Question

(1) Deduce the expressions relating pE to pH, the concentration of sulfate ion, and the partial pressure of hydrogen sulfide gas, given that for the half-reaction, pE° = —3.50 V when the pH is 7.0.

(2) Deduce the partial pressure of hydrogen sulfide when the sulfate ion concentration is 1Q^5 M and the pH is 6.0 for water that is in equilibrium with atmospheric oxygen.

Answer 1

1) First balance the half reaction that converts SO4^-2 to H2S in acidic conditions

10H⁺ + SO4^-2 + 8e^- H2S +4H2O

From the above reaction to 1 electron , devide by 8

10/8 H⁺ +1/8 SO4^-2 + 8/8e^- 1/8 H2S + 4/8 H2O

5/4 H⁺ +1/8 SO4^-2 + e^- 1/8 H2S + 1/2 H2O

Given pE⁰ =+3.50

Deduce the expression relating pE to pH

pE = pE⁰ -log(P_H28^1/8 /( SO4^-2)^1/8 (H⁺)^5/4)

      =pE⁰ - (5/4) (log[H⁺]) - (1/8) log (P_H28/[SO4^-2])

       =pE⁰ - (5/4) pH - (1/8) log (P_H28/[SO4^-2])

     pE = 3.50 - (5/4) pH - (1/8) log (P_H28/[SO4^-2])

2)

Rearrange the equation

-(1/8) log (P_H28/[SO4^-2]) = pE + pE⁰ + 5/4 pH

we already know that

pE = 20.75 - pH + (1/4) log P₀₂ (For any solution in equillibrium with atmospheric oxygen)

Given that pH = 6, P_o2 =0.21 , substitue the values

pE = 20.75 - 6 + (1/4) log (0.21)

      = 14.75 + 0.25(-0.677)

      =+ 14.32