Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Transcribed Image Text:The following cell notation contains two half-reactions.
Pt | Fe2+ (aq), Fe3+ (aq) || Sn4+ (aq), Sn²+ (aq) | Pt
Which of the following statement/equation correctly describes the two half-reactions?
anode: Sn4+ (aq)+2e→Sn²+ (aq); cathode: Fe2+(aq)-e→Fe³+(aq);
O anode: Fe²+ (aq)-e→Fe³+(aq); cathode: Sn²+ (aq)+2e→Sn4+ (aq)
anode: Fe³+ (aq)+e→Fe2+ (aq); cathode: Sn²+ (aq)+2e→Sn³+ (aq)
anode: Pt + e → Fe³+ (aq)+Fe2+ (aq); cathode: Sn4+ (aq)+Sn²+ (aq) +2e→Pt
O anode: Fe²+ (aq)-e→Fe³+ (aq); cathode: Sn4+ (aq)+2e-Sn²+ (aq)
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Follow-up Questions
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Follow-up Question
Using data provided in the table of Standard Potentials, calculate standard emf of the cell described in the last question. Please enter your answer with 2 decimals. for example, 1.056 is written as 1.06. Pay attention to the sign.
this is in par to this question thread
Transcribed Image Text:**Half-Reaction Table and Electrode Potentials**
This table presents various half-reactions along with their standard electrode potentials (E°) measured in volts (V). These potentials indicate the tendency of a chemical species to be reduced—the higher the value, the stronger the oxidizing agent. Conversely, lower values imply stronger reducing agents. The table is organized to show this transition from oxidizing to reducing agents.
### Table Details:
#### Half-Reactions and E° Values:
1. **F₂(g) + 2e⁻ → 2F⁻(aq)**
*E° = +2.87 V*
2. **Cl₂(g) + 2e⁻ → 2Cl⁻(aq)**
*E° = +1.36 V*
3. **MnO₂(g) + 4H⁺(aq) + 2e⁻ → Mn²⁺(aq) + 2H₂O(l)**
*E° = +1.23 V*
4. **NO₃⁻(aq) + 4H⁺(aq) + 3e⁻ → NO(g) + 2H₂O(l)**
*E° = +0.96 V*
5. **Ag⁺(aq) + e⁻ → Ag(s)**
*E° = +0.80 V*
6. **Fe³⁺(aq) + e⁻ → Fe²⁺(aq)**
*E° = +0.77 V*
7. **O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)**
*E° = +0.40 V*
8. **Cu²⁺(aq) + 2e⁻ → Cu(s)**
*E° = +0.34 V*
9. **2H⁺(aq) + 2e⁻ → H₂(g)**
*E° = 0.00 V*
10. **N₂(g) + 5H⁺(aq) + 4e⁻ → N₂H₇⁺(aq)**
*E° = −0.23 V*
11. **Fe²⁺(aq) + 2e⁻ → Fe(s)**
Solution
by Bartleby Expert
Follow-up Question
Transcribed Image Text:Based on the cell notation given in the last question, write the correct cell reaction?
O Fe³+ (aq) + Sn²+ (aq)
O Fe²+ (aq) + Fe3+ (aq)
O 2Fe2+ (aq) + Sn4+ (aq)
O 2Fe3+ (aq) + Sn²+ (aq)
Fe2+ (aq) + Sn4+ (aq)
Fe2+ (aq) + Sn4+ (aq)
› Sn4+ (aq) + Sn2+ (aq)
2Fe3+ (aq) + Sn2+ (aq)
2Fe2+ (aq) + Sn4+ (aq)
Fe3+ (aq) + Sn²+ (aq)
O
Solution
by Bartleby Expert
Follow-up Questions
Read through expert solutions to related follow-up questions below.
Follow-up Question
Using data provided in the table of Standard Potentials, calculate standard emf of the cell described in the last question. Please enter your answer with 2 decimals. for example, 1.056 is written as 1.06. Pay attention to the sign.
this is in par to this question thread
Transcribed Image Text:**Half-Reaction Table and Electrode Potentials**
This table presents various half-reactions along with their standard electrode potentials (E°) measured in volts (V). These potentials indicate the tendency of a chemical species to be reduced—the higher the value, the stronger the oxidizing agent. Conversely, lower values imply stronger reducing agents. The table is organized to show this transition from oxidizing to reducing agents.
### Table Details:
#### Half-Reactions and E° Values:
1. **F₂(g) + 2e⁻ → 2F⁻(aq)**
*E° = +2.87 V*
2. **Cl₂(g) + 2e⁻ → 2Cl⁻(aq)**
*E° = +1.36 V*
3. **MnO₂(g) + 4H⁺(aq) + 2e⁻ → Mn²⁺(aq) + 2H₂O(l)**
*E° = +1.23 V*
4. **NO₃⁻(aq) + 4H⁺(aq) + 3e⁻ → NO(g) + 2H₂O(l)**
*E° = +0.96 V*
5. **Ag⁺(aq) + e⁻ → Ag(s)**
*E° = +0.80 V*
6. **Fe³⁺(aq) + e⁻ → Fe²⁺(aq)**
*E° = +0.77 V*
7. **O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)**
*E° = +0.40 V*
8. **Cu²⁺(aq) + 2e⁻ → Cu(s)**
*E° = +0.34 V*
9. **2H⁺(aq) + 2e⁻ → H₂(g)**
*E° = 0.00 V*
10. **N₂(g) + 5H⁺(aq) + 4e⁻ → N₂H₇⁺(aq)**
*E° = −0.23 V*
11. **Fe²⁺(aq) + 2e⁻ → Fe(s)**
Solution
by Bartleby Expert
Follow-up Question
Transcribed Image Text:Based on the cell notation given in the last question, write the correct cell reaction?
O Fe³+ (aq) + Sn²+ (aq)
O Fe²+ (aq) + Fe3+ (aq)
O 2Fe2+ (aq) + Sn4+ (aq)
O 2Fe3+ (aq) + Sn²+ (aq)
Fe2+ (aq) + Sn4+ (aq)
Fe2+ (aq) + Sn4+ (aq)
› Sn4+ (aq) + Sn2+ (aq)
2Fe3+ (aq) + Sn2+ (aq)
2Fe2+ (aq) + Sn4+ (aq)
Fe3+ (aq) + Sn²+ (aq)
O
Solution
by Bartleby Expert
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