If the voltmeter was replaced with a resistor (that allows current to flow) and the current is measured to be a constant 1.250 A, it would take [Select] If the concentrations were changed to [Mg] -5.00 M and [Ag] -0.0100 M, the cell potential would then be [Select] to deposit 0.500 g of metal on the cathode.
If the voltmeter was replaced with a resistor (that allows current to flow) and the current is measured to be a constant 1.250 A, it would take [Select] If the concentrations were changed to [Mg] -5.00 M and [Ag] -0.0100 M, the cell potential would then be [Select] to deposit 0.500 g of metal on the cathode.
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
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
Just the last 2 blocks. Use table if needed in first pic.
![### Question 1
**Diagram Explanation:**
The diagram shows an electrochemical cell with the following components:
- **Voltmeter:** Measures the potential difference across the cell.
- **Salt bridge:** Maintains electrical neutrality by allowing ions to flow between the solutions.
- **Electrodes:**
- **Left:** Magnesium electrode in a solution of \( \text{Mg(NO}_3\text{)}_2 \).
- **Right:** Silver electrode in a solution of \( \text{AgNO}_3 \).
**Questions and Answers:**
1. **In the balanced cell reaction, the number of electrons transferred is:**
\( 2 \)
2. **The electrons would flow:**
From the Mg electrode to the Ag electrode.
3. **The standard (assuming 1 M concentrations) cell potential is:**
\( 1.76 \, \text{V} \)
4. **The change in standard Gibbs' Energy, \( \Delta G \), for the overall reaction would be:**
\( 609 \, \text{kJ/mol} \)
5. **The silver electrode gains mass by:**
\[ \text{Select appropriate amount (not provided)} \]
6. **If the voltmeter was replaced with a resistor (that allows current to flow) and the current is measured to be a constant 1.250 A, it would take:**
\[ \text{Select appropriate time (not provided)} \]
To deposit 0.500 g of metal on the cathode.
7. **If the concentrations were changed to \([ \text{Mg}^{2+} ] = 5.00 \, \text{M}\) and \([ \text{Ag}^+ ] = 0.0100 \, \text{M}\), the cell potential would then be:**
\[ \text{Select new cell potential (not provided)} \]
**Question 2**
\[ \text{Please refer to the next section for detailed content (not visible in the image).} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff3a38996-6782-40c2-af9f-f624a7021da3%2F3b878426-24e7-433c-9aae-99331ff80640%2F0kgzly_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 1
**Diagram Explanation:**
The diagram shows an electrochemical cell with the following components:
- **Voltmeter:** Measures the potential difference across the cell.
- **Salt bridge:** Maintains electrical neutrality by allowing ions to flow between the solutions.
- **Electrodes:**
- **Left:** Magnesium electrode in a solution of \( \text{Mg(NO}_3\text{)}_2 \).
- **Right:** Silver electrode in a solution of \( \text{AgNO}_3 \).
**Questions and Answers:**
1. **In the balanced cell reaction, the number of electrons transferred is:**
\( 2 \)
2. **The electrons would flow:**
From the Mg electrode to the Ag electrode.
3. **The standard (assuming 1 M concentrations) cell potential is:**
\( 1.76 \, \text{V} \)
4. **The change in standard Gibbs' Energy, \( \Delta G \), for the overall reaction would be:**
\( 609 \, \text{kJ/mol} \)
5. **The silver electrode gains mass by:**
\[ \text{Select appropriate amount (not provided)} \]
6. **If the voltmeter was replaced with a resistor (that allows current to flow) and the current is measured to be a constant 1.250 A, it would take:**
\[ \text{Select appropriate time (not provided)} \]
To deposit 0.500 g of metal on the cathode.
7. **If the concentrations were changed to \([ \text{Mg}^{2+} ] = 5.00 \, \text{M}\) and \([ \text{Ag}^+ ] = 0.0100 \, \text{M}\), the cell potential would then be:**
\[ \text{Select new cell potential (not provided)} \]
**Question 2**
\[ \text{Please refer to the next section for detailed content (not visible in the image).} \]
Transcribed Image Text:# Standard Reduction Potentials (25 °C)
This table lists the standard reduction potentials for various half-reactions at 25 °C. The reduction potentials are given in volts (V).
| Half-Reaction | E° (V) |
|--------------------------------------|---------|
| Co³⁺(aq) + e⁻ → Co²⁺(aq) | +1.92 |
| Au³⁺(aq) + 3e⁻ → Au(s) | +1.498 |
| Cl₂(g) + 2e⁻ → 2Cl⁻(aq) | +1.36 |
| Pd²⁺(aq) + 2e⁻ → Pd(s) | +0.915 |
| Ag⁺(aq) + e⁻ → Ag(s) | +0.80 |
| I₂(s) + 2e⁻ → 2I⁻(aq) | +0.54 |
| Cu²⁺(aq) + 2e⁻ → Cu(s) | +0.34 |
| Cu⁺(aq) + e⁻ → Cu(s) | +0.153 |
| 2H⁺(aq) + 2e⁻ → H₂(g) | 0.00 |
| Pb²⁺(aq) + 2e⁻ → Pb(s) | -0.13 |
| Ni²⁺(aq) + 2e⁻ → Ni(s) | -0.232 |
| Cd²⁺(aq) + 2e⁻ → Cd(s) | -0.40 |
| Fe²⁺(aq) + 2e⁻ → Fe(s) | -0.44 |
| Cr³⁺(aq) + 3e⁻ → Cr(s) | -0.740 |
| Zn²⁺(aq) + 2e⁻ → Zn(s) | -0.76 |
| Mn²⁺(aq) + 2e⁻ → Mn(s) | -1.185 |
| Ti²⁺(aq) + 2e⁻ → Ti(s) | -1.630
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