Using the voltaic cell below: A) Write the equation for this cell. B) Calculate standard cell potential (E°). C) Calculate the voltage for this cell. V Porous Barrier AI Sn 0.435 M AI3+ 0.00212 M Sn2+ D. In the above diagram, label the anode, cathode and draw an arrow showing the flow of electrons in the cell. E. Write the cell diagram.

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Chapter1: Chemical Foundations
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Please answer A:

### Using the Voltaic Cell

#### A) Write the equation for this cell.

---

#### B) Calculate standard cell potential (E°).

---

#### C) Calculate the voltage for this cell.

---

#### D) Diagram Analysis

**In the above diagram, label the anode, cathode, and draw an arrow showing the flow of electrons in the cell.**

The diagram depicts a voltaic cell with two half-cells connected by a porous barrier. The left half-cell contains an aluminum (Al) electrode submerged in a 0.435 M Al³⁺ solution. The right half-cell contains a tin (Sn) electrode submerged in a 0.00212 M Sn²⁺ solution.

At the top of the diagram, a voltmeter (V) is shown connecting the aluminum and tin electrodes.

---

#### E) Write the cell diagram.

---

### Diagram Analysis:

1. **Voltmeter (V)**: Positioned at the top, indicating measurement of potential difference between the two electrodes.
   
2. **Porous Barrier**: Positioned in the center, it separates the two half-cells while allowing the flow of ions to maintain electrical neutrality.
   
3. **Aluminum Electrode (Al)**: Located on the left, submerged in a solution of 0.435 M Al³⁺ indicating it likely functions as the anode (site of oxidation).
   
4. **Tin Electrode (Sn)**: Positioned on the right, submerged in 0.00212 M Sn²⁺ solution, indicating it functions as the cathode (site of reduction).

5. **Concentrations**:
   - For Al³⁺: 0.435 M
   - For Sn²⁺: 0.00212 M

---

### Notation:

- **Anode (Oxidation Site)**: Al is typically here, as it will lose electrons.
- **Cathode (Reduction Site)**: Sn often gains the electrons.

*Flow of Electrons*: Electrons will flow from the anode (Al) to the cathode (Sn).

---

### Cell Diagram:

\[ \text{Anode} | \text{Anode Solution} (concentration) || \text{Cathode Solution} (concentration) | \text{Cathode} \]
\[ \text{Al} | \text{Al}^{3+} (0.435 \text{ M}) || \text{
Transcribed Image Text:### Using the Voltaic Cell #### A) Write the equation for this cell. --- #### B) Calculate standard cell potential (E°). --- #### C) Calculate the voltage for this cell. --- #### D) Diagram Analysis **In the above diagram, label the anode, cathode, and draw an arrow showing the flow of electrons in the cell.** The diagram depicts a voltaic cell with two half-cells connected by a porous barrier. The left half-cell contains an aluminum (Al) electrode submerged in a 0.435 M Al³⁺ solution. The right half-cell contains a tin (Sn) electrode submerged in a 0.00212 M Sn²⁺ solution. At the top of the diagram, a voltmeter (V) is shown connecting the aluminum and tin electrodes. --- #### E) Write the cell diagram. --- ### Diagram Analysis: 1. **Voltmeter (V)**: Positioned at the top, indicating measurement of potential difference between the two electrodes. 2. **Porous Barrier**: Positioned in the center, it separates the two half-cells while allowing the flow of ions to maintain electrical neutrality. 3. **Aluminum Electrode (Al)**: Located on the left, submerged in a solution of 0.435 M Al³⁺ indicating it likely functions as the anode (site of oxidation). 4. **Tin Electrode (Sn)**: Positioned on the right, submerged in 0.00212 M Sn²⁺ solution, indicating it functions as the cathode (site of reduction). 5. **Concentrations**: - For Al³⁺: 0.435 M - For Sn²⁺: 0.00212 M --- ### Notation: - **Anode (Oxidation Site)**: Al is typically here, as it will lose electrons. - **Cathode (Reduction Site)**: Sn often gains the electrons. *Flow of Electrons*: Electrons will flow from the anode (Al) to the cathode (Sn). --- ### Cell Diagram: \[ \text{Anode} | \text{Anode Solution} (concentration) || \text{Cathode Solution} (concentration) | \text{Cathode} \] \[ \text{Al} | \text{Al}^{3+} (0.435 \text{ M}) || \text{
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