Question

A 100.0 mL solution of 0.0400 M Fe2+ in 1 M HCIO s titrated with 0.100 M Ce4+ resulting in the formation of Fe3+ and Ce3+. A Pt indicator electrode and a saturated calomel electrode are used to monitor the titration. Write the balanced titration reaction titration reaction:> Complete the two half-reactions that occur at the Pt indicator electrode. Write the half-reactions as reductions half-reaction: Ce e half-reaction: Fe e- E 0.767 V Select the two equations that can be used to determine the cell voltage at different points in the titration. E of the saturated calomel electrode is 0.241 V. 0.767 -0.0591dlog e 0.241 E 0.767 - 0.05916log 0.241 Fe3+] Fe21)-0.241 [Fe2+] рез*) )-0.241 E 1.70-0.059 1610g . E 1.70-0.059 1610g [Ce3+1 )-0.241 Ce4+ [Fe2+) ce1-0241 E 0.767 - 0.05916log E 0.767 - 0.05916log

Answer 1

& Tbh Reachm reachon eM - 0.241

The Ce⁴⁺ is reduced to Ce³⁺ in the net reaction. Hence the standard reduction potential of Ce⁴⁺|Ce³⁺ should be more than that of iron ion electrode.

Electrode potential with respect to SCE is equal to electrode potential with respect to SHE minus the electrode potential of SCE.

E_{half electrode} = E⁰_{half reduction} - (0.05926/n)log Q .