1. 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 1.0x10 8 М. Ag- Ag [Ag"] 1.0 M
1. 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 1.0x10 8 М. 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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![**Concentration Cell Example**
1. Consider the concentration cell below: Identify the anode, cathode, and the direction of electron flow. Calculate the cell potential at 25°C. The \([ \text{Ag}^+ ]\) in the right-hand beaker is \(1.0 \times 10^{-8}\) M.
![Diagram of a concentration cell with silver electrodes. The beaker on the left has Ag+ concentration of 1.0 M, and the beaker on the right has Ag+ concentration of 1.0 x 10^-8 M. Both electrodes are connected by a wire.]()
**Explanation of the Diagram:**
The diagram represents a concentration cell consisting of two beakers connected by a salt bridge. Each beaker contains a silver electrode (Ag) dipped in a solution containing silver ions (\(\text{Ag}^+\)). Here are the specific details:
- **Left Beaker**: The silver ion concentration (\([ \text{Ag}^+ ]\)) is \(1.0 \text{ M}\).
- **Right Beaker**: The silver ion concentration (\([ \text{Ag}^+ ]\)) is \(1.0 \times 10^{-8} \text{ M}\).
**Concepts to Identify:**
- **Anode**: In a concentration cell, the anode is where oxidation occurs (loss of electrons). It will have the lower concentration of ions. Here, the right electrode (with \(1.0 \times 10^{-8} \text{ M}\) of \(\text{Ag}^+\)) is the anode.
- **Cathode**: The cathode is where reduction occurs (gain of electrons). It will have the higher concentration of ions. Hence, the left electrode (with \(1.0 \text{ M}\) of \(\text{Ag}^+\)) is the cathode.
- **Direction of Electron Flow**: Electrons will flow from the anode to the cathode. Thus, they flow from the right beaker to the left beaker.
**Calculation of Cell Potential:**
To calculate the cell potential at 25°C, we use the Nernst equation:
\[ E_{\text{cell}} = E^\circ - \frac{0.0592}{n} \log \left( \frac{[\text{Ag}^+]_{\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7c13c2cd-7092-4d54-837b-d1e36936f099%2F08f0cfbd-43dd-45d4-8a11-94f4d280a6bd%2Fnt0kx6.jpeg&w=3840&q=75)
Transcribed Image Text:**Concentration Cell Example**
1. Consider the concentration cell below: Identify the anode, cathode, and the direction of electron flow. Calculate the cell potential at 25°C. The \([ \text{Ag}^+ ]\) in the right-hand beaker is \(1.0 \times 10^{-8}\) M.
![Diagram of a concentration cell with silver electrodes. The beaker on the left has Ag+ concentration of 1.0 M, and the beaker on the right has Ag+ concentration of 1.0 x 10^-8 M. Both electrodes are connected by a wire.]()
**Explanation of the Diagram:**
The diagram represents a concentration cell consisting of two beakers connected by a salt bridge. Each beaker contains a silver electrode (Ag) dipped in a solution containing silver ions (\(\text{Ag}^+\)). Here are the specific details:
- **Left Beaker**: The silver ion concentration (\([ \text{Ag}^+ ]\)) is \(1.0 \text{ M}\).
- **Right Beaker**: The silver ion concentration (\([ \text{Ag}^+ ]\)) is \(1.0 \times 10^{-8} \text{ M}\).
**Concepts to Identify:**
- **Anode**: In a concentration cell, the anode is where oxidation occurs (loss of electrons). It will have the lower concentration of ions. Here, the right electrode (with \(1.0 \times 10^{-8} \text{ M}\) of \(\text{Ag}^+\)) is the anode.
- **Cathode**: The cathode is where reduction occurs (gain of electrons). It will have the higher concentration of ions. Hence, the left electrode (with \(1.0 \text{ M}\) of \(\text{Ag}^+\)) is the cathode.
- **Direction of Electron Flow**: Electrons will flow from the anode to the cathode. Thus, they flow from the right beaker to the left beaker.
**Calculation of Cell Potential:**
To calculate the cell potential at 25°C, we use the Nernst equation:
\[ E_{\text{cell}} = E^\circ - \frac{0.0592}{n} \log \left( \frac{[\text{Ag}^+]_{\
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