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How to Find E°cell Values

  1. Nov 10, 2007 #1
    I needed to find the E°cell (always positive for a galvanic cell), based on the following (unbalanced) reaction:

    IO3-(aq) + Fe2+(aq) --> Fe3+(aq) + I2(s)

    So E°cell = Ecathode - Eanode

    I made the half reactions

    Fe3+ (aq) ---> Fe2+(aq)
    EØ = +0.77 V (i found this reaction in my table)

    IO3-(aq) ---> I2(s)
    balanced the 2nd one to

    12 H + 2IO3 + 5e- ---> I2 + 6H2O
    EØ = I cannot find this value in that table
    I FOUND IT ONLINE ON A WEBSITE TO BE EØ = +1.19V, HOW DID THEY GET THIS VALUE! Anyone know?
     
  2. jcsd
  3. Nov 11, 2007 #2
    2IO3-(aq) + 10Fe2+(aq) + 12H+ --> 10Fe3+(aq) + I2(s) + 6H2O;
    E° = +1.19 - 0.77 = +0.42V.

    About the reaction:

    (1) 2IO3-(aq) + 12 H+ + 10e- ---> I2(s) + 6H2O

    You can find its E° if you know the E° of other similar reactions from which you can get that one, for example:

    (2) IO3-(aq) + 6H+ +6e- --> I- + 3H2O;_______>E°(2) = +1.08V
    (3) I2(s) + 2e- --> 2I- ;____________________>E°(3) = +0.53V

    Reaction (1) is given by: 2*Reaction(2) - Reaction(3) and so:

    Delta G°(1) = 2*Delta G°(2) - Delta G°(3)

    Now you use the equation: Delta G° = -nF*Delta E° where n is the number of electrons of the reaction, so you have:

    -10F*Delta E°(1) = 2*[-6F*Delta E°(2)] - [-2F*Delta E°(3)]

    10*Delta E°(1) = 12*Delta E°(2) - 2*Delta E°(3)

    Delta E°(1) = (1/10)[12*Delta E°(2) - 2*Delta E°(3)] =

    = (1/5)[6*Delta E°(2) - Delta E°(3)] =

    = (1/5)[6*1.08 - 0.53] = +1.19V.
     
    Last edited: Nov 11, 2007
  4. Nov 15, 2007 #3
    wowi would never have that of that...:S


    how is the delta G of the first equation equal to the delta G if the second and third with those operations. Like how do we know to relate delta G?
     
  5. Nov 15, 2007 #4
    Do you agree on the fact that Reaction (1) is given by 2*Reaction(2) - Reaction(3) ?

    2*Reaction(2) = 2IO3-(aq) + 12H+ +12e- --> 2I- + 6H2O

    - Reaction(3) = 2I- --> I2(s) + 2e-

    Now you sum them:

    2IO3-(aq) + 12H+ +12e- + 2I- --> 2I- + 6H2O + I2(s) + 2e-

    you simplify canceling 2I- and 2e- on both members:

    2IO3-(aq) + 12H+ +10e- --> 6H2O + I2(s).


    About summing ΔG: if you have

    (1) A --> B you know that ΔG(1) = G(B) - G(A)

    (2) C --> D you know that ΔG(2) = G(D) - G(C)

    (3) A + C --> B + D you know that ΔG(3) = G(B + D) - G(A + C)

    But G(B + D) - G(A + C) = G(B) + G(D) - G(A) - G(C) =

    = G(B) - G(A) + G(D) - G(C) = ΔG(1) + ΔG(2)
     
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