Question

Consider the corrosion of elemental Fe(0) under anaerobic conditions coupled to the oxidation of protons according to:

Fe(0) + 2H+ -> Fe2+ + H2(g)

(a) Calculate the equilibrium constant K for this reaction at 25 o C.

(b) For T = 25 o C, pH = 10, [Fe2+] = 10-3 M, at what partial pressure of H2(g) does this reaction become unfavorable?

Answer 1

Answer – a) We are given oxidation-reduction reaction and from that we need to calculate the equilibrium constant.

We know the relationship between equilibrium constant and cell potential as follow –

E^ocell = RT / nF x ln K

For calculating the equilibrium constant we need to calculate E^ocall

Given, oxidation-reduction-

Fe + 2H⁺(aq) ----> Fe²⁺(aq) + H₂(g)

Oxidation half reaction –

Fe(s) -----> Fe²⁺(aq) + 2e^- E^o = + 0.44 V

Reduction half reaction -

2 H⁺(aq) + 2e^- ----> H₂(g)   E^o = 0.00

Overall reaction –

Fe(s) -----> Fe²⁺(aq) + 2e^- E^o = + 0.44 V

2 H⁺(aq) + 2e^- ----> H₂(g)   E^o = 0.00

_______________________________________________________________

Fe(s) + 2 H⁺(aq) --------> Fe²⁺(aq) + H₂(g)   E^o cell = 0.44

Now plug the values in the formula

E^ocell = RT / nF x ln K

0.44 V = 8.314 J. mol. C^-1 x 298 K / 2 x 96485 C/mol x ln K

ln K = 0.44 V x 2 x 96485 C.mol^-1 / 8.314 J. mol. C^-1 x 298 K

      = 34.27

Antiln from both side

K = 7.64 x 10¹⁴

The equilibrium constant for this reaction is 7.64 x 10¹⁴

b) Now we calculated equilibrium constant from that we can calculate pressure of H₂

We are given, pH = 10 , [Fe²⁺] = 1.0 x 10^-3 M , T = 25+273 = 298 K

Now we need to calculate [H⁺] from given pH

pH = - log [H⁺]

[H⁺] = 10 ^–pH

= 10 ^-10

= 1 x 10 ^-10M

We know equilibrium constant expression –

K = [Fe²⁺] P(H₂) / [H⁺]²

7.64 x 10¹⁴    = 1.0 x 10^-3 M x P(H₂) / (1 x 10 ^-10M)²

7.64 x 10¹⁴ x (1 x 10 ^-10M)² = 1.0 x 10^-3 M x P(H₂)

P(H₂) = 7.64 x 10¹⁴ x (1 x 10 ^-10M)² / 1.0 x 10^-3 M

         = 0.00764 atm

Partial pressure of hydrogen is must more than 0.00764 atm for reaction become unfavorable.