6. Calculate Ecell (in V to two decimal places) for an electrochemical cell based on the followir half-reactions at equilibrium. In addition, determine AG (in kJ mol¹ to two decimal place for the reaction under standard conditions (i.e. all concentrations are 1.0 M) and predict t magnitude of K. The standard reduction potential for the reduction reaction is 1.68 V. Oxidation: Cu(s) → Cu²+ (aq, 0.010 M) + 2e¯ Reduction: MnO4 (aq, 2.0 M)+ 4H+ (aq, 1.0 M) + 3e →MnO₂ (s) + 2H₂O(1) A: 1.41; -775.74, very larg
6. Calculate Ecell (in V to two decimal places) for an electrochemical cell based on the followir half-reactions at equilibrium. In addition, determine AG (in kJ mol¹ to two decimal place for the reaction under standard conditions (i.e. all concentrations are 1.0 M) and predict t magnitude of K. The standard reduction potential for the reduction reaction is 1.68 V. Oxidation: Cu(s) → Cu²+ (aq, 0.010 M) + 2e¯ Reduction: MnO4 (aq, 2.0 M)+ 4H+ (aq, 1.0 M) + 3e →MnO₂ (s) + 2H₂O(1) A: 1.41; -775.74, very larg
Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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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![### Calculation of Ecell for an Electrochemical Cell
**Problem Statement:**
1. **Calculate Ecell (in V to two decimal places)** for an electrochemical cell based on the following half-reactions at equilibrium.
2. **Determine ΔG° (in kJ/mol to two decimal places).**
3. **Predict the magnitude of the equilibrium constant, K.**
**Conditions:**
- All concentrations are 1.0 M.
- The standard reduction potential for the reduction reaction is 1.68 V.
**Half-Reactions:**
- **Oxidation:** \( \text{Cu(s)} \rightarrow \text{Cu}^{2+}(\text{aq, 0.010 M}) + 2e^- \)
- **Reduction:** \( \text{MnO}_4^-(\text{aq, 2.0 M}) + 8\text{H}^+(\text{aq, 1.0 M}) + 3e^- \rightarrow \text{MnO}_2(\text{s}) + 2\text{H}_2\text{O(l)} \)
**Given Solution:**
- \( E_{\text{cell}} = 1.41 \, \text{V} \)
- \( \Delta G^\circ = -775.74 \, \text{kJ/mol} \)
- \( K \) is very large.
These calculations are fundamental for understanding the thermodynamics of electrochemical cells and predicting the direction and spontaneity of the reactions involved.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8e77b376-6a5c-4963-870c-f49a6e984799%2F463160d6-1328-428f-b698-d960843c8e0f%2Ff4xx56d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculation of Ecell for an Electrochemical Cell
**Problem Statement:**
1. **Calculate Ecell (in V to two decimal places)** for an electrochemical cell based on the following half-reactions at equilibrium.
2. **Determine ΔG° (in kJ/mol to two decimal places).**
3. **Predict the magnitude of the equilibrium constant, K.**
**Conditions:**
- All concentrations are 1.0 M.
- The standard reduction potential for the reduction reaction is 1.68 V.
**Half-Reactions:**
- **Oxidation:** \( \text{Cu(s)} \rightarrow \text{Cu}^{2+}(\text{aq, 0.010 M}) + 2e^- \)
- **Reduction:** \( \text{MnO}_4^-(\text{aq, 2.0 M}) + 8\text{H}^+(\text{aq, 1.0 M}) + 3e^- \rightarrow \text{MnO}_2(\text{s}) + 2\text{H}_2\text{O(l)} \)
**Given Solution:**
- \( E_{\text{cell}} = 1.41 \, \text{V} \)
- \( \Delta G^\circ = -775.74 \, \text{kJ/mol} \)
- \( K \) is very large.
These calculations are fundamental for understanding the thermodynamics of electrochemical cells and predicting the direction and spontaneity of the reactions involved.
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