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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+= 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

Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter18: Electrochemistry
Section: Chapter Questions
Problem 18.103QE
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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
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
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
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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