FORLAD 4 Consider the redox reaction where Ag+ (aq) is oxidized to Ago (s) (E° 0x = -1.77 v) and Helo₂ (aq) is reduced to Helo (aq) (E°red = + 1.67v). a) write the Oxidation half reaction and reduction half reaction b) Calculate the equilibrium constant at 25°c hint (log k = (n. E°) - 0.0592)
FORLAD 4 Consider the redox reaction where Ag+ (aq) is oxidized to Ago (s) (E° 0x = -1.77 v) and Helo₂ (aq) is reduced to Helo (aq) (E°red = + 1.67v). a) write the Oxidation half reaction and reduction half reaction b) Calculate the equilibrium constant at 25°c hint (log k = (n. E°) - 0.0592)
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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![### Redox Reaction Analysis
In this exercise, we will consider the redox reaction where Hg (in its oxidation state) and \( \text{HClO}_2 \) (in its reduction state) participate. Here are the given standard electrode potentials for each species:
- \( E^\circ_{\text{ox}} = -1.67 \text{ V} \) for \( \text{Hg} \)
- \( E^\circ_{\text{red}} = +1.67 \text{ V} \) for \( \text{HClO}_2 \)
To complete this analysis, we are asked to do the following:
#### (a) Write the Oxidation and Reduction Half-Reactions
First, identify the oxidation and reduction half-reactions separately.
- The oxidation half-reaction involves \( \text{Hg} \) being oxidized:
\[
\text{Hg} \rightarrow \text{Hg}^{2+} + 2e^{-} \quad (E^\circ_{\text{ox}} = -1.67 \text{ V})
\]
- The reduction half-reaction involves \( \text{HClO}_2 \) being reduced:
\[
\text{HClO}_2 + 2e^{-} \rightarrow \text{HCl} + \text{{other products}} \quad (E^\circ_{\text{red}} = +1.67 \text{ V})
\]
#### (b) Calculate the Equilibrium Constant at 25°C
Using the Nernst equation and the relationship between the standard electrode potential and the equilibrium constant, we can determine the equilibrium constant \( K \). The relevant formula is:
\[
\log K = \frac{n \cdot E^\circ_{\text{cell}}}{0.0592}
\]
where \( n \) is the number of moles of electrons transferred in the balanced equation and \( E^\circ_{\text{cell}} \) is the standard cell potential.
To calculate \( E^\circ_{\text{cell}} \):
\[
E^\circ_{\text{cell}} = E^\circ_{\text{red}} - E^\circ_{\text{ox}} = +1.67 \text{ V} - (-1.67 \text{ V}) = +3.34 \text{ V}
\]
Since \( n = 2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2704aedc-cd59-4651-b157-0d8c71ae74f0%2F2b1f8266-7700-4a92-a5ed-c598e6720fc2%2Fbx5kf8o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Redox Reaction Analysis
In this exercise, we will consider the redox reaction where Hg (in its oxidation state) and \( \text{HClO}_2 \) (in its reduction state) participate. Here are the given standard electrode potentials for each species:
- \( E^\circ_{\text{ox}} = -1.67 \text{ V} \) for \( \text{Hg} \)
- \( E^\circ_{\text{red}} = +1.67 \text{ V} \) for \( \text{HClO}_2 \)
To complete this analysis, we are asked to do the following:
#### (a) Write the Oxidation and Reduction Half-Reactions
First, identify the oxidation and reduction half-reactions separately.
- The oxidation half-reaction involves \( \text{Hg} \) being oxidized:
\[
\text{Hg} \rightarrow \text{Hg}^{2+} + 2e^{-} \quad (E^\circ_{\text{ox}} = -1.67 \text{ V})
\]
- The reduction half-reaction involves \( \text{HClO}_2 \) being reduced:
\[
\text{HClO}_2 + 2e^{-} \rightarrow \text{HCl} + \text{{other products}} \quad (E^\circ_{\text{red}} = +1.67 \text{ V})
\]
#### (b) Calculate the Equilibrium Constant at 25°C
Using the Nernst equation and the relationship between the standard electrode potential and the equilibrium constant, we can determine the equilibrium constant \( K \). The relevant formula is:
\[
\log K = \frac{n \cdot E^\circ_{\text{cell}}}{0.0592}
\]
where \( n \) is the number of moles of electrons transferred in the balanced equation and \( E^\circ_{\text{cell}} \) is the standard cell potential.
To calculate \( E^\circ_{\text{cell}} \):
\[
E^\circ_{\text{cell}} = E^\circ_{\text{red}} - E^\circ_{\text{ox}} = +1.67 \text{ V} - (-1.67 \text{ V}) = +3.34 \text{ V}
\]
Since \( n = 2
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