For each of the metals listed in the following table, compute the Pilling- Bedworth ratio. Also, on the basis of this value, specify whether you would expect the oxide scale that forms on the surface to be protective, and then justify your decision. Density data for both the metal and its oxide are also tabulated. \begin{tabular}{lccc} \hline & Metal Metal & Density \(\left(\mathrm{g} / \mathrm{cm}^{3}\right)\) & Metal Oxide & Oxide Density $\left(\mathrm{g} / \mathrm{cm}^{\mathbf{7}}\right)$ \\ \hline \(\mathrm{Zr}\) & \(6.51\) & \(\mathrm{ZrO}_{2}\) & \(5.89\) \\ \(\mathrm{Sn}\) & \(7.30\) & \(\mathrm{SnO}_{2}\) & \(6.95\) \\ \(\mathrm{Bi}\) & \(9.80\) & \(\mathrm{Bi}_{2} \mathrm{O}_{3}\) & \(8.90\) \\ \hline \end{tabular}

Step by step solution

01

Understanding the Pilling-Bedworth ratio formula

The Pilling-Bedworth ratio (PB) is calculated using the following formula: PB = \(\frac{V_\text{oxide}}{n \cdot V_\text{metal}}\) where \(V_\text{oxide}\) is the molar volume of the metal oxide, \(V_\text{metal}\) is the molar volume of the metal, and \(n\) is the number of metal atoms per formula unit of metal oxide. Note that molar volume can be calculated as \(\frac{Molar\, mass}{Density}\).

02

Calculate the Pilling-Bedworth ratio for Zr and ZrO₂

For Zr: Density of Zr = 6.51 g/cm³ Molar mass of Zr = 91.22 g/mol For ZrO₂: Density of ZrO₂ = 5.89 g/cm³ Molar mass of ZrO₂ = 123.22 g/mol (91.22 g/mol for Zr + 16 g/mol for each oxygen atom) Since there is 1 Zr atom per formula unit of ZrO₂, we have n = 1. Calculate the PB ratio for Zr and ZrO₂ using the formula: PB = \(\frac{\frac{123.22\,g/mol}{5.89\,g/cm^3}}{1 \cdot \frac{91.22\,g/mol}{6.51\,g/cm^3}} \approx 1.45\)

03

Calculate the Pilling-Bedworth ratio for Sn and SnO₂

For Sn: Density of Sn = 7.30 g/cm³ Molar mass of Sn = 118.71 g/mol For SnO₂: Density of SnO₂ = 6.95 g/cm³ Molar mass of SnO₂ = 150.71 g/mol (118.71 g/mol for Sn + 16 g/mol for each oxygen atom) Since there is 1 Sn atom per formula unit of SnO₂ , we have n = 1. Calculate the PB ratio for Sn and SnO₂ using the formula: PB = \(\frac{\frac{150.71\,g/mol}{6.95\,g/cm^3}}{1 \cdot \frac{118.71\,g/mol}{7.30\,g/cm^3}} \approx 1.07\)

04

Calculate the Pilling-Bedworth ratio for Bi and Bi₂O₃

For Bi: Density of Bi = 9.80 g/cm³ Molar mass of Bi = 208.98 g/mol For Bi₂O₃: Density of Bi₂O₃ = 8.90 g/cm³ Molar mass of Bi₂O₃ = 465.96 g/mol (2 * 208.98 g/mol for Bi + 3 *16 g/mol for oxygen) Since there are 2 Bi atoms per formula unit of Bi₂O₃, we have n = 2. Calculate the PB ratio for Bi and Bi₂O₃ using the formula: PB = \(\frac{\frac{465.96\,g/mol}{8.90\,g/cm^3}}{2 \cdot \frac{208.98\,g/mol}{9.80\,g/cm^3}} \approx 1.33\)

05

Determine if the oxide scale is protective

The oxide scale is considered protective if the Pilling-Bedworth ratio is between 1 and 2. Based on our calculations: 1. For Zr and ZrO₂: PB ≈ 1.45, so the oxide scale is protective. 2. For Sn and SnO₂: PB ≈ 1.07, so the oxide scale is protective. 3. For Bi and Bi₂O₃: PB ≈ 1.33, so the oxide scale is protective. All three metals have a Pilling-Bedworth ratio between 1 and 2, which indicates that the oxide scales forming on their surfaces are protective.

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