termine E°, AG" , and K for the overall reaction from the balanced half-reactions and their standard reduction potentials. 4 Fe3+ + 2H,0 2 4 Fe²+ + O, + 4 H+ Fe3+ + e-= Fe2+ 0, + 2 H* + 2 e¯ H,0 E° = 0.771 V E' = 1.229 V
termine E°, AG" , and K for the overall reaction from the balanced half-reactions and their standard reduction potentials. 4 Fe3+ + 2H,0 2 4 Fe²+ + O, + 4 H+ Fe3+ + e-= Fe2+ 0, + 2 H* + 2 e¯ H,0 E° = 0.771 V E' = 1.229 V
Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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![**Determining \( E^\circ \), \( \Delta G^\circ \), and \( K \) for the Overall Reaction**
To find the standard cell potential (\( E^\circ \)), Gibbs free energy change (\( \Delta G^\circ \)), and equilibrium constant (\( K \)) for the overall reaction, we begin with the given balanced half-reactions and their standard reduction potentials.
The overall reaction is:
\[ 4 \text{Fe}^{3+} + 2 \text{H}_2\text{O} \rightleftharpoons 4 \text{Fe}^{2+} + \text{O}_2 + 4 \text{H}^+ \]
**Half-Reactions and Standard Reduction Potentials:**
1. **Reduction of Iron:**
\[ \text{Fe}^{3+} + e^- \rightleftharpoons \text{Fe}^{2+} \]
\[ E^\circ = 0.771 \, \text{V} \]
2. **Reduction of Oxygen:**
\[ \frac{1}{2} \text{O}_2 + 2 \text{H}^+ + 2 e^- \rightleftharpoons \text{H}_2\text{O} \]
\[ E^\circ = 1.229 \, \text{V} \]
These half-reactions represent the basic redox processes taking place in the overall reaction. The standard reduction potentials \( E^\circ \) are crucial for calculating the standard cell potential of the full reaction.
**Calculation Overview:**
- By using these half-reactions and their potentials, one can calculate the overall standard cell potential (\( E^\circ \)).
- Using \( E^\circ \), the Gibbs free energy change (\( \Delta G^\circ \)) can be found with the relation:
\[ \Delta G^\circ = -nFE^\circ \]
where \( n \) is the number of moles of electrons transferred and \( F \) is Faraday's constant.
- The equilibrium constant (\( K \)) can be determined from \( \Delta G^\circ \) using:
\[ \Delta G^\circ = -RT \ln K \]
These calculations allow us to understand the energetic and equilibrium characteristics of the redox reaction.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3ac3995c-903b-421c-80b6-6f6642a05fa9%2F03f8bf84-635e-45b5-be65-e56752a1dc3e%2Fclp6ffb_processed.png&w=3840&q=75)
Transcribed Image Text:**Determining \( E^\circ \), \( \Delta G^\circ \), and \( K \) for the Overall Reaction**
To find the standard cell potential (\( E^\circ \)), Gibbs free energy change (\( \Delta G^\circ \)), and equilibrium constant (\( K \)) for the overall reaction, we begin with the given balanced half-reactions and their standard reduction potentials.
The overall reaction is:
\[ 4 \text{Fe}^{3+} + 2 \text{H}_2\text{O} \rightleftharpoons 4 \text{Fe}^{2+} + \text{O}_2 + 4 \text{H}^+ \]
**Half-Reactions and Standard Reduction Potentials:**
1. **Reduction of Iron:**
\[ \text{Fe}^{3+} + e^- \rightleftharpoons \text{Fe}^{2+} \]
\[ E^\circ = 0.771 \, \text{V} \]
2. **Reduction of Oxygen:**
\[ \frac{1}{2} \text{O}_2 + 2 \text{H}^+ + 2 e^- \rightleftharpoons \text{H}_2\text{O} \]
\[ E^\circ = 1.229 \, \text{V} \]
These half-reactions represent the basic redox processes taking place in the overall reaction. The standard reduction potentials \( E^\circ \) are crucial for calculating the standard cell potential of the full reaction.
**Calculation Overview:**
- By using these half-reactions and their potentials, one can calculate the overall standard cell potential (\( E^\circ \)).
- Using \( E^\circ \), the Gibbs free energy change (\( \Delta G^\circ \)) can be found with the relation:
\[ \Delta G^\circ = -nFE^\circ \]
where \( n \) is the number of moles of electrons transferred and \( F \) is Faraday's constant.
- The equilibrium constant (\( K \)) can be determined from \( \Delta G^\circ \) using:
\[ \Delta G^\circ = -RT \ln K \]
These calculations allow us to understand the energetic and equilibrium characteristics of the redox reaction.
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