[Electrochemistry] Reduction potential, keq and kd calculations

[HOMO]

Homework Statement

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

Homework Equations

keq = [products]/[reactants]
kd = [unbound]/[bound]

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.

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

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

but I don't know how to get Kd from there or if I even did part A right?

Any help is appreciated

 

[nate808]

.

Hello there!

For part A, you are correct in assuming that the complex is the oxidant and the iron(II) is the reductant. However, when adding the two half-reactions, you should flip the second reaction and its potential, not just switch the sign. This is because the half-reactions are written in the opposite direction to show the flow of electrons. So the correct addition would be:

E \cdot Fe^{3+} + e \rightarrow E \cdot Fe^{2+} \hspace{15pt} E = -0.75 V
Fe^{3+} (aq) + e \rightarrow Fe^{2+} (aq) \hspace{15pt} E = 0.22 V
addition yields:
E \cdot Fe^{3+} + Fe^{3+} \rightarrow E \cdot Fe^{2+} + Fe^{2+} \hspace{15pt} E = -0.53 V

For part B, the equilibrium expression would be:
keq = \frac{[E \cdot Fe^{2+}][Fe^{2+}]}{[E \cdot Fe^{3+}][Fe^{3+}]}

The equilibrium constant (K_{eq}) would be equal to the keq value, and the dissociation constant (K_d) would be the inverse of the equilibrium constant (K_d = \frac{1}{K_{eq}}).

I hope this helps! Let me know if you have any further questions.