Chapter 13.10, Problem 52AAP

To determine

The current density associated with corrosion rate.

Answer to Problem 52AAP

The current density associated with corrosion rate is $4.52\times {10}^{-7}{\text{A/cm}}^2$.

Explanation of Solution

Write the expression for corrosion current.

  $I=\frac{wnF}{tM}$                                                                                                       (I)

Here, the corrosion mass is $w$, the corrosion time is $t$, 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 $F$, $F=96500\text{C/mol}\text{or}96500\text{A}\cdot \text{s/mol}$.

Write the formula for current density $\left(i\right)$.

  $i=\frac{I}{A}$                                                                                                           (II)

Here, the current flow is $I$, the area is $A$.

Refer Equation (II).

Substitute $iA$ for $I$ in Equation (I).

  $\begin{array}{l}iA=\frac{wnF}{tM}\\ i=\frac{wnF}{AtM}\\ i=\left(\frac{w}{At}\right)\left(\frac{nF}{M}\right)\end{array}$                                                                                           (III)

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

Conclusion:

It is given that corroding rate $2.40\text{mdd}$ is in the unit of mg of metal per square decimeter per day.

The unit conversion of $\text{mdd}$ to ${\text{g/cm}}^2\cdot \text{day}$ is as follows.

  $\begin{array}{c}\text{corrosion rate in}{\text{g/cm}}^2\cdot \text{day}=\frac{2.40\text{mg}}{{\text{dm}}^2\cdot \text{day}}\times \frac{1\text{g}}{1000\text{mg}}\times \frac{{\text{dm}}^2}{100{\text{cm}}^2}\\ =2.4\times {10}^{-5}{\text{g/cm}}^2\cdot \text{day}\end{array}$

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

The anodic reaction of tin is as follows:

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

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

Refer Figure 2.3, “Periodic Table of Elements”

The atomic mass of tin is, $M_{\text{Sn}}=118.70\text{g/mol}$.

Substitute $2.4\times {10}^{-5}{\text{g/cm}}^2\cdot \text{day}$ for $w/At$, $2$ for $n$, $96500\text{A}\cdot \text{s/mol}$ for $F$, $2419200\text{s}$ for $t$, and $118.70\text{g/mol}$ for $M$ in Equation (III).

  $\begin{array}{c}i=\left(2.4\times {10}^{-5}{\text{g/cm}}^2\cdot \text{day}\right)\left(\frac{2\times 96500\text{A}\cdot \text{s/mol}}{118.70\text{g/mol}}\right)\\ =\left(2.4\times {10}^{-5}\frac{\text{g}}{{\text{cm}}^2\cdot \text{day}}\times \frac{1\text{day}}{24\text{hr}}\times \frac{1\text{hr}}{3600\text{s}}\right)\left(\frac{2\times 96500\text{A}\cdot \text{s/mol}}{118.70\text{g/mol}}\right)\\ =\left(2.7777\times {10}^{-10}\frac{\text{g}}{{\text{cm}}^2\cdot \text{s}}\right)\left(1625.9477\text{A}\cdot \text{s/g}\right)\\ =4.5165\times {10}^{-7}{\text{A/cm}}^2\end{array}$

  $\simeq 4.52\times {10}^{-7}{\text{A/cm}}^2$

Thus, the current density associated with corrosion rate is $4.52\times {10}^{-7}{\text{A/cm}}^2$.