Question

In a potentiometric titration of iron with cerium, the reaction is Fe2+ + Ce4+ --> Ce3+ + Fe3+.

After collecting data it was found that 0.8925 grams of iron could be titrated with 20.95 milliliters of 0.04945 Molarity cerium(IV) to reach an end point with a potential of 0.806 Volts.

Using the Nernst equation    [Ecell = {E^o (Fe) - 0.0592 log {(21.05 - Volume added)/Volume added}} - Eref],    calculate values of formal electrode potentials for reduction of iron and cerium.

Answer 1

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Now there are many reference electrodes but usually used one is standared calomel electrod (SCE) with electrode potential of'0.242 v' i.e East = 0.242 v & Given rightarrow Edegree_Fe = 0.806v Ecell = 0.806-0.0592 log[21.05-20.95/20.95]-0.242 = (0.806-0.242)-0.592(-2.3211) = (0.806-0.242) + 0.137 Ecell = 0.701