close
    how to balance ionic equations

    Pb(s) + Ag(aq)   Pb2+(aq) + Ag(s)

    0  Views: 364 Answers: 1 Posted: 6 years ago

    1 Answer

    we must first learn about electrolytes.


    Pb2+ + 2 e- → Pb (s) ξo = -0.13 V
    Ag+ + 1 e- → Ag (s) ξo = 0.80 V

    What is the voltage, at 298 K, of this voltaic cell starting with the
    following non-standard concentrations:

    [Pb2+] (aq) = 0.08 M
    [Ag+] (aq) = 0.5 M




    Use the Nernst equation:

    ξ = ξo - (RT/nF) ln Q




    ξo=0.80-(-0.13)=0.93V

    first I balanced the equation:

    2(Ag+ + 1 e- → Ag (s))
    Pb(s)→ Pb2+ + 2 e-
    --------------------
    2Ag+ + Pb(s) --> Pb2+ + 2Ag(s)

    Q = [products]^p/[reactants]^r
    so Q = 0.08/0.5^2 = 0.32

    i used ξ = ξo - (RT/nF) ln Q
    ξ = 0.93V - ((8.314)(298K)/(2)(96500)) ln0.32 = 0.9446V

    I also used E=Eo-(0.05916/n)logQ
    E = 0.93V - (0.05916/2)log0.32= 0.9446.... hoping if you can double check this for me. thanks!



    Top contributors in Chemistry category

     
    ROMOS
    Answers: 66 / Questions: 0
    Karma: 3315
     
    country bumpkin
    Answers: 42 / Questions: 0
    Karma: 2580
     
    Colleen
    Answers: 112 / Questions: 0
    Karma: 2220
     
    Benthere
    Answers: 2 / Questions: 0
    Karma: 1440
    > Top contributors chart

    Unanswered Questions

    442233
    questions
    711202
    answers
    520178
    users