Question

Environmental issues arising from arsenic toxicity are frequently associated with the oxidation of pyrite, which generates significant SO4^2- and acid. Write a balanced half-cell reaction for the oxidate of HS^- to SO4^2-. Assume aHS^-=aSO4^2- (a stands for activity) and plot a phase boundary between the two ions on an Eh-pH diagram. PLEASE HELP! I dont understand how to do this at all.

Thermodynamic data Delta Gf (KJ/mol)

Al^3+                                                    -485.0

Ni^2+                                                    -45.6

NH4^+                                                  -79.31

NO3^-                                                   -108.74

SO4^2-                                                 -177.75

HS^-                                                      2.93

H2O                                                       -237.129

Answer 1

You have the reaction:

$4H_2O+HS^{-}\rightarrow SO_4^{2-}+9H^++8e^{-}$

Using the energies of formation you can calculate the energy of reaction

$\Delta G_{rxn}^{o}=\Delta G^{f}_{SO_4^{2-}}+9\Delta G^{f}_{H^+}+8\Delta G^{f}_{e^{-}}-4\Delta G^{f}_{H_2O}-\Delta G^{f}_{HS^-}$

The energy for the electron is negligible and for the H+ is 0 thus, the energy for water is -237.129 kJ/mol, for SO4 -744.53 kJ/mol and for HS- 12.05kJ/mol, I am using from my own data because you are mixing some in kcal and other is kJ

$\Delta G_{rxn}^{o}=-744.53-4\times(-237.129)-(12.05)=191.936kJ/mol$

Now you can calculate the potential:

$\Delta G_{rxn}^{o}=-nFE$

$E^o=-\frac{\Delta G^{o}_{rxn}}{nF}=-\frac{191936J/mol}{(8)(96485C/mol)}=-0.2487V$

Using the nernst equation:

$E=E^{o}-\frac{0.0592}{n}\log Q$

n is the electrons transfered

$E=-0.2487-\frac{0.0592}{8}\log \frac{[SO_4^{2-}][H^+]^9}{[HS^{-}]}$<-----You can rearrange the [H+] concentration into pH, and assuming the [SO4]=[HS]

$E=-0.2487-0.0074\log [H^+]^9$<-----using the logarithm rules, the 9 is the coefficient of the log and form the definition pH=-log[H+], yields:

$E=-0.2487+0.0666pH$<------This is the equation of the line that divides the two ions in a Eh-pH diagram

Eh-pH 0.65 0.55 0.45 HS 0.35 2 0.25 0.15 0.05 0.05 0.15 0.25 SO42 0 1 23 4 5 67 8 910 11 12 13 14 pH