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Electrochemistry - Voltaic Cell

  1. Apr 3, 2015 #1
    • 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.
     
  2. jcsd
  3. Apr 9, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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