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.