Table 11. 4 Equilibrium relationships in Fe0-Feii-feiii-h2O system Reaction Constant Equation




Дата канвертавання19.04.2016
Памер7.94 Kb.
Table 11.4 Equilibrium relationships in Fe0-FeII-FeIII-H2O system*

Reaction Constant Equation

_________________________________________________________________________________________


Fe3+ + e- = Fe2+ pεº = 13.03 pε = 13.03 + log{Fe3+}/{Fe2+}

Fe2+ + 2e- = Fe0 pεº = −7.45 pε = −7.45 + ½log{Fe2+}

Fe3+ + 3H2O = Fe(OH)3(s) + 3H+ logK = −3.191

Fe(OH)3(s) = Fe3+ + 3OH- logKs0 = −38.80

Fe3+ + H2O = FeOH2+ + H+ logK1 = −2.19 pH = 2.19 + log{FeOH2+}/{Fe3+}

Fe3+ + 2H2O = Fe(OH)2+ + 2H+ logK = −4.59

FeOH2+ + H2O = Fe(OH)2+ + H+ logK2 = −2.40 pH = 2.40 + log{Fe(OH)2+}/{FeOH2+}

Fe(OH)3(s) + H+ = Fe(OH)2+ + H2O logK = −1.39 pH = −1.39 – log{Fe(OH)2+}

Fe(OH)3(s) + H2O = Fe(OH)4- + H+ logK = −18.40 pH = 18.4 + log{Fe(OH)4-}

Fe(OH)3(s) + 3H+ + e- = Fe2+ + 3H2O pεº = 16.22 pε = 16.22 – log{Fe2+} – 3pH

FeOH2+ + H+ + e- = Fe2+ + H2O pε = 15.22 – pH – log{Fe2+}/{FeOH2+}

Fe(OH)3(s) + H+ + e- = Fe(OH)2(s) + H2O pεº = 2.66 pε = 2.66 – pH

Fe2+ + H2O = FeOH+ + H+ logK = −9.40 pH = 9.40 + log{FeOH+}/{Fe2+}

FeOH+ + H2O = Fe(OH)2(s) + H+ logK = −4.16 pH = 4.16 – log{FeOH+}

Fe(OH)2(s) + H2O = Fe(OH)3- + H+ logK = −15.43 pH = 15.43 + log{Fe(OH)3-}

Fe(OH)2(s) + 2H+ + 2e- = Fe0(s) + 2H2O pεº = −1.34 pε = −1.34 – pH



_________________________________________________________________

* Values calculated to be consistent with MINEQL+ v.4.6 and Table 11.1; solid phase for FeIII assumed to be ferrihydrite; that for FeII assumed to be amorphous Fe(OH)2(s).


База данных защищена авторским правом ©shkola.of.by 2016
звярнуцца да адміністрацыі

    Галоўная старонка