Chapter Home Pes) Calcolate the eauiü briem (castant for te dispro portionaticn of ne T ron (II) :on at a soom temperature F2ig3+2e>2F Tag) E°(V)=+2-87 → Fers) +2 Fes"caq) Feztra2> Fecs) E CU)=0.41 3 Fe2+caq>
Chapter Home Pes) Calcolate the eauiü briem (castant for te dispro portionaticn of ne T ron (II) :on at a soom temperature F2ig3+2e>2F Tag) E°(V)=+2-87 → Fers) +2 Fes"caq) Feztra2> Fecs) E CU)=0.41 3 Fe2+caq>
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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![**Homework: Chapter 20**
**20. (03) Calculate the equilibrium constant for the disproportionation of the iron (II) ion at room temperature.**
\[ 3 \text{Fe}^{2+} (\text{aq}) \rightarrow \text{Fe}(\text{s}) + 2 \text{Fe}^{3+} (\text{aq}) \]
**Reactions and Standard Electrode Potentials:**
1. \[ \text{Fe}_{2}(\text{s}) + 2 \text{e}^- \rightarrow 2 \text{Fe} (\text{aq}), \quad E^\circ (\text{V}) = +2.07 \]
2. \[ \text{Fe}^{2+} (\text{aq}) + 2 \text{e}^- \rightarrow \text{Fe}(\text{s}), \quad E^\circ (\text{V}) = -0.44 \]
3. \[ \text{Fe}^{3+} (\text{aq}) + \text{e}^- \rightarrow \text{Fe}^{2+} (\text{aq}), \quad E^\circ (\text{V}) = 0.77 \]
---
**Explanation:**
This text is a handwritten homework problem about calculating the equilibrium constant for the disproportionation of iron (II) ions at room temperature. It includes a chemical reaction and a series of electrochemical reactions with their corresponding standard electrode potentials.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6ef85c79-8e8f-4d9a-890a-b67fe0c4488a%2F8ab97c14-49dd-496b-8da6-500d0a8860d8%2Fticqa3r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Homework: Chapter 20**
**20. (03) Calculate the equilibrium constant for the disproportionation of the iron (II) ion at room temperature.**
\[ 3 \text{Fe}^{2+} (\text{aq}) \rightarrow \text{Fe}(\text{s}) + 2 \text{Fe}^{3+} (\text{aq}) \]
**Reactions and Standard Electrode Potentials:**
1. \[ \text{Fe}_{2}(\text{s}) + 2 \text{e}^- \rightarrow 2 \text{Fe} (\text{aq}), \quad E^\circ (\text{V}) = +2.07 \]
2. \[ \text{Fe}^{2+} (\text{aq}) + 2 \text{e}^- \rightarrow \text{Fe}(\text{s}), \quad E^\circ (\text{V}) = -0.44 \]
3. \[ \text{Fe}^{3+} (\text{aq}) + \text{e}^- \rightarrow \text{Fe}^{2+} (\text{aq}), \quad E^\circ (\text{V}) = 0.77 \]
---
**Explanation:**
This text is a handwritten homework problem about calculating the equilibrium constant for the disproportionation of iron (II) ions at room temperature. It includes a chemical reaction and a series of electrochemical reactions with their corresponding standard electrode potentials.
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