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.
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.
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
Related questions
Question
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F115ad5cc-4a4a-4beb-be95-bcb8ae603251%2F39c41b18-4492-4e8e-b391-1332487b02aa%2Fkcv8p7l.png&w=3840&q=75)
Transcribed Image Text: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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F115ad5cc-4a4a-4beb-be95-bcb8ae603251%2F39c41b18-4492-4e8e-b391-1332487b02aa%2Fvbpdub6.png&w=3840&q=75)
Transcribed Image Text: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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