1. Consider the concentration cell below: Identify the anode, cathode, and the electron flow. Calculate the cell potential at 25 C. The [Ag*] in the right-hand beaker is 1.0x10 8 M. I Ag- Ag [Ag ]= 1.0 M
1. Consider the concentration cell below: Identify the anode, cathode, and the electron flow. Calculate the cell potential at 25 C. The [Ag*] in the right-hand beaker is 1.0x10 8 M. I Ag- Ag [Ag ]= 1.0 M
Chemistry
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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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Concentration Cell Analysis
#### Problem Statement:
Consider the concentration cell below: Identify the anode, cathode, and the direction of electron flow. Calculate the cell potential at 25°C. The \([Ag^+]\) in the right-hand beaker is \(1.0 \times 10^{-8} \, M\).
#### Diagram Explanation:
The diagram represents a concentration cell involving silver electrodes immersed in different concentrations of \( Ag^+ \) ions:
- **Left-hand Beaker**: Contains \( Ag \) electrode immersed in a silver nitrate solution where \([Ag^+] = 1.0 \, M\).
- **Right-hand Beaker**: Contains \( Ag \) electrode immersed in a silver nitrate solution where \([Ag^+] = 1.0 \times 10^{-8} \, M\).
The two beakers are connected by a salt bridge that allows for the movement of ions to maintain electrical neutrality.
#### Identification:
- **Anode (Oxidation)**: The anode is where oxidation occurs (loss of electrons). In this cell, it corresponds to the electrode in the low concentration solution.
- Therefore, the right-hand beaker (\([Ag^+] = 1.0 \times 10^{-8} \, M\)) is the **anode**.
- **Cathode (Reduction)**: The cathode is where reduction occurs (gain of electrons). It corresponds to the electrode in the high concentration solution.
- Therefore, the left-hand beaker (\([Ag^+] = 1.0 \, M\)) is the **cathode**.
- **Direction of Electron Flow**: Electrons will flow from the anode to the cathode. In this case, from the right-hand beaker to the left-hand beaker.
#### Calculation of Cell Potential:
Using the Nernst equation for a concentration cell:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln \frac{[Ag^+]_{\text{anode}}}{[Ag^+]_{\text{cathode}}} \]
Given:
- \( E^\circ_{\text{cell}} = 0 \) (for identical electrodes)
- \( R = 8.314 \, J \cdot K](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7c13c2cd-7092-4d54-837b-d1e36936f099%2Fa97d6fbe-00b4-427f-8bdc-d301fc0981f3%2F5m8d4o.jpeg&w=3840&q=75)
Transcribed Image Text:### Concentration Cell Analysis
#### Problem Statement:
Consider the concentration cell below: Identify the anode, cathode, and the direction of electron flow. Calculate the cell potential at 25°C. The \([Ag^+]\) in the right-hand beaker is \(1.0 \times 10^{-8} \, M\).
#### Diagram Explanation:
The diagram represents a concentration cell involving silver electrodes immersed in different concentrations of \( Ag^+ \) ions:
- **Left-hand Beaker**: Contains \( Ag \) electrode immersed in a silver nitrate solution where \([Ag^+] = 1.0 \, M\).
- **Right-hand Beaker**: Contains \( Ag \) electrode immersed in a silver nitrate solution where \([Ag^+] = 1.0 \times 10^{-8} \, M\).
The two beakers are connected by a salt bridge that allows for the movement of ions to maintain electrical neutrality.
#### Identification:
- **Anode (Oxidation)**: The anode is where oxidation occurs (loss of electrons). In this cell, it corresponds to the electrode in the low concentration solution.
- Therefore, the right-hand beaker (\([Ag^+] = 1.0 \times 10^{-8} \, M\)) is the **anode**.
- **Cathode (Reduction)**: The cathode is where reduction occurs (gain of electrons). It corresponds to the electrode in the high concentration solution.
- Therefore, the left-hand beaker (\([Ag^+] = 1.0 \, M\)) is the **cathode**.
- **Direction of Electron Flow**: Electrons will flow from the anode to the cathode. In this case, from the right-hand beaker to the left-hand beaker.
#### Calculation of Cell Potential:
Using the Nernst equation for a concentration cell:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln \frac{[Ag^+]_{\text{anode}}}{[Ag^+]_{\text{cathode}}} \]
Given:
- \( E^\circ_{\text{cell}} = 0 \) (for identical electrodes)
- \( R = 8.314 \, J \cdot K
Expert Solution
Step 1
The Nernst equation is shown below:
Where;
Ecell = electrode potential at any condition
E0cell = standard electrode potential
R = gas constant
T = temperature
n = no. of electrons transferred
F = Faraday's constant
Q = reaction quotient
Step 2
Given information:
E0Ag+/Ag = 0.80 V
The concentration of Ag+ in right hand compartment = 1.0×10-8 M
The concentration of Ag+ in left hand compartment = 1.0 M
Step 3
In galvanic cell; the left-hand compartment is taken as anode and the right-hand compartment is taken as cathode.
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