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

1. Nov 26, 2012

HOMO

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

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

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 $E \cdot Fe^{3+} (aq)$ by $Fe^{2+}$" 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.

$E \cdot Fe^{3+} + e \rightarrow E \cdot Fe^{2+} \hspace{15pt} E = -0.75 V$
$Fe^{2+}(aq) \rightarrow Fe^{3+} (aq) + e \hspace{15pt} E = -0.22 V$
$E \cdot Fe^{3+} + Fe^{2+} \rightarrow E \cdot Fe^{2+} + Fe^{3+} \hspace{15pt} E = -0.97 V$
B) the keq therefore should be $\frac{[E][Fe^{3+}]}{[E \cdot Fe^{3+}][Fe^{2+}]}$