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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.

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2 Answers 2

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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.

enter image description here

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    $\begingroup$ @OscarLanzi yes I suppose you could have O− radicals, especially at room temperature but this is not likely to provide cooper pairs below $T_C$. My point was that oxygen does not have to be 2- and thus copper does not have to be 3+ to balance. I have edited things. $\endgroup$
    – A.K.
    Aug 11, 2018 at 3:00
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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.

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  • $\begingroup$ For the purpose of this question, I'm just providing the equations from the experiment and the textbook which says, "Copper(I) is not an oxidizing equivalent: It cannot oxidize iodide to iodine." As for the math that you provided, what was the reasoning for multiplying II by 2, III by 3, and the avg OS by 3? $\endgroup$
    – John Snow
    Feb 21, 2015 at 19:19
  • $\begingroup$ It won't oxidize iodide, but it will make a precipitate with it. You multiple 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. $\endgroup$
    – EJC
    Feb 21, 2015 at 19:58
  • $\begingroup$ if you edit your question to include that information, I can give you the upvote. $\endgroup$
    – John Snow
    Feb 23, 2015 at 20:13
  • $\begingroup$ Also, just checking here, I wound up doing the math and found the formula to be $\frac{2x+3y}{x+y}=2.3$. I had trouble solving your equation $\endgroup$
    – John Snow
    Feb 23, 2015 at 21:05
  • $\begingroup$ Why replacing a number with variables? Express one variable from the first equation and replace it with the expression you got in the second equation. $\endgroup$
    – EJC
    Feb 23, 2015 at 21:08

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