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Homework Help: [Electrochemistry] Reduction potential, keq and kd calculations

  1. Nov 26, 2012 #1
    1. The problem statement, all variables and given/known data
    A) Calculate the reduction potential for the reduction of [itex]E \cdot Fe^{3+} (aq)[/itex] by [itex] Fe^{2+}[/itex]
    B)What is the equilibrium expression, equilbrium constant, and [itex]K_{d}[/itex] for [itex]E \cdot Fe^{3+} (aq)[/itex]

    [itex] Fe^{3+} (aq) + e \rightarrow Fe^{2+}(aq) \hspace{15pt} E = 0.22 V[/itex]
    [itex]E \cdot Fe^{3+} + e \rightarrow E \cdot Fe^{2+} \hspace{15pt} E = -0.75 V [/itex]
    [itex]E \cdot Fe^{2+} (aq) \Leftrightarrow E + Fe^{2+} (aq) \hspace{15pt} K_{d}= 3.7 x 10^{-9} M[/itex]

    2. Relevant equations
    keq = [products]/[reactants]
    kd = [unbound]/[bound]

    3. The attempt at a solution
    A) I'm assuming when the question states "the reduction of [itex]E \cdot Fe^{3+} (aq)[/itex] by [itex] Fe^{2+}[/itex]" it means that the complex is the oxidant and the iron(II) is the reductant.

    Then I setup the following know that it is a reduction I want both my oxidant and my reductant on the same side (the left) and switched the signs of the reduction potentials accordingly.

    [itex]E \cdot Fe^{3+} + e \rightarrow E \cdot Fe^{2+} \hspace{15pt} E = -0.75 V [/itex]
    [itex] Fe^{2+}(aq) \rightarrow Fe^{3+} (aq) + e \hspace{15pt} E = -0.22 V[/itex]
    addition yields:
    [itex]E \cdot Fe^{3+} + Fe^{2+} \rightarrow E \cdot Fe^{2+} + Fe^{3+} \hspace{15pt} E = -0.97 V [/itex]

    B) the keq therefore should be [itex]\frac{[E][Fe^{3+}]}{[E \cdot Fe^{3+}][Fe^{2+}]}[/itex]

    but I dont know how to get Kd from there or if I even did part A right?

    Any help is appreciated
  2. jcsd
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