Chapter 13.10, Problem 48AAP

(a)

To determine

The corrosion current.

Answer to Problem 48AAP

The corrosion current is $0.5\text{A}$.

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 $F$, $F=96500\text{C/mol}\text{or}96500\text{A}\cdot \text{s/mol}$.

Conclusion:

Consider the steel as iron and the anodic reaction of iron is expressed as follows.

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

The anodic reaction of iron is as follows:

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

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

Refer Figure 2.3, “Periodic Table of Elements”

The atomic mass of iron is, $M_{\text{Fe}}=55.85\text{g/mol}$.

The corrosion time is 4 weeks.

  $\begin{array}{c}t=4\text{weeks}\times \frac{7\text{days}}{1\text{week}}\times \frac{24\text{hr}}{1\text{day}}\times \frac{3600\text{s}}{1\text{hr}}\\ =2419200\text{s}\end{array}$

Substitute $360\text{g}$ for $w$, $2$ for $n$, $96500\text{A}\cdot \text{s/mol}$ for $F$, $2419200\text{s}$ for $t$, and $55.85\text{g/mol}$ for $M$ in Equation (I).

  $\begin{array}{c}I=\frac{\left(360\text{g}\right)\left(2\right)\left(96500\text{A}\cdot \text{s/mol}\right)}{\left(2419200\text{s}\right)\left(55.85\text{g/mol}\right)}\\ =\frac{69480000}{135112320}\text{A}\\ =0.5142\text{A}\\ \simeq 0.5\text{A}\end{array}$

Thus, the corrosion current is $0.5\text{A}$.

(b)

To determine

The corrosion density associated with corrosion current.

Answer to Problem 48AAP

The corrosion density associated with corrosion current is $7.94\times {10}^{-5}{\text{A/cm}}^2$.

Explanation of Solution

The corrosion density associated with corrosion current is nothing but the current

density $\left(i\right)$.

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

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

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

Conclusion:

The total surface area of the tank is expressed as follows.

  $\begin{array}{c}A=4\left(L\times H\right)+\left(W\times L\right)\\ =4\left(30\text{cm}\times \text{45}\text{cm}\right)+\left(30\text{cm}\times \text{30}\text{cm}\right)\\ =5400{\text{cm}}^2+900{\text{cm}}^2\\ =6300{\text{cm}}^2\end{array}$

Substitute $0.5\text{A}$ for $I$ and $6300{\text{cm}}^2$ for $A$ in Equation (II).

  $\begin{array}{c}i=\frac{0.5\text{A}}{6300{\text{cm}}^2}\\ =7.9365\times {10}^{-5}{\text{A/cm}}^2\\ =7.94\times {10}^{-5}{\text{A/cm}}^2\end{array}$

Thus, the corrosion density associated with corrosion current is $7.94\times {10}^{-5}{\text{A/cm}}^2$.