Chapter 13.10, Problem 54AAP

To determine

The thickness of the metal that is corroded in 210 days.

Answer to Problem 54AAP

The thickness of the metal that is corroded in 210 days is $0.003\text{mm}$.

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)

Write the formula to calculate the corrosion depth.

  $\text{corrsion depth }=\frac{w/At}{\rho }\left(t\right)$                                                                        (III)

Here, the corrosion mass is $w$, the area is $A$, the time taken is $t$, and the density of metal is $\rho$. In other words the corrosion rate per area is $w/At$.

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 $3.45\times {10}^{-7}{\text{A/cm}}^2$ 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).

  $\begin{array}{c}\frac{w}{At}=\frac{\left(3.45\times {10}^{-7}{\text{A/cm}}^2\right)\left(65.38\text{g/ mol}\right)}{\left(2\right)\left(96500\text{A}\cdot \text{s/mol}\right)}\\ =\frac{2.2556\times {10}^{-5}}{193000}{\text{g/cm}}^2\cdot \text{s}\\ =\text{1}\text{.1687}\times {\text{10}}^{-10}{\text{g/cm}}^2\cdot \text{s}\end{array}$

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

Substitute $\text{1}\text{.1687}\times {\text{10}}^{-10}{\text{g/cm}}^2\cdot \text{s}$ for $w/At$, $7.13{\text{g/cm}}^3$ for $\rho$, and 210 days for $t$ in Equation (III).

  $\begin{array}{c}\text{corrsion depth }=\frac{\text{1}\text{.1687}\times {\text{10}}^{-10}{\text{g/cm}}^2\cdot \text{s}}{7.13{\text{g/cm}}^3}\left(210\text{days}\right)\\ =3.4422\times {10}^{-9}\frac{\text{cm}\cdot \text{day}}{\text{s}}\times \frac{24\text{hr}}{1\text{day}}\times \frac{3600\text{s}}{1\text{hr}}\\ =2.9741\times {10}^{-4}\text{cm}\times \frac{10\text{mm}}{1\text{cm}}\\ =2.9741\times {10}^{-3}\text{mm}\end{array}$

     $\begin{array}{c}\text{=}\text{0}\text{.0029741}\text{mm}\\ \simeq 0.003\text{mm}\end{array}$

Thus, the thickness of the metal that is corroded in 210 days is $0.003\text{mm}$.