Question

Problem: The formal potential of the Fe(III)/Fe(II) couple in three acids at pH=0 were measured as:

HCl: +524 mV vs. Hg/Hg₂Cl₂ in saturated KCl (SCE)

H₂SO₄: +39 mV vs. Hg/Hg₂SO₄ in saturated K₂SO₄

H₃PO₄: +438 mV vs. SHE

The potentials of the saturated calomel and saturated mercurous sulfate reference electrodes are +246 mV and +640 mV vs. SHE, respectively.

a) Explain how would you measure these formal potentials.

b) In which medium is Fe(III) the strongest oxidizing agent? explain.

Answer 1

a) Prepare solutions in each of acids , pH=0 with equal conc of Fe(III) and Fe(II) say (5 mM) . Place each solution ito a cell with an inert working electrode and the reference electrode. Remove O2 with N2 bubbling throught solution . Measure the Potential difference between two eletrodes with a high impedance digital voltmeter

b) The formal potentials must be converted into values versus a single reference electrode .

                Ec Vs   SHE/mV

Hcl   524+246 = 770

H2SO4 39+640 = 679

H3PO4   438

Fe(III) is stronger oxidising agent in HCl since this is medium where formal potential is most positive