Question
Consider the titration of 25.0 mL of 0.1000 M Fe2+with 0.1000 M Ce4+. If a platinum electrode is used toindicate the cell potential versus a standard hydrogen electrode,what will the cell potential be after the addition of 20.0 mL ofCe4+ solution? Ce4+ + e- -->Ce3+ Eo= +1.44V Fe3+ + e- -->Fe2+ Eo= +0.77V 25oC 2.3RT/F = 0.059 I know the answer is supposed to be +0.81 V, but I really wantto know how to solve this. Thanks. Consider the titration of 25.0 mL of 0.1000 M Fe2+with 0.1000 M Ce4+. If a platinum electrode is used toindicate the cell potential versus a standard hydrogen electrode,what will the cell potential be after the addition of 20.0 mL ofCe4+ solution? Ce4+ + e- -->Ce3+ Eo= +1.44V Fe3+ + e- -->Fe2+ Eo= +0.77V 25oC 2.3RT/F = 0.059 I know the answer is supposed to be +0.81 V, but I really wantto know how to solve this. Thanks.
Explanation / Answer
Reaction at anode : Fe3+ +e- --> Fe2+ Eo= +0.77 V Reaction at cathode: Ce4+ + e- -->Ce3+ Eo= +1.44V Ce4+ + e- -->Ce3+ Eo= +1.44V Overall reaction : Ce4+ +Fe2+ --> Ce3+ + Fe3+ ; Eocell = +1.44V - 0.77V = 0.67V Initial moles of Fe2+ = (25mL)(0.1M) =2.5m mol Moles of Ce4+ addded = (20mL)(0.1M) = 2.0mmol Moles of Fe2+ left = 2.5m mol -2.0m mol = 0.5m mol [Fe2+ ] = 0.5m mol / (25 + 20)mL =0.0111M [Fe3+] = 2.0m mol / (25 +20)mL =0.0444M [Ce3+ ] = 2.0m mol / (25 +20)mL =0.0444M After mixing the two solution, w can find Fe2+ , Fe3+ , Ce3+ Ecell = Eocell - (0.059V /1)log{[Ce3+ ][Fe3+]/[Fe2+][Ce4+]} = +0.67V -(0.059V)log{(0.0444)(0.0444) / (0.0111)} = +0.71V