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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
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
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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}^+]_{\
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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