In the 1980’s a class of ceramic materials was discovered that exhibits superconductivity at the unu — Physical Chemistry Chemistry Question
Ceramic Materials (YBCO)
In the 1980’s a class of ceramic materials was discovered that exhibits superconductivity at the unusually high temperature of 90 K. One such material contains yttrium, barium, copper and oxygen and is called “YBCO”. It has a nominal composition of YBa2Cu3O7, but its actual composition is variable according to the formula YBa2Cu3O7-δ (0 < δ < 0.5).
One unit cell of the idealized crystal structure of YBCO is shown below. Identify which circles correspond to which elements in the structure.
a
b
c
The true structure is actually orthorhombic (a ≠ b ≠ c), but it is approximately tetragonal, with a ≈ b ≈ (c/3).
Model Answer
4.1
a
b
c
= Cu
= O
= Ba
= Y
A sample of YBCO with δ = 0.25 was subjected to X-ray diffraction using CuKα radiation (λ = 154.2 pm). The lowest-angle diffraction peak was observed at 2 θ = 7.450º. Assuming that a = b = (c/3), calculate the values of a and c.
Model Answer
4.2 sin θ = nλ / 2d
d = (1)(154.2 pm) / 2 sin(3.725 º)
d = 1187 pm
lowest-angle = > d = longest axis = c
c = 1187 pm
a = c / 3 = 396 pm
a = 396 pm
Estimate the density of this sample of YBCO (with δ = 0.25) in g cm-3. If you were unable to calculate the values for a and c from part 4.2, then use a = 500 pm and c = 1500 pm.
Model Answer
4.3 Vunit cell = a × b × c = 3a3 = 3 (396 pm)3 = 1.863 ⋅ 10–22 cm3
munit cell = (88.91 + 2 × 137.33 + 3 × 63.55 + 6.75 × 16.00) / NA
munit cell = 662.22 g mol–1 / 6.0221 ⋅ 1023 mol–1 = 1.100 ⋅ 10–21 g
density = 1.100 ⋅ 10–21 g / 1.863 ⋅ 10–22 cm3 = 5.90 g cm–3
When YBCO is dissolved in aqueous HCl (c = 1.0 mol dm–3) bubbles of gas are observed (identified as O2 by gas chromatography). After boiling for 10 min to expel the dissolved gases, the solution reacts with excess KI solution, turning yellow-brown. This solution can be titrated with thiosulfate solution to a starch endpoint. If YBCO is added under Ar directly to a solution in which concentrations of both KI and HCl are equal to 1.0 mol dm–3, the solution turns yellow-brown but no gas evolution is observed.
Write a balanced net ionic equation for the reaction when solid YBa2Cu3O7–δ dissolves in aqueous HCl with evolution of O2.
Model Answer
4.4 YBa2Cu3O7-δ (s) + 13 H+(aq) →
→ Y3+(aq) + 2 Ba2+(aq) + 3 Cu2+(aq) + (0.25 [1 – 2δ]) O2(g) + 6.5 H2O(l)
Write a balanced net ionic equation for the reaction when the solution from 4.4 reacts with excess KI in acidic solution after the dissolved oxygen is expelled.
Model Answer
4.5 2 Cu2+(aq) + 5 I– (aq) → 2 CuI(s) + I3– (aq)
or
2 Cu2+(aq) + 4 I–(aq) → 2 CuI(s) + I2(aq)
Write a balanced net ionic equation for the reaction when the solution from 4.5 is titrated with thiosulfate (S2O32–).
Model Answer
4.6 I3– (aq) + 2 S2O32– (aq) → 3 I–(aq) + S4O62– (aq)
or
I2(aq) + 2 S2O32– (aq) → 2 I–(aq) + S4O62– (aq)
Write a balanced net ionic equation for the reaction when solid YBa2Cu3O7–δ dissolves in aqueous HCl containing excess KI in an Ar atmosphere.
Model Answer
4.7 YBa2Cu3O7-δ (s) + (14 – 2 δ) H+(aq) + (9 – 3 δ) I–(aq) →
→ Y3+(aq) + 2 Ba2+(aq) + 3 CuI(s) + (7 – δ) H2O(l) + (2 – δ) I3– (aq)
or
YBa2Cu3O7-δ (s) + (14 – 2 δ) H+(aq) + (7 – 2 δ) I–(aq) →
→ Y3+(aq) + 2 Ba2+(aq) + 3 CuI(s) + (7 – δ) H2O(l) + (2 – δ) I2(aq)
Two identical samples of YBCO with an unknown value of δ were prepared. The first sample was dissolved in 5 cm3 of aqueous HCl (c = 1.0 mol dm–3), evolving O2. After boiling to expel gases, cooling, and addition of 10 cm3 of KI solution (c = 0.7 mol dm–3) under Ar, titration with thiosulfate to the starch endpoint required 1.542 ⋅ 10–4 mol thiosulfate. The second sample of YBCO was added under Ar directly to 7 cm3 of a solution in which c(KI) = 1.0 mol dm–3 and c(HCl) = 0.7 mol dm–3. Titration of this solution required 1.696 ⋅ 10–4 mol thiosulfate to reach the endpoint.
Calculate the amount of substance of Cu (in mol) in each of these samples of YBCO.
Model Answer
4.8 n(Cu) = n(thiosulfate) in the first titration
n(Cu) = 1.542 ⋅ 10–4 mol
Calculate the value of δ for these samples of YBCO.
Model Answer
4.9 Total n(Cu) = 1.542 ⋅ 10–4 mol
n(CuIII) = (1.696 ⋅ 10–4 mol) – (1.542 ⋅ 10–4 mol) = 1.54 ⋅ 10–5 mol
Thus: 90 % of Cu is present as Cu(II) and 10 % as Cu(III).
From charge balance:
2(7 – δ) = 3 + (2 × 2) + 3 × [(0.90 × 2) + (0.10 × 3)] = 13.30
δ = 0.35
Alternatively, using the balanced equations in (d):
In the 1st titration, each mol YBCO corresponds 1.5 mol I3– and 3 mol S2O32–
In the 2nd titration, each mol YBCO corresponds (2 – δ) mol I3– and (4 – 2 δ) mol S2O32–
Thus,
3 / (4 – 2 δ) = 1.5 / (2 – δ) = (1.542 ⋅ 10–4 mol) / (1.696 ⋅ 10–4 mol)
2 – δ = 1.650
δ = 0.35