Analytical Chem istry - DePauw University

486 Analytical Chemistry 2.0So l u t i o nThe half-reactions for Fe 2+ and MnO 4 – areFe( aq) → Fe ( aq)+ e2+ 3+ −− + −2+MnO ( aq) + 8H ( aq)+ 5e → Mn ( aq) + 4H O()l42for which the Nernst equations are2+oFeE = E3+ 2+−0. 05916log [ ]Fe / Fe3+[ Fe ]2+o 0.05916 [ Mn ]E = E− 2+− logMnO Mn− +4 /5 [ MnO ][ H ]Before adding these two equations together we must multiply the secondequation by 5 so that we can combine the log terms; thus2+ 2+oo[ Fe ][ Mn ]6E = E3+ 2+ + 5E− 2+−0. 05916logFe / FeMnO4/ Mn3+ − +[ Fe ][ MnO ][ H ]At the equivalence point we know that2+ −[ Fe ] = 5×[ MnO ]3+ 2+[ Fe ] = 5×[ Mn ]Substituting these equalities into the previous equation and rearranginggives us a general equation for the potential at the equivalence point.− 2+56E = oE + o3 25E−[MnO ][ Mn ]420 05916eq + + − +. logFe / FeMnO4/ Mn2+ − +5[ Mn ][ MnO ][ H ]EEeqeqE=E=Eeq+ 5Eoo3+ 2+ − 2+Fe / FeMnO4/ Mn .6+ 5E−oo3+ 2+ − 2+Fe / FeMnO4/ Mn .E=6+ 5Eoo3+ 2+ − 2+Fe / FeMnO / Mn6440 0591660 05916×8+6841log[ H + ]84−0.07888pH4log[ H + ]88Instead of standard state potentials, youcan use formal potentials.Our equation for the equivalence point has two terms. The first term isa weighted average of the titrand’s and the titrant’s standard state potentials,in which the weighting factors are the number of electrons in theirrespective half-reactions. The second term shows that E eq for this titration