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 
$$\begin{align}
\ce{Cu^{3+} + 2I- &amp;-&gt; Cu+ +I2}\\
\ce{Cu^{2+} +I- &amp;-&gt; Cu+ + \frac{1}{2}I2}
\end{align}$$

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.
Your fundamental assumption is that all of the oxygen is $\ce{O^2-}$. In oxide ceramics, however, it is possible for oxygen atoms to dissolve into the matrix. zirconia,  titania, and bismuth oxide can all easily accumulate non-stoichiometric oxygen. In fact this ability to "dissolve" oxygen into a solid matrix is how fuel cells are able to operate.

You are correct that the yttrium is $\ce{Y^{3+}}$, barium is $\ce{Ba^{2+}}$, and copper is $\ce{Cu^{2+}}$, but in YBCO the oxygen can form $\ce{O^{(2x-2)-}_x}$ sites as part of $\ce{Cu2O_x}$ ($1\leq x \leq 4$) clusters. Note: for $x = 2\ \ce{O^{2-}2}$ is not inherently a peroxide ion, you need cooper pairs of electrons to have superconductivity (see image below). Thus another way to present the structure of YBCO may be as $\ce{YBa2Cu3O5(O2)}$ or rather $\ce{YBa2Cu3O_{5+x}(O2)_{1-x}}$ or even $\ce{YBa2Cu3O_{3+3x}\nu^0_x(O4)_{1-x}}$ $(0 \leq x \leq 1)$ as the ceramic can be non-stoichiometric and produce oxygen vacancies which affect the superconductivity.