material chemistry metals and corrosion The corrosion of iron in aerated aqueous
ID: 1052334 • Letter: M
Question
material chemistry
metals and corrosion
The corrosion of iron in aerated aqueous media is schematically depicted in the figure below. Using the following information, show that the absorption of oxygen is the dominant cathodic reaction sustaining the corrosion of iron in neutral water at equilibrium with air. The corrosion potential on iron under these conditions is Ecorr = -0.44 V. Take the following steps to solve this problem. Discuss the meaning of anodic and cathodic reactions in a corrosion process. Write down the chemical equations of the anodic and cathodic half reactions for the corrosion process shown in the figure. Using the Nernst equation, E = E degree - (2.3 RT/nF) log (a_red/a_ox), calculate the equilibrium potential of the iron oxidation reaction, assuming a standard reduction potential of E degree = -0.44 V and an Fe^2+ activity of a(Fe^2+) = 10^-6. Using the Nernst equation, calculate the equilibrium potential of the H^+ reduction reaction, assuming p(H_2(g)) = 1 bar and pH = 7. Calculate the anodic current density for iron dissolution, i_anodic, by applying the Tafel equation: eta_anodic = E_corr - E_anodic = b log (i_anodic/i_0) where eta_anodic is the anodic overpotential and E_anodic is the equilibrium potential. The Tafel slope equals b = 0.06 V and the exchange current density is i_0 = 10^-11 A cm^-2. Calculate the cathodic current density due to discharge of hydrogen, i_cathodic, by applying the Tafel equation. The Tafel slope equals b = 0.12 V and the exchange current density at pH = 7 is i_0 = 10^-10 A cm^-2. Use the electroneutrality condition: Sigma ianodic = Sigma icathodic to calculate the current density due to oxy gen reduction. Discuss the results of your calculations.Explanation / Answer
(a) cathodic reaction means the reactant is reduced in this reaction (reduction occurs at cathode). In other words, the reactant will aceept electron in this process.
Oxidation occurs at anode. So , an anodic reaction means that reactant is oxidised.
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(b) Cathodic reaction : O2 + 2H^+ + 4e^- ==> 2H2O
anodic reaction : 2Fe ==> 2Fe^2+ 4e^-
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E = E^0 - 2.3RT/nFlog [a(red)/a(ox)]
= -0.44 - 0.059/2 log[a(O2)/a(Fe)]
= -0.44 - 0.059 /2 log[1/10^-6] = 0.263 V [a(o2)=1 as it is gaseous]
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E= E^0 -RT/nF*pH
= 0.00- 0.059/1*7 = 0.0084 V
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