Calculate pE of an acid-mine water sample with [Fe3+]=3.00x10^-3 M and [Fe2+]=5.00x10^-5 M and pH=2.80. Also calculate Ehalf cell for Fe3+e->Fe2+ and the hydronium ion concentration.
Fe3+e->Fe2+ Ered=+0.771 pE=Ecal/0.05916 Ecell=Ecell-0.05916/n(log)reduced/oxidized) pH=-log (H3O) pE=pE-(1/n)(log(reduced/oxidized))
Fe3+e->Fe2+ Ered=+0.771 pE=Ecal/0.05916 Ecell=Ecell-0.05916/n(log)reduced/oxidized) pH=-log (H3O) pE=pE-(1/n)(log(reduced/oxidized))