2Hg²+ (aq) + 2e Hg₂²+ (aq) + 2e¯ calculate AG Hg₂+ (aq) Mac Hg₂ 2+ (aq) 2Hg(1) = 0.92 V Eº = 0.85 V 0 for the following process at 25°C: Hg2+ (aq) + Hg(1) kJ/mol
2Hg²+ (aq) + 2e Hg₂²+ (aq) + 2e¯ calculate AG Hg₂+ (aq) Mac Hg₂ 2+ (aq) 2Hg(1) = 0.92 V Eº = 0.85 V 0 for the following process at 25°C: Hg2+ (aq) + Hg(1) kJ/mol
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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![### Electrochemical Reactions and Gibbs Free Energy Calculation
#### Given Reactions:
1. \( 2 \text{Hg}^{2+} (\text{aq}) + 2 \text{e}^- \rightarrow \text{Hg}_2^{2+} (\text{aq}) \)
- Standard electrode potential, \( E^\circ = 0.92 \, \text{V} \)
2. \( \text{Hg}_2^{2+} (\text{aq}) + 2 \text{e}^- \rightarrow 2 \text{Hg} (\text{l}) \)
- Standard electrode potential, \( E^\circ = 0.85 \, \text{V} \)
### Task:
**Calculate \(\Delta G^\circ\) for the following process at 25°C:**
\[ \text{Hg}_2^{2+} (\text{aq}) \rightarrow \text{Hg}^{2+} (\text{aq}) + \text{Hg} (\text{l}) \]
### Explanation:
To calculate the Gibbs Free Energy (\(\Delta G^\circ\)), you will use the standard electrode potentials from the given half-reactions. Apply the formula:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]
where:
- \( n \) is the number of moles of electrons transferred.
- \( F \) is the Faraday constant (approximately \( 96485 \, \text{C/mol} \)).
- \( E^\circ_{\text{cell}} \) is the standard cell potential, which can be calculated by combining the given electrode potentials.
**Note:** Enter your calculated \(\Delta G^\circ\) value in the box provided, in units of \(\text{kJ/mol}\).
**Box: [ _____ ] kJ/mol**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faed2933e-b67f-42a0-9fac-b6dce0e828b0%2F44cea6ca-1f86-42e5-88e6-ef80e64fbdf7%2F6dsikn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Electrochemical Reactions and Gibbs Free Energy Calculation
#### Given Reactions:
1. \( 2 \text{Hg}^{2+} (\text{aq}) + 2 \text{e}^- \rightarrow \text{Hg}_2^{2+} (\text{aq}) \)
- Standard electrode potential, \( E^\circ = 0.92 \, \text{V} \)
2. \( \text{Hg}_2^{2+} (\text{aq}) + 2 \text{e}^- \rightarrow 2 \text{Hg} (\text{l}) \)
- Standard electrode potential, \( E^\circ = 0.85 \, \text{V} \)
### Task:
**Calculate \(\Delta G^\circ\) for the following process at 25°C:**
\[ \text{Hg}_2^{2+} (\text{aq}) \rightarrow \text{Hg}^{2+} (\text{aq}) + \text{Hg} (\text{l}) \]
### Explanation:
To calculate the Gibbs Free Energy (\(\Delta G^\circ\)), you will use the standard electrode potentials from the given half-reactions. Apply the formula:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]
where:
- \( n \) is the number of moles of electrons transferred.
- \( F \) is the Faraday constant (approximately \( 96485 \, \text{C/mol} \)).
- \( E^\circ_{\text{cell}} \) is the standard cell potential, which can be calculated by combining the given electrode potentials.
**Note:** Enter your calculated \(\Delta G^\circ\) value in the box provided, in units of \(\text{kJ/mol}\).
**Box: [ _____ ] kJ/mol**
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