MISSED THIS? Read Section 20.6 (Page). A voltaic cell employs the following redox reaction: 2 Fe³+ (aq) + 3 Mg (s) → 2 Fe (s) + 3 Mg²+ (aq) Calculate the cell potential at 25 °C under each of the following conditions. ▼ standard conditions Express your answer in units of volts. Ecell = Submit Part B Ecell = [Fe³+] = 2.5x10-3 M: [Mg²+] = 2.05 M Express your answer in units of volts. Submit [ΠΙ ΑΣΦ Part C Request Answer ΠΫΠΙ ΑΣΦ K Request Answer = 2.05 M [Mg²+] = 2.5x10-3 M [Fe³+] = Express your answer in units of volts. ? ? V V
MISSED THIS? Read Section 20.6 (Page). A voltaic cell employs the following redox reaction: 2 Fe³+ (aq) + 3 Mg (s) → 2 Fe (s) + 3 Mg²+ (aq) Calculate the cell potential at 25 °C under each of the following conditions. ▼ standard conditions Express your answer in units of volts. Ecell = Submit Part B Ecell = [Fe³+] = 2.5x10-3 M: [Mg²+] = 2.05 M Express your answer in units of volts. Submit [ΠΙ ΑΣΦ Part C Request Answer ΠΫΠΙ ΑΣΦ K Request Answer = 2.05 M [Mg²+] = 2.5x10-3 M [Fe³+] = Express your answer in units of volts. ? ? V V
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Educational Content: Calculating Cell Potential in a Voltaic Cell
#### Overview
This exercise involves calculating the cell potential for a given voltaic cell reaction at 25 °C. The reaction used is:
\[ 2 \text{Fe}^{3+} (\text{aq}) + 3 \text{Mg} (\text{s}) \rightarrow 2 \text{Fe} (\text{s}) + 3 \text{Mg}^{2+} (\text{aq}) \]
You are required to calculate the cell potential under various conditions.
---
#### Part A: Standard Conditions
- **Problem**: Calculate the cell potential under standard conditions.
In this setup, you'll compute the value of \(\text{E}_{\text{cell}}\) in volts. Use standard reduction potentials and the Nernst equation where needed to determine the potential.
---
#### Part B: Given Concentrations
- **Conditions**:
- \([\text{Fe}^{3+}] = 2.5 \times 10^{-3} \, \text{M}\)
- \([\text{Mg}^{2+}] = 2.05 \, \text{M}\)
- **Task**: Express the cell potential (\(\text{E}_{\text{cell}}\)) in volts with the given concentrations.
Use the Nernst equation to account for non-standard conditions, considering the activity of ions in solution.
---
#### Part C: Reversed Concentrations
- **Conditions**:
- \([\text{Fe}^{3+}] = 2.05 \, \text{M}\)
- \([\text{Mg}^{2+}] = 2.5 \times 10^{-3} \, \text{M}\)
- **Task**: Again, express \(\text{E}_{\text{cell}}\) in volts with these concentrations.
The reversal in concentration compared to Part B will cause a change in the calculated cell potential. Apply the Nernst equation as required.
---
### Calculation Guidance
- **Nernst Equation**: Used to find cell potential under non-standard conditions:
\[ \text{E}_{\text{cell}} = \text{E}^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q \]
Where:
- \(\text{E}^\circ_{\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5a897ece-3308-4566-8055-790e7cb4e78a%2F2e78bf12-9c5d-4926-8cbe-e6f8eca157fa%2Fab92odo_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Educational Content: Calculating Cell Potential in a Voltaic Cell
#### Overview
This exercise involves calculating the cell potential for a given voltaic cell reaction at 25 °C. The reaction used is:
\[ 2 \text{Fe}^{3+} (\text{aq}) + 3 \text{Mg} (\text{s}) \rightarrow 2 \text{Fe} (\text{s}) + 3 \text{Mg}^{2+} (\text{aq}) \]
You are required to calculate the cell potential under various conditions.
---
#### Part A: Standard Conditions
- **Problem**: Calculate the cell potential under standard conditions.
In this setup, you'll compute the value of \(\text{E}_{\text{cell}}\) in volts. Use standard reduction potentials and the Nernst equation where needed to determine the potential.
---
#### Part B: Given Concentrations
- **Conditions**:
- \([\text{Fe}^{3+}] = 2.5 \times 10^{-3} \, \text{M}\)
- \([\text{Mg}^{2+}] = 2.05 \, \text{M}\)
- **Task**: Express the cell potential (\(\text{E}_{\text{cell}}\)) in volts with the given concentrations.
Use the Nernst equation to account for non-standard conditions, considering the activity of ions in solution.
---
#### Part C: Reversed Concentrations
- **Conditions**:
- \([\text{Fe}^{3+}] = 2.05 \, \text{M}\)
- \([\text{Mg}^{2+}] = 2.5 \times 10^{-3} \, \text{M}\)
- **Task**: Again, express \(\text{E}_{\text{cell}}\) in volts with these concentrations.
The reversal in concentration compared to Part B will cause a change in the calculated cell potential. Apply the Nernst equation as required.
---
### Calculation Guidance
- **Nernst Equation**: Used to find cell potential under non-standard conditions:
\[ \text{E}_{\text{cell}} = \text{E}^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q \]
Where:
- \(\text{E}^\circ_{\
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