Assume the homogeneous (uniform) corrosion of iron in a deaerated acid solution

Question

Assume the homogeneous (uniform) corrosion of iron in a deaerated acid solution at pH =4. Plot polarization curves for iron and hydrogen. Estimate Ecorr, icorr, and the corrosion rate in microns per year.

Given:

         For Iron:

                  i0 = 10-8 A/cm2

                  ba = bc = 50 mV

                  aFe2+ = 10-6

         For hydrogen

                  i0 = 10-6 A/cm2

                  ba = bc = 100 mV

                  iL = 10-3 A/cm2

What is the minimum current density required for the cathodic protection of Fe?

Explanation / Answer

The question is not very clear. Let’s see.

The solution is deaerated, so you can suppose that O2 is missing as dissolved gas. But H2 is there (introduced by bubbling).

The anodic reaction in the electrochemical cell is

Fe = Fe2+ + 2 e.

At the moment of your plot, some Fe2+ is in solution at a concentration/activity 10-6 M

The standard reduction potential ( from a Table) is – 0.44 V   versus SHE

(half reaction Fe2+(aq) + 2 e–  = Fe(s)).

From Nernst equation, the actual potential of Fe electrode is

E = - 0.44 – (0.059/2)log([Fe2+]/[Fe])=

    = - 0.44 – 0.0295log([10-6/1)=

    =   - 0.44 + 0.0295x6 =

    = - 0.263 V versus SHE

This is an equilibrium potential. For any higher potential the reaction goes in this direction Fe(s) = Fe2+(aq) + 2 e– , i.e. corrosion.

So, assume that it is the corrosion potential.

Ecorr = - 0.263 V versus SHE

At the corrosion potential the current is i0 =1x 10-8 A/cm2

icorr = 1x 10-8 A/cm2 (in fact a current density)

( Note 1A= 1C/s)

For a 1 cm2 Fe sample, the current quantity for 1 year is

Q = I x t =

    = 10-8 A x 365 days x 24 h/day x 3600 s/h =

   = 3.1536x10-1 C per cm2 = 3 x10-1 C per cm2

For 1 mol Fe to be reacted……………you need 96500/2 C

X …………………………………… 3.1536x10-1 C

X = 6.5 x 10-6 mol Fe is corroded in 1 year

6.5 x 10-6 mol x 55.8 g/mol = 3.7 g Fe corroded in 1 year per 1 cm2

This is the corrosion rate:

3.7 g Fe year-1 cm-2

For a cathodic protection you have to stop the process, imposing a reverse current with the same current density:    1x 10-8 A/cm2

With these data:

For Iron:

                  i0 = 10-8 A/cm2

                  ba = bc = 50 mV

                  Ecorr =- 0.263 V versus SHE

and looking to a general polarisation curve, you can plot/imagine the polarisation curve for Fe, including a Tafel plot.

The polarisation curve pass by i=0 at E = - 0.263 V

But I suppose that the task is to record it in fact/experimentaly.

For the cathodic process you may consider many reactions

2H+ +2e = H2

2H2O + O2 +4e = 4 HO-

At least partially the catodic reaction involves hydrogen

2H+ +2e = H2          i0 = 10-6 A/cm2

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