Question

For the half reactions, determine if hydrogen sulfide can be oxidized by hydrogen peroxide.

H2S → S + 2H+ + 2e-     Eo = -0.14 V

H2O2 + 2H+ + 2e-   → 2H2O Eo = 1.776 V

Also, determine the equilibrium constant for the reaction. Assume a water temperature of 25oC.

Answer 1

The two half cell reactions are

Oxidation reaction

H2S = S + 2H+ + 2e-

Eox = -0.14 V

Oxidation number of S in H2S = - 2

S is oxidized in the reaction from - 2 to 0

Reduction reaction

H2O2 + 2H+ + 2e- = 2H2O

Ered = 1.776 V

Oxidation number of O in H2O2 = - 1

Oxidation number of O in H2O = - 2

O is reduced from - 1 to - 2

Overall cell reaction

H2S + H2O2 = S + 2H2O

H2S is oxidized by H2O2 and gives S and water

E° = Eox + Ered

= - 0.14 + 1.776

= 1.636 V

Free energy G° = - nFE°

= - 2 x 96500 C/mol x 1.636 V

= - 315748 J/mol

Equilibrium constant

K = exp (-G°/RT)

= exp (315748/8.314*298)

= exp(127.44)

= 2.22 x 10^55