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Q1 : A piece of corroded steel plate was found in a submerged ocean vessel. It

was estimated that the original area of the plate was 10 in2 and the

approximately 2.6 kg had corroded away during the submersion. Assuming a

corrosion penetration rate of 200 mpy for this alloy in seawater, estimate the

time of submersion in years. The density of steel is 7.9 g/cm3.

mpy = 534W/DAT

t =(534) 2.6 x 10

6mg

7.9 g / cm3 10 in.2 (200 mpy) 

= 8.8 x 104

 h = 10 yr 

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. Zinc experiences corrosion in an acid solution according to the reaction;

Zn + 2H+  Zn2+  + H2

he rates of both oxidation and reduction half-reactions are controlled by

ctivation polarisation and the data are given in TABLE Q2b. 

TABLE Q2b

Zinc Hydrogen

Reduction

Potential

E (Zn/ Zn2+)=

0.763 V

E (H+/ H2) = 0 V

Exchange

Current Density

i o = 10-7 A/cm2  i o = 10-10 A/cm2 

Tafel Slope    = +0.09    = 0.08 

alculate:

. the rate of oxidation of Zinc in mol/cm2-s.

[6 marks

i. the value of the corrosion potential.

[4 marks]

For hydrogen reduction

 

  

 

 H oi

i log+E=E H)/H(HH 2+

     

 And for Zn oxidation

 

  

 

 Znoi

in  logZ+E=EZn )(zn/zn2      

Setting EH

 = EZn

 and solving for log i (log icorr 

) leads to

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 Ni H  Zn   oo H  Zn H  H 

 H  Ni

i Zni E  E  loglog1

 =ilog)()/(corr  /

22

        

 

  

 

   

710 10log)09.0(106log)08.0()763.0(0)08.0(09.0

1 =  

  x  

= -3.924

Or2-4-3.924

corr  A/cm101.19x=10=i 

 And from Equation

 F n

icorr  =r

s-mol/cm10x6.17=)/500,96)(2(

-/1019.1 = 210-

24

mol C 

cm sC  x 

 

(b) Now it becomes necessary to compute the value of the corrosionpotential, E

corr . This is possible by using either of the above equations for

EH or EZn and substituting for i the value determined above for icorr . Thus

 

  

 

 H o

corr 

i

i log+E=E H)/H(Hc 2+

     

V0.486=/10

/1019.1logV)0.08(+0=

210

24

 

  

 

cm A

cm A x