Chapter 13.10, Problem 79SEP

To determine

The oxide-to-metal volume (Pilling-Bedworth) ratios for the oxidation of the metals listed and comment whether oxides are protective or not.

Explanation of Solution

Write the formula for Pilling-Bedworth $\left(\text{P}\text{.B}\text{.}\right)$ ratio.

  $\text{P}\text{.B}\text{.}\text{ratio}=\frac{\text{volume of oxide produced by oxidation}}{\text{volume of metal consumed by oxidation}}$        (I)

In general, the Pilling-Bedworth ratios are determined for the 100 g mass of the metal.

Write the formula for volume from density.

  $\text{Volume}=\frac{\text{mass}\left(m\right)}{\text{density}\left(\rho \right)}$        (II)

The Pilling-Bedworth ratio alone will not predict the protective nature of the oxide film. The P.B. ratio is the additional factor to say that the metal’s oxide film is protective. The P.B. ration must be 1 or very close 1 to make the metal’s oxide film as a protective film.

Conclusion:

For Tungsten $\left(\text{W}\right)$:

The oxidation reaction tungsten is,

  $\text{W}+\frac{3}{2}\text{O}\to {\text{WO}}_3$

The molecular mass of the tungsten $\left(\text{W}\right)$ is $183.84\text{g/mol}$.

The molecular mass of the tungsten oxide $\left({\text{WO}}_3\right)$ is $231.84\text{g/mol}$.

Calculate the volume of the tungsten using equation (II).

  $\begin{array}{c}{V}_{\text{W}}=\frac{m_{\text{W}}}{{\rho }_{\text{W}}}\\ =\frac{100\text{g}}{19.35{\text{g/cm}}^3}\\ =5.1679{\text{cm}}^3\end{array}$

The mass of ${\text{WO}}_3$ produced by the oxidation of $100\text{g}$ tungsten $\left(\text{W}\right)$ is found by,

  $\begin{array}{l}\frac{\text{100}\text{g}}{\text{molecular mass of tungsten}}=\frac{m_{{\text{WO}}_3}}{\text{molecular mass of}{\text{WO}}_3}\\ \frac{100\text{g}}{183.84\text{g/mol}}=\frac{m_{{\text{WO}}_3}}{231.84\text{g/mol}}\\ m_{{\text{WO}}_3}=\frac{100\text{g}}{183.84\text{g/mol}}\times 231.84\text{g/mol}\\ m_{{\text{WO}}_3}=126.1096\text{g}\end{array}$

Calculate the volume of tungsten oxide $\left({\text{WO}}_3\right)$ using equation (II).

  $\begin{array}{c}{V}_{{\text{WO}}_3}=\frac{m_{{\text{WO}}_3}}{{\rho }_{{\text{WO}}_3}}\\ =\frac{126.1096\text{g}}{12.11{\text{g/cm}}^3}\\ =10.4137{\text{cm}}^3\end{array}$

Calculate the Pilling-Bedworth ratio between tungsten and its oxide using equation (I).

  $\begin{array}{c}\text{P}\text{.B}\text{.}\text{ratio}=\frac{V_{{\text{WO}}_3}}{V_{\text{W}}}\\ =\frac{10.4137{\text{cm}}^3}{5.1679{\text{cm}}^3}\\ =2.01\end{array}$

Here, the P.B. ratio is $2.01>1$.

Thus, the tungsten oxide $\left({\text{WO}}_3\right)$ film is non protective.

For Sodium $\left(\text{Na}\right)$:

The oxidation reaction sodium is,

  $\text{2Na}+\frac{1}{2}{\text{O}}_2\to {\text{Na}}_2\text{O}$

The molecular mass of the sodium $\left(\text{Na}\right)$ is $22.9897\text{g/mol}$.

The molecular mass of the sodium oxide $\left({\text{Na}}_2\text{O}\right)$ is $61.9789\text{g/mol}$.

Calculate the volume of the sodium using equation (II).

  $\begin{array}{c}{V}_{\text{W}}=\frac{m_{\text{W}}}{{\rho }_{\text{W}}}\\ =\frac{100\text{g}}{0.967{\text{g/cm}}^3}\\ =103.41{\text{cm}}^3\end{array}$

The mass of ${\text{Na}}_2\text{O}$ produced by the oxidation of $100\text{g}$ sodium $\left(\text{Na}\right)$ is found by,

  $\begin{array}{l}\frac{\text{100}\text{g}}{\text{2}\times \text{molecular mass of Na}}=\frac{m_{{\text{Na}}_2\text{O}}}{\text{molecular mass of}{\text{Na}}_2\text{O }}\\ \frac{100\text{g}}{2\times 22.9897\text{g/mol}}=\frac{m_{{\text{Na}}_2\text{O}}}{61.9789\text{g/mol}}\\ m_{{\text{Na}}_2\text{O}}=\frac{100\text{g}}{2\times 22.9897\text{g/mol}}\times 61.9789\text{g/mol}\\ m_{{\text{Na}}_2\text{O}}=134.7971\text{g}\end{array}$

Calculate the volume of sodium oxide $\left({\text{Na}}_2\text{O}\right)$ using equation (II).

  $\begin{array}{c}{V}_{{\text{Na}}_2\text{O}}=\frac{m_{{\text{Na}}_2\text{O}}}{{\rho }_{{\text{Na}}_2\text{O}}}\\ =\frac{134.7971\text{g}}{2.27{\text{g/cm}}^3}\\ =59.3819{\text{cm}}^3\end{array}$

Calculate the Pilling-Bedworth ratio between tungsten and its oxide using equation (I).

  $\begin{array}{c}\text{P}\text{.B}\text{.}\text{ratio}=\frac{V_{{\text{Na}}_2\text{O}}}{V_{\text{Na}}}\\ =\frac{59.3819{\text{cm}}^3}{103.41{\text{cm}}^3}\\ =0.57\end{array}$

Here, the P.B. ratio is $0.57<1$.

Thus, the sodium oxide $\left({\text{Na}}_2\text{O}\right)$ film is non protective.

Similarly, the Pilling-Bedworth ratios for the other metals listed in the given table is calculated and tabulated in below Table 1.

S. No.	Metal	P.B. ratio	Description
1	Tungsten, W	2.01>1	Non protective
2	Sodium, Na	0.57<1	Non protective
3	Halnium, Hf	1.62>1	May protective
4	Copper, Cu	1.74>1	May protective
5	Manganese, Mn	1.70>1	May protective
6	Tin, Sn	1.15>1, $1.15\approx 1$	Protective