1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electrochemical Cells (involving partial pressures and pH)

  1. Nov 4, 2014 #1
    1. The problem statement, all variables and given/known data
    A voltaic cell utilises the following reaction:

    4 Fe2+(aq) + O2(g) + 4 H+(aq) ---> 4 Fe3+(aq) + 2 H2O(l)

    What is the emf of this cell when [Fe2+] = 1.3 M, [Fe3+] = 0.010 M, PO2 = 0.51 bar and the pH of the solution in the cathode is 3.5.

    The standard reduction potentials for each half-reaction.

    Fe3+(aq)+ e- ---> Fe2+(aq) Eored = +0.77 V

    O2(g) + 4 H+(aq) + 4 e- ---> 2 H2O(l) Eored = +1.23 V

    2. Relevant equations
    ## E = E^o - \frac{RT}{nF}ln|Q| ##

    Q=[Products][Reactants]


    3. The attempt at a solution
    I'm not exactly sure how to incorporate the pH and Partial pressure of oxygen into this equation, but I had a go.

    ## \frac{P}{RT} = C ##

    ## \frac{51}{(8.314)(298)} = C ##

    ## {0.027M = C} ##

    C being the concentration of Oxygen:

    Then proceeded to find the concentration of hydrogen
    ## pH = -log_{10}|[H^+]| ##

    ## 3.5 = -log_{10}|[H^+]| ##

    ## 10^{-3.5} = [H^+] ##

    ## 0.00032M = [H^+] ##

    Then I found the half reactions for the redox reaction:

    ## 4Fe^{2+} \rightarrow 4Fe^{3+} + 4e^- ## ---- Oxidation (+0.77V)

    ## 4e^- + O_2 + 4H^+ \rightarrow 2H_2O ## ---- Reduction (+1.23V)

    Hence,
    ## E^o = E_{reduction} - E_{oxidation} ##

    ## E^o = (1.23) - (0.77) ##

    ## E^o = 0.46V ##

    Now to find Q (I think this is where my error could be, I'm not sure what to do with the water, but I figured the concentration may be 1?).

    ## Q = \frac{[products]}{[reactants]} ##

    ## Q = \frac{[Fe^{3+}]}{[H^+]^4[O_2][Fe^{2+}]^4} ##

    ## Q = \frac{(1*10^{-8})}{(0.00032)^4(0.027)(1.3)^4} ##

    ## Q = 1250 ## (rounded, obviously)

    Now using the equation:
    ## E = E^o - \frac{RT}{nF}ln|Q| ##

    ## E = 0.46 - \frac{(8.314)(298)}{(4)(96485)}ln|1250| ##

    ## E = 0.46 - 0.0457 ##

    ## E = 0.41V ##

    however, this is incorrect. The correct answer is:
    0.37V
     
  2. jcsd
  3. Nov 4, 2014 #2

    Borek

    User Avatar

    Staff: Mentor

    What is the Nernst equation for the formal potential of this half cell?
     
  4. Nov 4, 2014 #3
    Ok...

    ## E = E^o - \frac{RT}{nF} ln| Q | ##

    ## Q = \frac{[H_2O]}{[O_2][H^+]^4} ##

    ## Q = \frac{1^2}{(0.00032)^4(0.027)} ##

    ## ln|Q| = 35.8 ##

    so:

    ## E = (1.23) - \frac{(8.314)(298)}{(4)(96485)} (35.8) ##

    ## E = (1.23) - 0.230 = 1 ##

    Hmm, and If I did the same with the other half cell?

    I get ## E = 0.801 ##, but I digress...
     
    Last edited: Nov 4, 2014
  5. Nov 4, 2014 #4

    Borek

    User Avatar

    Staff: Mentor

    Q should use partial pressure of the gas, not the concentration.
     
  6. Nov 4, 2014 #5
    Righto, I converted it to a Kpa, I just has a feeling that Bar is not a the correct unit of measurement.

    ## Q = \frac{1}{(51)^4(0.027)} ##

    ## ln|Q| = -11.9 ##

    ## E = 1.23 - \frac{(8.314)(298)}{(4)(96485)}(-11.9) ##

    ## E = 1.306V ##

    Quick question, why do I use the partial pressure of Oxygen as opposed to concentration (## M ##), doesn't the ##Q## have to be dimensionless?
     
  7. Nov 5, 2014 #6

    Borek

    User Avatar

    Staff: Mentor

    To confuse you further, is Q for

    HA <-> H+ + A-

    dimensionless?

    Q should be constructed using not concentrations, but dimensionless activities. Activity of a substance - for a diluted solution - is approximately equal its concentration, so we typically omit the activity coefficients assuming it equals 1 - but in effect we also introduce dimensionality into the equation.
     
  8. Nov 5, 2014 #7
    I see, I always assumed it to be dimensionless (as a ratio of concentration), since it's analogous to equilibrium constant.

    I'm still not sure exactly sure how to calculate ##Q##
     
  9. Nov 5, 2014 #8

    Borek

    User Avatar

    Staff: Mentor

    Standard state of a gas is 1 bar, so just putting 0.51 as a partial pressure of oxygen should work.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Electrochemical Cells (involving partial pressures and pH)
  1. Electrochemical cell (Replies: 1)

  2. Electrochemical Cell (Replies: 1)

Loading...