# Corrosion Worksheet Solutions ```Corrosion Worksheet Solutions.
1. Anodic half cell reaction:- Zn  Zn2+ +2eCathodic half cell reaction:- 2H+ + 2e-  H2
Fully balanced reaction :- Zn + 2HCl  Zn(Cl)2 + H2
2. Anodic half cell reaction:- Fe  Fe2+ +2eCathodic half cell reaction:- 2H+ + 2e-  H2
Fully balanced reaction:- Fe + H2SO4  FeSO4 + H2
3
CR = 0.13 (icorr x EW)
d
icorr units are A/cm2, CR units are mils per year.
Fe a=55.845, d= 7.874, z=2.
a)CR=(0.13*0.001*1000*(55.845/2))/7.874
CR =0.4610
b)CR=(0.13*1.2*1000*(55.845/2))/7.874
CR =553.2017
c) CR=(0.13*16*1000*(55.845/2))/7.874
CR= 7.3760e+003
d) CR=(0.13*146*1000*(55.845/2))/7.874
CR =6.7306e+004
4. Al a=26.98, d=2.7, z=3.
a) CR=(0.13*0.05*(26.98/3))/2.7
CR = 0.022
b) CR=(0.13*3*(26.98/3))/2.7
CR = 1.299
c) CR=(0.13*15*(26.98/3))/2.7
CR = 6.496
d) CR=(0.13*125*(26.98/3))/2.7
CR = 54.125
5. Na – a= 22.99, Cl – a= 35.45,NaCl, and only one electron involved
NaCl= 58.44, so
a) 1N is 58.44 gms of NaCl in 1 liter of water,
b) 0.5N is 58.44/2 or 29.22 gms of NaCl in 1 liter of water,
c) 0.25N is 58.44/4 or 14.61 gms of NaCl in 1 liter of water.
6. Al – 26.98, Cl – = 35.45, Al(Cl)3, so three electrons involved, so N= molwt/3.
Al(Cl)3 = 133.33 (1N is 133.3/3=44.443)
a) 10N is 444.43 gms of Al(Cl)3 in one liter of water.
b) 0.1N is 4.4443 gms of Al(Cl)3 in one liter of water.
c) 0.5N is 22.22 gms of Al(Cl)3 in one liter of water.
7. E = Eo + 2.3(RT/zF) log aCu2+, z=2.
Cu/Cu2+ Eo is 0.34V(SHE)
E = 0.34 +( (2.3*8.314*348)/(2*96845.3399))*log(10)
E= 0.3744V(SHE).
8. Zn/Zn2+ Eo is -0.76V(SHE), z=2.
a) 0.1N
E = -0.76+ 2.3(RT/zF) log (0.1)
E = -0.789V(SHE)
b) 10N
E = -0.76 + 2.3(RT/zF) log(10)
E = -0.7306V(SHE).
9. Pourbaix diagrams indicate where areas of active corrosion, immunity and passive
behavior occur as a function of potential and pH. Therefore, materials like
titanium and its alloys become good selections for surgical implant materials due
to their extensive passive region indicated by the Pourbaix diagram.
10. The area on the Pourbaix diagram between the oxygen line and the hydrogen line
is where water is stable and would continuously permit the cathodic reaction of
dissolved oxygen being reduced to hydroxyl ions. Therefore the best materials to
choose from these diagrams will have extensive passivity over this region.
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