Chapter 13.10, Problem 58AAP

To determine

The time required for the pit to corrode through the wall.

Answer to Problem 58AAP

The time required for the pit to corrode through the wall is $232.2\text{days}$.

Explanation of Solution

Write the expression for corrosion rate.

  $\frac{w}{t}=\frac{IM}{nF}$                                                                                                        (I)

Here, the corrosion rate is $w/t$, the current flow is $I$, atomic mass of the metal is $M$, number of electrons per atoms produced or consumed in the process is $n$, and the Faraday’s constant is $F$, $F=96500\text{C/mol}\text{or}96500\text{A}\cdot \text{s/mol}$.

The current density is expressed as,

  $\begin{array}{l}i=\frac{I}{A}\\ I=iA\end{array}$

Substitute $iA$ for $I$ in Equation (I) and rearrange it to obtain $t$.

  $\begin{array}{l}\frac{w}{t}=\frac{\left(iA\right)M}{nF}\\ t=\frac{wnF}{iAM}\end{array}$                                                                                                  (II)

Conclusion:

It is assumed that the pit is in the shape of cylinder.

The density of the aluminum is $2.70{\text{g/cm}}^3$.

The mass of the cylindrical aluminum pit $\left(w\right)$  is expressed as follows.

 $\begin{array}{c}w={\rho }_{\text{Al}}\times \text{Volume of the pit}\\ ={\rho }_{\text{Al}}\times \frac{\pi }{4}{d}^{2}h\\ =2.70{\text{g/cm}}^3\times \frac{\pi }{4}\times {\left(0.70\text{mm}\right)}^2\times 0.90\text{mm}\\ =2.70{\text{g/cm}}^3\times 0.3464{\text{mm}}^3\times \frac{1{\text{cm}}^3}{1000{\text{mm}}^3}\end{array}$

 $\begin{array}{c}=9.3517\text{g}\\ w\simeq \text{9}\text{.352}\text{g}\end{array}$

The top surface area of the cylindrical pit is expressed as follows.

  $\begin{array}{c}A=\frac{\pi }{4}\times {\left(0.70\text{mm}\right)}^2\\ =0.3848{\text{mm}}^2\times \frac{1{\text{cm}}^2}{100{\text{mm}}^2}\\ =3.848\times {10}^{-3}{\text{cm}}^2\end{array}$

Refer Table 13.1, “Standard electrode potentials at 25 °C”.

The anodic reaction of aluminum is as follows:

  ${\text{Al}}^{3+}+3e^{-}\to \text{Al}$

Here, the number of electrons produced is, $n=3$.

Refer Figure 2.3, “Periodic Table of Elements”

The atomic mass of aluminum is, $M_{\text{Al}}=26.98\text{g/mol}$.

Substitute $9.352\times {10}^{-4}\text{g}$ for $w$, $3$ for $n$, $96500\text{A}\cdot \text{s/mol}$ for $F$, $1.30\times {10}^{-4}{\text{A/cm}}^2$ for $i$, $3.848\times {10}^{-3}{\text{cm}}^2$ for $A$, and $26.98\text{g/mol}$ for $M$ in Equation (II).

  $\begin{array}{c}t=\frac{\left(9.352\times {10}^{-4}\text{g}\right)\left(3\right)\left(96500\text{A}\cdot \text{s/mol}\right)}{\left(1.30\times {10}^{-4}{\text{A/cm}}^2\right)\left(3.848\times {10}^{-3}{\text{cm}}^2\right)\left(26.98\text{g/mol}\right)}\\ =20060082.06\text{s}\times \frac{1\text{day}}{24\text{hr}}\times \frac{1\text{hr}}{3600\text{s}}\\ =232.1768\text{days}\\ \simeq 232.2\text{days}\end{array}$

Thus, the time required for the pit to corrode through the wall is $232.2\text{days}$.