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Solution: Δt Δx I M Zρ × I →

Chromium should be used to coat the cobalt plate, as chromium is more reactive than cobalt and will therefore protect the cobalt from corrosion when exposed to corrosive environments. Electroplating offers advantages over hot-dipping for metal coating processes, including greater flexibility, more uniform thickness deposition, and better thickness control. Using the formula provided and known values for iron's density, atomic weight, and the thickness loss of a steel plate over 10 years in seawater, the average corrosion current density is calculated to be 2.55 μA/Cm2.
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0% found this document useful (0 votes)
78 views1 page

Solution: Δt Δx I M Zρ × I →

Chromium should be used to coat the cobalt plate, as chromium is more reactive than cobalt and will therefore protect the cobalt from corrosion when exposed to corrosive environments. Electroplating offers advantages over hot-dipping for metal coating processes, including greater flexibility, more uniform thickness deposition, and better thickness control. Using the formula provided and known values for iron's density, atomic weight, and the thickness loss of a steel plate over 10 years in seawater, the average corrosion current density is calculated to be 2.55 μA/Cm2.
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Solution

1. A cobalt plate tended to protect it by coating method, which element do you


select, chromium or Tin? And why?

- We should coat it with (Chromium) to protect it from corrosion because it is


more reactive (less noble) than Cobalt. so it will protect Cobalt when exposed
to corrosive environment.

2. What are the advantages of electro plating over hot-dipping process?


- These advantages are respect of flexibility, uniformity and control of thickness
of film.

3. A steel plate has corroded on both sides in seawater. After 10 years a thickness
reduction of 3 mm is measured. Calculate the average corrosion current
density Take into consideration that the dissolution reaction is mainly Fe = Fe
2+ + 2e–, and that the density and the atomic weight of iron are 7.8 g/cm3
and 56, respectively.
Δt I corr M
- =3268 ×
Δx Zρ
I corr 56
0.3 = 3268 × → Icorr = 2.55 µA/Cm2
2∗7.8

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