# Electrochemistry - Voltaic Cell

Please post this type of questions in HW section using the template.
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.