Answered: A 100.0 mL solution of 0.0200 M Fe+ in…

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter17: Electrochemistry

Section: Chapter Questions

Problem 59QAP

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A 100.0 mL100.0 mL solution of 0.0200 M Fe3+0.0200 M Fe3+ in 1 M HClO41 M HClO4 is titrated with 0.100 M Cu+0.100 M Cu+, resulting in the formation of Fe2+Fe2+ and Cu2+Cu2+. A PtPt indicator electrode and a saturated Ag∣∣AgClAg|AgCl electrode are used to monitor the titration.

Write the balanced titration reaction.

titration reaction:

Fe3++Cu+⟶Fe2++Cu2+Fe3++Cu+⟶Fe2++Cu2+

Complete the two half‑reactions that occur at the PtPt indicator electrode. Write the half‑reactions as reductions.

half‑reaction:

?∘=0.161 V

half‑reaction:

?∘=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 Ag∣∣AgClAg|AgClelectrode is 0.197 V.0.197 V.

?=0.767 V−0.05916×log([Cu2+][Cu+])−0.197 V

E=0.767 V−0.05916×log([Cu2+][Cu+])−0.197 V

?=0.767 V−0.05916×log([Fe3+][Fe2+])−0.197 V

E=0.767 V−0.05916×log([Fe3+][Fe2+])−0.197 V

?=0.767 V−0.05916×log([Cu+][Cu2+])−0.197 V

E=0.767 V−0.05916×log([Cu+][Cu2+])−0.197 V

?=0.161 V−0.05916×log([Cu+][Cu2+])−0.197 V

E=0.161 V−0.05916×log([Cu+][Cu2+])−0.197 V

?=0.161 V−0.05916×log([Fe3+][Fe2+])−0.197 V

E=0.161 V−0.05916×log([Fe3+][Fe2+])−0.197 V

?=0.767 V−0.05916×log([Fe2+][Fe3+])−0.197 V

E=0.767 V−0.05916×log([Fe2+][Fe3+])−0.197 V

?=0.161 V−0.05916×log([Cu2+][Cu+])−0.197 V

E=0.161 V−0.05916×log([Cu2+][Cu+])−0.197 V

?=0.161 V−0.05916×log([Fe2+][Fe3+])−0.197 V

E=0.161 V−0.05916×log([Fe2+][Fe3+])−0.197 V

Calculate the values of E for the cell after each of the given volumes of the Cu+Cu+ titrant have been added.

1.50 mL?=

10.0 mL?=

18.5 mL?=

20.0 mL?=

22.0 mL?=

40.0 mL?=

A 100.0 mL solution of 0.0200 M Fe+ in 1 M HCIO, is titrated with 0.100 M Cu*, resulting in the formation of Fe2+ and
Cu2+. A Pt indicator electrode and a saturated Ag|AgCl electrode are used to monitor the titration.
Write the balanced titration reaction.
titration reaction: Fe3+ + Cu+
Fe2+ + Cu?+
Complete the two half-reactions that occur at the Pt indicator electrode. Write the half-reactions as reductions.
half-reaction:
E° = 0.161 V
half-reaction:
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 Ag AgCl
electrode is 0.197 V.
E = 0.767 V – 0.05916 x log ()
[Cu²*]
[Cu*]
[Fe*]
E = 0.767 V – 0.05916 × log (
[Fe²*]
- 0.197 V
- 0.197 V
2+]
[Cu*]
E = 0.767 V – 0.05916 × log ( ) -
[Cu²*]
[Cu*]
E = 0.161 V – 0.05916 × log (ar21
-) - 0.197 V
0.197 V
E = 0.161 V – 0.05916 × log( Fe
[Fe2*]
[Fe2+]
E = 0.767 V – 0.05916 × log ( r -
Fe+]
-0.197 V
- 0.197 V
[Cu²*]
E = 0.161 V – 0.05916 × log (
(Cu*]
[Fe2+]
E = 0.161 V – 0.05916 × log ( i) -
[Fe**]
0.197 V
0.197 V
Calculate the values of E for the cell after each of the given volumes of the Cut titrant have been added.

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