Chapter 13.10, Problem 56AAP

To determine

The time required to uniformly corrode (zinc coating) the steel to a thickness of $0.030\text{mm}$.

Answer to Problem 56AAP

The time required to uniformly corrode the steel to a thickness of $0.030\text{mm}$ is $15.2\text{years}$.

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).

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

Here, the corrosion rate per area is $w/At$.

Write the formula for time required for corrosion of required depth.

  $t=\frac{\rho \times \text{required depth}}{\text{corrosion rate per area}}$                                                                          (III)

Conclusion:

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

The anodic reaction of zinc is as follows:

  ${\text{Zn}}^{2+}+2e^{-}\to \text{Zn}$

Here, the number of electrons produced is, $n=2$.

Refer Figure 2.3, “Periodic Table of Elements”

The atomic mass of zinc is, $M_{\text{Zn}}=65.38\text{g/mol}$.

Substitute $1.32\times {10}^{-7}{\text{A/cm}}^2$ for $i$, $65.38\text{g/ mol}$ for $M$, $2$ for $n$, and $96500\text{A}\cdot \text{s/mol}$ for $F$ in Equation (II).

  $\begin{array}{c}\frac{w}{At}=\frac{\left(1.32\times {10}^{-7}{\text{A/cm}}^2\right)\left(65.38\text{g/ mol}\right)}{\left(2\right)\left(96500\text{A}\cdot \text{s/mol}\right)}\\ =4.4716\times {10}^{-11}{\text{g/cm}}^2\cdot \text{s}\\ \simeq 4.472\times {10}^{-11}{\text{g/cm}}^2\cdot \text{s}\end{array}$

The density of the zinc is $7.13{\text{g/cm}}^3$.

Substitute $7.13{\text{g/cm}}^3$ for $\rho$, $0.030\text{mm}$ for $\text{required depth}$, and $4.472\times {10}^{-11}{\text{g/cm}}^2\cdot \text{s}$ for $\text{corrosion rate per area}$ in Equation (II).

  $\begin{array}{c}t=\frac{7.13{\text{g/cm}}^3\times 0.030\text{mm}}{4.472\times {10}^{-11}{\text{g/cm}}^2\cdot \text{s}}\\ \text{=}\frac{7.13{\text{g/cm}}^3\times 0.030\text{mm}\times \frac{\text{1}\text{cm}}{10\text{mm}}}{4.472\times {10}^{-11}{\text{g/cm}}^2\cdot \text{s}}\\ =478353819.6\text{s}\times \frac{1\text{hr}}{3600\text{s}}\times \frac{1\text{day}}{24\text{hr}}\\ =5536.5025\text{days}\times \frac{1\text{year}}{365\text{days}}\end{array}$

  $\begin{array}{c}=15.1685\text{years}\\ \simeq \text{15}\text{.2}\text{years}\end{array}$

Thus, the time required to uniformly corrode (zinc coating) the steel to a thickness of $0.030\text{mm}$ is $15.2\text{years}$.