For a particular redox reaction, Cr is oxidized to CrO2- and Fe³+ is reduced to Fe²+. Complete and balance the equation for this reaction in basic solution. The phases are optional.
For a particular redox reaction, Cr is oxidized to CrO2- and Fe³+ is reduced to Fe²+. Complete and balance the equation for this reaction in basic solution. The phases are optional.
General Chemistry - Standalone book (MindTap Course List)
11th Edition
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter19: Electrochemistry
Section: Chapter Questions
Problem 19.123QP: An aqueous solution of an unknown salt of gold is electrolyzed by a current of 2.75 amps for 3.39...
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![### Redox Reaction Balancing in Basic Solution
**Problem Statement:**
For a particular redox reaction, chromium (Cr) is oxidized to chromate ion \(\text{CrO}_4^{2-}\) and iron (Fe\(^3+\)) is reduced to iron(II) ion (Fe\(^2+\)). Complete and balance the equation for this reaction in a basic solution. The phases are optional.
**Equation to Balance:**
\[ \text{Cr} + 6\text{Fe}^{3+} + 8\text{OH}^- \longrightarrow \text{CrO}_4^{2-} + 6\text{Fe}^{2+} + 4\text{H}_2\text{O} \]
**Solution:**
The provided equation is not correctly balanced for the redox reaction in a basic solution as indicated by the "Incorrect" label.
Ensure the final balanced equation satisfies the following:
1. **Mass Balance**: The same number of each type of atom on both sides of the equation.
2. **Charge Balance**: The total charge should be the same on both sides of the equation.
**Detailed Steps for Balancing:**
1. **Identify Oxidation and Reduction Half-Reactions:**
- **Oxidation**: Cr is being oxidized to CrO₄²⁻.
- **Reduction**: Fe³⁺ is being reduced to Fe²⁺.
2. **Balance Each Half-Reaction Separately:**
- **Oxidation Half-Reaction:**
\[ \text{Cr} \rightarrow \text{CrO}_4^{2-} \]
To balance oxygen, add H₂O to the right side:
\[ \text{Cr} \rightarrow \text{CrO}_4^{2-} + 4\text{H}_2\text{O} \]
To balance hydrogen, add OH⁻ to both sides:
\[ \text{Cr} + 8\text{OH}^- \rightarrow \text{CrO}_4^{2-} + 4\text{H}_2\text{O} \]
Balance the charges by adding electrons to the left side:
\[ \text{Cr} + 8\text{OH}^- \rightarrow \text{CrO}_4^{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3f33cc2f-0221-41a8-a3a2-cc96bc7960a2%2Ff8cab5d8-22e2-425c-aa1a-988011acdd97%2Fj750ge_processed.png&w=3840&q=75)
Transcribed Image Text:### Redox Reaction Balancing in Basic Solution
**Problem Statement:**
For a particular redox reaction, chromium (Cr) is oxidized to chromate ion \(\text{CrO}_4^{2-}\) and iron (Fe\(^3+\)) is reduced to iron(II) ion (Fe\(^2+\)). Complete and balance the equation for this reaction in a basic solution. The phases are optional.
**Equation to Balance:**
\[ \text{Cr} + 6\text{Fe}^{3+} + 8\text{OH}^- \longrightarrow \text{CrO}_4^{2-} + 6\text{Fe}^{2+} + 4\text{H}_2\text{O} \]
**Solution:**
The provided equation is not correctly balanced for the redox reaction in a basic solution as indicated by the "Incorrect" label.
Ensure the final balanced equation satisfies the following:
1. **Mass Balance**: The same number of each type of atom on both sides of the equation.
2. **Charge Balance**: The total charge should be the same on both sides of the equation.
**Detailed Steps for Balancing:**
1. **Identify Oxidation and Reduction Half-Reactions:**
- **Oxidation**: Cr is being oxidized to CrO₄²⁻.
- **Reduction**: Fe³⁺ is being reduced to Fe²⁺.
2. **Balance Each Half-Reaction Separately:**
- **Oxidation Half-Reaction:**
\[ \text{Cr} \rightarrow \text{CrO}_4^{2-} \]
To balance oxygen, add H₂O to the right side:
\[ \text{Cr} \rightarrow \text{CrO}_4^{2-} + 4\text{H}_2\text{O} \]
To balance hydrogen, add OH⁻ to both sides:
\[ \text{Cr} + 8\text{OH}^- \rightarrow \text{CrO}_4^{2-} + 4\text{H}_2\text{O} \]
Balance the charges by adding electrons to the left side:
\[ \text{Cr} + 8\text{OH}^- \rightarrow \text{CrO}_4^{
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