---
title: "Corrosion protection and corrosion rate"
subject: "Material Science"
chapter: 8
key: "semester_3/material_science:corrosion-protection-and-rate-numerical"
source: https://www.bctengineer.com/q/semester_3/material_science/corrosion-protection-and-rate-numerical
site: BCT Engineering
---

# Corrosion protection and corrosion rate

Material Science · Chapter 8: Environmental Effects

- Asked in: year not given, 4+4 marks
- Video: [Corrosion Rate - Definitions and Units](https://www.youtube.com/watch?v=A4kk2HD2WQg) — ETH Corrosion
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/material_science/corrosion-protection-and-rate-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_3/material_science/chapter-8

## Question

(a) Describe four methods of protecting metals from corrosion. (b) A mild steel coupon (density 7.87 g/cm$^3$) with exposed area 12 cm$^2$ lost 0.45 g after being immersed in an acid for 240 hours. Calculate the corrosion penetration rate in mm/year.

## Short answer

Four protection methods: coatings, cathodic protection, inhibitors, and alloy selection. Corrosion rate: weight loss 0.45 g, density 7.87 g/cm³, area 12 cm², time 240 h. Thickness lost = 0.45/(7.87×12) = 0.004765 cm = 0.04765 mm. CPR = 0.04765×(8760/240) = 1.74 mm/year.

## Answer

#### (a) Corrosion protection methods
1. **Protective coatings**: paint, enamel, plastic, or metallic coatings. *Galvanising* (zinc on steel) protects both as a barrier and sacrificially, since zinc is anodic to iron even if scratched. Tin plating protects only as a barrier.
2. **Cathodic protection**: make the structure the cathode.
   - *Sacrificial anode*: connect Mg, Zn or Al blocks to buried pipes or ship hulls; they corrode instead of the steel.
   - *Impressed current*: a DC source drives current from an inert anode to the structure.
3. **Inhibitors**: small additions (chromates, phosphates, amines) to the environment that form a protective film or slow the electrode reaction.
4. **Material selection and alloying**: use stainless steel, aluminium, or Cu alloys; design to avoid crevices, galvanic couples, and stagnant water.
   - Other: anodising of aluminium, controlling temperature/humidity.

#### (b) Corrosion rate
Volume of metal lost:

$$
V = \frac{W}{\rho} = \frac{0.45}{7.87} = 0.05718\ \text{cm}^3
$$

Thickness lost over 240 h:

$$
x = \frac{V}{A} = \frac{0.05718}{12} = 0.004765\ \text{cm} = 0.04765\ \text{mm}
$$

Hours in a year $= 8760$, so scaling to a year:

$$
CPR = 0.04765 \times \frac{8760}{240} = 0.04765 \times 36.5 = 1.74\ \text{mm/year}
$$

This is the same as the standard formula $CPR = \dfrac{87.6\,W}{\rho A t}$ with $W$ in mg, $\rho$ in g/cm$^3$, $A$ in cm$^2$, $t$ in hours: $\dfrac{87.6 \times 450}{7.87 \times 12 \times 240} = 1.74$ mm/yr.

**Answer: corrosion penetration rate = 1.74 mm/year (about 68 mpy).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why does galvanising protect steel better than tin plating when the coating is scratched?
   A. Zinc is harder than tin
   B. Zinc is anodic to iron and corrodes sacrificially, while tin protects only as a barrier
   C. Tin is more reactive than iron
   D. Zinc forms no oxide
   Answer: B. Zinc is more active than iron and protects exposed steel sacrificially; tin is cathodic to iron and only a barrier. (Key point: Galvanising vs tin plating)

2. How does a sacrificial anode protect a buried steel pipe?
   A. Mg, Zn or Al is connected to it and corrodes instead, making the pipe the cathode
   B. It paints the pipe
   C. It removes oxygen from the soil
   D. It makes the pipe the anode
   Answer: A. The more active metal supplies electrons and corrodes, so the structure is the cathode. (Key point: Cathodic protection)

3. How is the penetration rate found from the coupon data?
   A. Mass lost divided by time only
   B. Volume lost = W/rho, thickness = V/A, then scaled to a year (8760 h)
   C. Mass lost multiplied by density
   D. Area divided by mass lost
   Answer: B. 0.45/7.87 = 0.0572 cm^3; over 12 cm^2 it is 0.04765 mm in 240 h, and x 36.5 gives about 1.74 mm/year. (Key point: Corrosion rate method)

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