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Nernst Equation

  1. Apr 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Use the Nernst equation and data from Appendix D in the textbook to calculate E cell for each of the following cells.

    Mg(s)|Mg+2 (0.012 M) || [Al(OH)4]- (0.25M), OH- (0.048M)|Al(s)


    2. Relevant equations

    Ecell = Ecell(standard) - 0.0591/n * log Q


    3. The attempt at a solution

    Half reaction of Mg: -2.356
    Half reaction of [Al(OH)4]-: -2.310

    Ecell (standard) = -2.310-(-2.356) = .046 V

    Equation written out in spontaneous form would be...

    3Mg + 2[Al(OH)4]- -> 3Mg+2 + 8OH- + 2Al

    Then solving for Q... which is [product]/[reactant] would be...

    Q = ([.048]^8 * [.012]^3)/[.25]^2 = 7.79E-16

    Number of moles of electrons transferred is 2.

    Plug everything in...

    Ecell = .046 - (.0591/2)*log (7.79E-16)

    Which gives me

    Ecell = 0.492 V

    However, the program I'm entering this into says it's incorrect.

    Any pointers on where I went wrong?

    Thanks in advance.
     
  2. jcsd
  3. Apr 14, 2009 #2

    symbolipoint

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    Homework Helper
    Education Advisor
    Gold Member

    Recheck how many electrons transfer for the Aluminim.
     
  4. Apr 14, 2009 #3
    Hrm, +3 -> 0. So 3 electrons being transferred, unless you count the coefficient as well.

    Would it be 6 electrons transferred?
     
  5. Apr 14, 2009 #4

    Borek

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    Staff: Mentor

    Calculate each half cell separately, then combine them.
     
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