Calculate the potential at the equivalence point

[il postino]

Homework Statement

A KI solution is titrated using a K2Cr2O7 standard at a pH of 4. Calculate the potential at the equivalence point

Relevant Equations

Nernst equation

Data
$E^{0}_{I_2/I-}= 536mV$
$E^{0}_{Cr_2O_7^{2-}/Cr^{3+}}= 1330mV$

$I_{2} + 2e- \Leftrightarrow 2I^{-}$
$Cr_2O_7^{2-} + 14H^+ + 6e^- \Leftrightarrow 2Cr^{3+} + 7H_2O$

Total Reaction

$6I^{-} + Cr_2O_7^{2-} + 14H^+ \Leftrightarrow 3I_{2} + 2Cr^{3+} + 7H_2O$

$(E_{eq}=536-\frac{59.2}{2}log(I^{-})^{2})*2$
$(E_{eq}=1330-\frac{59.2}{6}log(\frac{(Cr^{3+})^{2}}{(Cr_2O_7^{2-})(H^+)^{14}})*6$

Thus,

$(8*E_{eq}=9052-59.2log(\frac{(I^{-})^{2}(Cr^{3+})^{2}}{(Cr_2O_7^{2-})(H^+)^{14}}$

I know the concentration of H⁺ es $1.10^{-4}M$

According to the reaction, at the equivalence point:

$(I^{-})^{2}=6*(Cr_2O_7^{2-})$

But replacement in expression I can't cancel.
The expression is based on $(Cr_2O_7^{2-})$

I don't know how to continue.
Can you help me?
Thank you!

 

[nate808]

Hello,

First of all, I would like to commend you on your thorough analysis and calculation of the equilibrium potential for the two reactions. Your approach is correct and you have a good understanding of the principles involved.

To continue, we need to use the Nernst equation, which relates the equilibrium potential to the concentrations of the reactants and products. It is given by:

$E_{eq} = E^{0} - \frac{RT}{nF}ln(Q)$

Where E_{eq} is the equilibrium potential, E^{0} is the standard potential, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred in the reaction, F is Faraday's constant, and Q is the reaction quotient.

In your case, n = 2 for the first reaction and n = 6 for the second reaction. Therefore, using the values given in the forum post, we can write the following equations:

$E_{eq,1} = 536mV - \frac{RT}{2F}ln(I^{-})$
$E_{eq,2} = 1330mV - \frac{RT}{6F}ln(\frac{(Cr^{3+})^{2}}{(Cr_2O_7^{2-})(H^+)^{14}})$

At the equivalence point, we know that the concentrations of I^{-} and (Cr_2O_7^{2-}) are equal, as you have correctly stated. Therefore, we can write the following:

$E_{eq,1} = E_{eq,2}$
$536mV - \frac{RT}{2F}ln(I^{-}) = 1330mV - \frac{RT}{6F}ln(\frac{(Cr^{3+})^{2}}{(I^{-})^{2}(H^+)^{14}})$

Simplifying this equation, we get:

$ln(I^{-}) = \frac{3}{4}ln(\frac{(Cr^{3+})^{2}}{(H^+)^{14}}) - \frac{794}{RT}$

Now, we can use the given concentration of H+ to calculate the concentration of I^{-} at the equivalence point. From the reaction equation, we know that the concentration of I^{-} is equal to the concentration of (Cr_2O_7^{2-