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. I Ag- Ag [Ag']=1.0 M

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### Concentration Cells and Electrochemical Potential: An Example

#### 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 x 10⁻⁸ M.

#### Detailed Diagram Description:

The diagram depicts a concentration cell consisting of two beakers connected by a salt bridge. Each beaker contains a silver electrode (denoted as Ag), immersed in solutions of different concentrations of Ag⁺ ions.

- **Left Beaker:** Contains an Ag electrode and a solution with Ag⁺ concentration of 1.0 M.
- **Right Beaker:** Contains an Ag electrode and a solution with Ag⁺ concentration of 1.0 x 10⁻⁸ M.

#### Key Points:
- **Anode and Cathode Identification:**
  - Anode (oxidation will occur): The electrode in the less concentrated solution (right beaker, [Ag⁺] = 1.0 x 10⁻⁸ M).
  - Cathode (reduction will occur): The electrode in the more concentrated solution (left beaker, [Ag⁺] = 1.0 M).

- **Direction of Electron Flow:**
  - Electrons flow from the anode to the cathode. In this cell, electrons will flow from the right beaker to the left beaker.

- **Cell Potential Calculation:**
  - Use the Nernst equation to calculate the potential of the concentration cell:
    \[
    E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln \frac{C_2}{C_1}
    \]
  - For silver electrodes, \( E^\circ_{\text{cell}} = 0 \) since the same material is in both electrodes.
  - Simplified Nernst equation for concentration cell:
    \[
    E_{\text{cell}} = - \frac{RT}{nF} \ln \frac{[Ag^+]_{\text{right}}}{[Ag^+]_{\text{left}}}
    \]
  - At 25°C (298 K), \( \frac{RT}{F} \approx 0.0257 \) V:
    \[
    E_{\text{cell
Transcribed Image Text:### Concentration Cells and Electrochemical Potential: An Example #### 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 x 10⁻⁸ M. #### Detailed Diagram Description: The diagram depicts a concentration cell consisting of two beakers connected by a salt bridge. Each beaker contains a silver electrode (denoted as Ag), immersed in solutions of different concentrations of Ag⁺ ions. - **Left Beaker:** Contains an Ag electrode and a solution with Ag⁺ concentration of 1.0 M. - **Right Beaker:** Contains an Ag electrode and a solution with Ag⁺ concentration of 1.0 x 10⁻⁸ M. #### Key Points: - **Anode and Cathode Identification:** - Anode (oxidation will occur): The electrode in the less concentrated solution (right beaker, [Ag⁺] = 1.0 x 10⁻⁸ M). - Cathode (reduction will occur): The electrode in the more concentrated solution (left beaker, [Ag⁺] = 1.0 M). - **Direction of Electron Flow:** - Electrons flow from the anode to the cathode. In this cell, electrons will flow from the right beaker to the left beaker. - **Cell Potential Calculation:** - Use the Nernst equation to calculate the potential of the concentration cell: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln \frac{C_2}{C_1} \] - For silver electrodes, \( E^\circ_{\text{cell}} = 0 \) since the same material is in both electrodes. - Simplified Nernst equation for concentration cell: \[ E_{\text{cell}} = - \frac{RT}{nF} \ln \frac{[Ag^+]_{\text{right}}}{[Ag^+]_{\text{left}}} \] - At 25°C (298 K), \( \frac{RT}{F} \approx 0.0257 \) V: \[ E_{\text{cell
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