Chapter 13.10, Problem 49AAP

To determine

The mass of the zinc corroded in grams per minute.

Answer to Problem 49AAP

The mass of the zinc corroded in grams per minute is $\text{1}\text{.0651}\times {\text{10}}^{-2}\text{g/min}$.

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

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.

  $\begin{array}{c}A=\pi dh+{\frac{\pi d}{4}}^2\\ =\pi \left(50\text{cm}\right)\left(45\text{cm}\right)+\frac{\pi {\left(50\text{cm}\right)}^2}{4}\\ =7068.5835{\text{cm}}^2+1963.4954{\text{cm}}^2\\ =9032.0789{\text{cm}}^2\end{array}$

Substitute $5.8\times {10}^{-5}{\text{A/cm}}^2$ for $i$ and $9032.0789{\text{cm}}^2$ for $A$ in Equation (II).

  $\begin{array}{l}5.8\times {10}^{-5}{\text{A/cm}}^2=\frac{I}{9032.0789{\text{cm}}^2}\\ I=\left(5.8\times {10}^{-5}{\text{A/cm}}^2\right)\left(9032.0789{\text{cm}}^2\right)\\ I=0.5238\text{A}\\ I=0.524\text{A}\end{array}$

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

The anodic reaction of iron 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 $0.524\text{A}$ for $I$, $65.38\text{g/ mol}$ for $M$, $2$ for $n$, and $96500\text{A}\cdot \text{s/mol}$ for $F$ in Equation (I).

  $\begin{array}{c}\frac{w}{t}=\frac{\left(0.524\text{A}\right)\left(65.38\text{g/ mol}\right)}{\left(2\right)\left(96500\text{A}\cdot \text{s/mol}\right)}\\ =\frac{34.2591}{193000}\text{g/s}\\ =\text{1}\text{.7751}\times {\text{10}}^{-4}\frac{\text{g}}{\text{s}}\times \frac{60\text{s}}{\text{1}\text{min}}\\ \simeq \text{1}\text{.0651}\times {\text{10}}^{-2}\text{g/min}\end{array}$

Thus, the mass of the zinc corroded in grams per minute is $\text{1}\text{.0651}\times {\text{10}}^{-2}\text{g/min}$.