Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electrode potential ?

  1. Nov 14, 2011 #1
    I don't know why, but I just can't figure this out.

    "Calculate the electrode potential for a silver filled electrode in a solution of 0.00550 M AgNO3 solution."

    I got that

    Ag+ + e- --> Ag(s) had an E* = .80V

    I got that the answer is E=.80 - .0592 log (1/.00550M)= .6655 V

    But why do you put 1/.00550M in the parenthesis?

    My first thought was that AgNO3 + e- --> Ag(s) + NO3-
    and that the concentreation of AgNO3 is the same as NO3.

    so that it would be log(1) , which is wrong.

    I just don't get the concept I guess.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 14, 2011 #2


    User Avatar

    Staff: Mentor

    It all depends on how you write the Nernst equation. There are two conventions:

    [tex]E = E_0 + \frac {RT}{nF} ln \frac{[Ox]}{[Red]}[/tex]


    [tex]E = E_0 - \frac {RT}{nF} ln \frac{[Red]}{[Ox]}[/tex]

    (personally I prefer the first one, but that's just because I was taught this way).

    Note that both equations are equivalent thanks to the sign change.

    Oxidized form of silver is Ag+, reduced form is a metallic Ag - with activity equal to 1. So you can write the equation either as

    [tex]E = 0.88 + 0.059\ log([Ag^+])[/tex]


    [tex]E = 0.88 - 0.059\ log\big(\frac 1 {[Ag^+]}\big)[/tex]

    Concentration of Ag+ is 0.00550 M, concentration of NO3- doesn't matter.
  4. Nov 14, 2011 #3
    oh! okay that makes sense. I don't know why but I always confuse myself with these electrode problems.

    Thank you!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook