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Chapter 17: Problem 14

Using the results of Problem \(17.13,\) compute the corrosion penetration rate, in mpy, for the corrosion of iron in \(\mathrm{HCl}\) (to form \(\mathrm{Fe}^{2}\) ions if the corrosion current density is \(8 \times 10^{-5} \mathrm{A} / \mathrm{cm}^{2}\)

Short Answer

Expert verified
Question: Calculate the corrosion penetration rate for the corrosion of iron in HCl with a corrosion current density of \(8 \times 10^{-5} \mathrm{~A/cm}^2\). Answer: The corrosion penetration rate for the corrosion of iron in HCl is approximately \(7.24 \mathrm{~mils/year}\).

Step by step solution

01

Calculate the Corrosion Current per cm²

To calculate the corrosion current per cm², we simply use the given current density value: \(I = i \cdot A\) For \(A = 1 \mathrm{cm}^2\), \(I = 8 \times 10^{-5} \mathrm{~A}\)
02

Determine molecular weight and valence of iron

From the periodic table, we know the molecular weight of iron (Fe) is \(55.845 \mathrm{g/mol}\), and since it forms \(\mathrm{Fe}^{2+}\) ions, the valence of iron is 2.
03

Calculate the corrosion mass using Faraday's Law

To find the mass of corroded iron, we can make use of Faraday's Law: \(w = \frac{I \times t}{z \times F}\) For one year, \(t = 1 \mathrm{~year} = 365 \times 24 \times 3600 \mathrm{~s}\) Substitute the variables and constant into the equation: \(w = \frac{8 \times 10^{-5} \mathrm{~A} \times 365 \times 24 \times 3600 \mathrm{~s}}{2 \times 96485 \mathrm{~C/mol}}\) \(w = 0.145 \mathrm{g}\) (approximately)
04

Calculate the corrosion thickness

We know the density of iron \(\rho = 7.86 \mathrm{~g/cm^3}\), we also have the mass \(w\) and area \(A\) of corroded iron. So we can now determine the thickness of the corroded layer: \(V = \frac{w}{\rho} = \frac{0.145 \mathrm{~g}}{7.86 \mathrm{~g/cm^3}} = 1.84 \times 10^{-2} \mathrm{cm^3}\) Thickness (in cm) \(= \frac{V}{A} = \frac{1.84 \times 10^{-2} \mathrm{cm^3}}{1 \mathrm{cm^2}} = 1.84 \times 10^{-2} \mathrm{cm}\)
05

Convert thickness to mils per year

Now, we'll convert the thickness value from cm to mils (1 mil = 0.001 inch, and 1 inch = 2.54 cm): Thickness (in mils) \(= 1.84 \times 10^{-2} \mathrm{cm} \times \frac{1 \mathrm{~inch}}{2.54 \mathrm{~cm}} \times \frac{1000 \mathrm{~mils}}{1 \mathrm{~inch}} = 7.24 \mathrm{~mils}\) Hence, the corrosion penetration rate for the corrosion of iron in HCl is approximately \(7.24 \mathrm{~mils/year}\).

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Most popular questions from this chapter

A piece of corroded metal alloy plate was found in a submerged ocean vessel. It was estimated that the original area of the plate was \(800 \mathrm{cm}^{2}\) and that approximately \(7.6 \mathrm{kg}\) had corroded away during the submersion. Assuming a corrosion penetration rate of \(4 \mathrm{mm} / \mathrm{yr}\) for this alloy in seawater, estimate the time of submersion in years. The density of the alloy is \(4.5 \mathrm{g} / \mathrm{cm}^{3}\)

Briefly explain why, for a small anode-tocathode area ratio, the corrosion rate will be higher than for a large ratio.

(a) What are inhibitors? (b) What possible mechanisms account for their effectiveness?

Briefly explain why cold-worked metals are more susceptible to corrosion than noncoldworked metals.

An electrochemical cell is constructed such that on one side a pure Zn electrode is in contact with a solution containing \(\mathrm{Zn}^{2+}\) ions at a concentration of \(10^{-2} M\). The other cell half consists of a pure \(\mathrm{Pb}\) electrode immersed in a solution of \(\mathrm{Pb}^{2+}\) ions that has a concentration of \(10^{-4} M .\) At what temperature will the potential between the two electrodes be \(+0.568 \mathrm{V} ?\)
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