inorganic chemistry - How to calculate the oxidation state and the number of equivalents oxidized with YBa₂Cu₃O₇

I've hit some trouble with these calculations and need help sorting out confusion.

In the lab we synthesized $\ce{YBa2Cu3O7}$, per the lab manual $\ce{Y, Ba, O}$ have the usual charges of +3, +2, and -2 respectively. Spectroscopic studies show that no copper (III) centers are present in the material, but rather there are missing electrons from the copper-oxygen bonds. For the purpose of the titration though, the missing electrons are thought of as coming from the copper center and we're assuming there are copper III centers present.

We used idodine to tritrate, and there is the assumption that copper is doing the oxidizing. The text states, "Copper(I) is not an oxidizing equivalent: It cannot oxidize iodide to iodine."

The relevant equations are $$\begin{align} \ce{Cu^{3+} + 2I- &-> Cu+ +I2}\\ \ce{Cu^{2+} +I- &-> Cu+ + \frac{1}{2}I2} \end{align}$$

I've calculated the theoretical OS to be $$\ce{Y} (3^+ )+\ce{Ba} (2^+ * 2)+\ce{O} (2^-*7)= \frac{7}{3 \ \ce{Cu}}\ce{Cu}^{2.3+}$$

We have to calulate the theoretical number of equivalents of $\ce{I-}$ oxidized by 1 g of the sample.

Would this be correct for the oxidation state? If so would that mean that the total equivalents of Iodine are an addition of $\ce{Cu(II)}$ and $\ce{Cu(III)}$ (the two equations)? Also I'm feeling brain dead, because I can't figure out the fraction of $\ce{Cu(II)}$ and $\ce{Cu(III)}$ that would average to 2.3.

Answer

You forgot one relevant equation: $\ce{Cu+ + I- -> CuI}$ which uses up some iodide as copper precipitates with it.

Getting the fractions is easy. You know that they must add to 3 and that that charge balance must apply. I marked the fractions as $x$ for copper(II) and $y$ for copper (III) and set up this system of equations which is easy to solve: $$\begin{align} x+y&=3\\ 2x+3y&=3\times\text{average oxidation state} \end{align}$$

You multiply the fraction with the charge of the species and the average oxidation state by three because in one mole of the compound you have three moles of copper, so that the equation represents a charge balance.