cell = 3.172 V and finding Q with the given concentrations above yields E25C = 3.142V So b and c are eliminated I'm not quite sure how to proceed from there. Is it possible for Ecell at equilibrium to be 0? If not, A is eliminated as well. Thank you for the help, and sorry if I've broken any etiquette. It's my first time posting here.