You are performing a titration of 25.00 mL of 0.0100 M Sn2+ in 1 M HCl with 0.0200 M Fe3+ resulting in the formation of Sn+ and Fe2+. A Pt indicator electrode and a saturated calomel electrode (SCE) reference 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: Sn+= half-reaction: Fe3+= E' = 0.139 V E=0.732 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. [Sn4+] 0.05916 E = 0.139- log -0.241 0.05916 Sn4+1 E 0.139- log Sn2+1 })] -0.241 [Sn2+] E 0.139 0.05916log -0.241 E 0.139 0.05916log [Sn4+] [Sn2+] -0.241 E=0.732 0.05916log [Fe2+1 -0.241 [Fe3+] E= 0.732 0.05916log | Fe3+ -0.241 [Fe2+] E = 0.732- log -0.241 2 [Fe3+] 0.05916 [Fe2+] E=0.732- 0.05916 [Fe3+] -log 2 [Fe2+] -0.241
You are performing a titration of 25.00 mL of 0.0100 M Sn2+ in 1 M HCl with 0.0200 M Fe3+ resulting in the formation of Sn+ and Fe2+. A Pt indicator electrode and a saturated calomel electrode (SCE) reference 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: Sn+= half-reaction: Fe3+= E' = 0.139 V E=0.732 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. [Sn4+] 0.05916 E = 0.139- log -0.241 0.05916 Sn4+1 E 0.139- log Sn2+1 })] -0.241 [Sn2+] E 0.139 0.05916log -0.241 E 0.139 0.05916log [Sn4+] [Sn2+] -0.241 E=0.732 0.05916log [Fe2+1 -0.241 [Fe3+] E= 0.732 0.05916log | Fe3+ -0.241 [Fe2+] E = 0.732- log -0.241 2 [Fe3+] 0.05916 [Fe2+] E=0.732- 0.05916 [Fe3+] -log 2 [Fe2+] -0.241
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![You are performing a titration of 25.00 mL of 0.0100 M Sn²+ in 1 M HCl with 0.0200 M Fe³+ resulting in the formation of
Sn+ and Fe2+. A Pt indicator electrode and a saturated calomel electrode (SCE) reference 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: Sn+=
half-reaction:
Fe³+=
0
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.
E = 0.139
0.05916
2
-log
E 0.139 0.05916log
E = 0.732-
E = 0.732 0.05916log
0.05916
2
-log
31
[Sn²+]
[Sn++]
[Sn²+]
[Sn++]
[Fe²+]
[Fe³+]
[Fe²+
[Fe3+]
- 0.241
- 0.241
- 0.241
- 0.241
E = 0.139-
0.05916.
log
E 0.139 0.05916log
E = 0.732 -
E = 0.732 0.05916log
[Sn++]
Sn2+1
91
[Sn++]
[Sn²+]
[Fe²+]
0.05916 [Fe³+]
-log
[Fe²+]
Calculate the cell potential (E) after each of the given volumes of the Fe³+ titrant have been added.
E = 0.139 V
E = 0.732 V
0.241
- 0.241
- 0.241
-0.241](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe6263ceb-2148-4d04-a273-847311b190a4%2F01523dfb-b535-438b-accf-d70708f369f3%2Ft6c7lkfl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:You are performing a titration of 25.00 mL of 0.0100 M Sn²+ in 1 M HCl with 0.0200 M Fe³+ resulting in the formation of
Sn+ and Fe2+. A Pt indicator electrode and a saturated calomel electrode (SCE) reference 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: Sn+=
half-reaction:
Fe³+=
0
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.
E = 0.139
0.05916
2
-log
E 0.139 0.05916log
E = 0.732-
E = 0.732 0.05916log
0.05916
2
-log
31
[Sn²+]
[Sn++]
[Sn²+]
[Sn++]
[Fe²+]
[Fe³+]
[Fe²+
[Fe3+]
- 0.241
- 0.241
- 0.241
- 0.241
E = 0.139-
0.05916.
log
E 0.139 0.05916log
E = 0.732 -
E = 0.732 0.05916log
[Sn++]
Sn2+1
91
[Sn++]
[Sn²+]
[Fe²+]
0.05916 [Fe³+]
-log
[Fe²+]
Calculate the cell potential (E) after each of the given volumes of the Fe³+ titrant have been added.
E = 0.139 V
E = 0.732 V
0.241
- 0.241
- 0.241
-0.241
Transcribed Image Text:Calculate the cell potential (E) after each of the given volumes of the Fe³+ titrant have been added.
5.00 mL
12.50 mL
24.00 mL
25.00 mL
26.00 mL
26.00 mL
50.00 mL
E =
E =
E =
E =
E =
E =
E =
V
V
V
V
V
V
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