6. The following redox titration of 30.00 mL of 0.0800 M Fe? with 0.100 M Ce\" i

Question

6. The following redox titration of 30.00 mL of 0.0800 M Fe? with 0.100 M Ce" is being monitored with a silver-silver chloride reference electrode Titration Reaction: Fe2+(aq) + Ce4+(aq) Fe"(aq) + Ce-(aq) Monitoring cell: Ag(s), Agci(s)| Cr(saturated) I titration reaction (a) PtGs) (a) (6 pts) There are two "split" Nernst Equations that would be used to calculate the monitoring cell potential. Derive both equations. Tell which is used before the equivalence point and which is used after it. (b) (4 points) Derive the Nernst equation for the monitoring cell that would be used AT the equivalence point. Then use it to calculate the cell potential at the equivalence point. (e) (4 points) Calculate the cell potential for the titration after 20.00 mL of 0.100 M Ce have been added.

Explanation / Answer

Titration reaction

Ce4+ + Fe2+ Ce3+ + Fe3+

Ag/AgCl ref. electrode

a) At the cathode

Ce4+ + e- <---> Ce3+         E° = 1.70 V

Fe3+ + e- <---> Fe2+           E° = 0.767 V

At the anode

Cl- + Ag(s) <----> AgCl(s) + e- E = 0.197

Overall cell reaction

Ce4+ + Cl- + Ag(s) <---> AgCl(s) + Ce3+      (1)

Fe3+ + Cl- + Ag(s) <----> AgCl(s) + Fe2+      (2)

Cell reaction (1) is used after equivalence point and Cell reaction (2) is used before equivalence point.

b)

Titrate 30.00 ml of 0.0800 M Fe2+ with 0.100 M Ce4+

At equivalence point volume of Ce4+ = 24.00 ml

E cathode = 1.70 - 0.05916 log ([Ce3+]/[Ce4+]) = 0.797 - 0.05916 log([Fe2+]/[Fe3+])

E anode = 0.197 V

Ecell = E cathode + E anode

E cathode = 1.70 - 0.05916 log ([Ce3+]/[Ce4+]) = 0.797 - 0.05916 log([Fe2+]/[Fe3+])

2Eo cathode = Eo cathode (Ce) + Eo cathode (Fe) - 0.05916 log ([Ce3+][Fe2+]/[Ce4+][Fe3+])

At the equivalence point [Ce3+] = [Fe2+], and [Ce4+] = [Fe3+]

Eo cathode = (Ecathode°(Ce) + Ecathode°(Fe))/2

      = (1.70 + 0.797)V/2 = 1.25 V

Ecell = 1.25 – 0.197 = 1.05 V

c)

Ecell when 20.00 ml 0.100 M Ce4+added

[Fe2+] = {(30.00 ml)*(0.0800 M) – (20.00 ml)*(0.1000 M)} / (50.00 ml)

           = {0.0024 – 0.002}/0.05L

           = 0.008 M

[Fe3+] = (20.00 ml)*(0.1000 M) / (50.00 ml)

          = 0.04 M

Ecell = 0.797 - 0.05916 log(0.008/0.04) - 0.197

Ecell = 0.797 - 0.05916 (-0.69897) - 0.197

Ecell = 0.797 + 0.0413 - 0.197 = 0.641 V

Related Questions

Navigate

• Browse (All)
• Browse 6
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.