By passage of 1 F of electricity:

Options

A. 1 mol of Cu is deposited
B. 0.5 mol of Mg iş deposited
C. 5.6 L of `O_2` gas evolved at anode
D. 9 g of Al is deposited

Answer

To solve the problem of what is deposited by the passage of 1 Faraday of electricity, we will analyze each statement provided in the question regarding different substances.

### Step-by-Step Solution:

1. **Understanding Faraday's Law of Electrolysis**:
- Faraday's laws state that the amount of substance deposited or liberated during electrolysis is directly proportional to the quantity of electricity passed. One Faraday (1 F) corresponds to the charge of one mole of electrons (approximately 96485 coulombs).

2. **Analyzing Copper (Cu)**:
- The half-reaction for copper deposition is:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]
- This indicates that 2 moles of electrons are required to deposit 1 mole of copper.
- Therefore, with 1 Faraday (1 mole of electrons), the amount of copper deposited is:
\[ \text{Moles of Cu} = \frac{1 \text{ mole of electrons}}{2 \text{ electrons/mole of Cu}} = 0.5 \text{ moles of Cu} \]
- **Conclusion**: 0.5 moles of copper is deposited, so the statement regarding 1 mole of copper is incorrect.

3. **Analyzing Magnesium (Mg)**:
- The half-reaction for magnesium deposition is:
\[ \text{Mg}^{2+} + 2e^- \rightarrow \text{Mg} \]
- Similar to copper, 2 moles of electrons are required to deposit 1 mole of magnesium.
- Thus, with 1 Faraday:
\[ \text{Moles of Mg} = \frac{1 \text{ mole of electrons}}{2 \text{ electrons/mole of Mg}} = 0.5 \text{ moles of Mg} \]
- **Conclusion**: 0.5 moles of magnesium is deposited, so this statement is correct.

4. **Analyzing Oxygen (O2)**:
- The half-reaction for oxygen production from water is:
\[ 4 \text{OH}^- \rightarrow 2 \text{H}_2\text{O} + \text{O}_2 + 4e^- \]
- This indicates that 4 moles of electrons are required to produce 1 mole of O2.
- Therefore, with 1 Faraday:
\[ \text{Moles of O2} = \frac{1 \text{ mole of electrons}}{4 \text{ electrons/mole of O2}} = 0.25 \text{ moles of O2} \]
- To find the volume of O2 produced at STP:
\[ \text{Volume} = 0.25 \text{ moles} \times 22.4 \text{ L/mole} = 5.6 \text{ L} \]
- **Conclusion**: 5.6 liters of O2 is produced, so this statement is correct.

5. **Analyzing Aluminum (Al)**:
- The half-reaction for aluminum deposition is:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]
- Here, 3 moles of electrons are required to deposit 1 mole of aluminum.
- Thus, with 1 Faraday:
\[ \text{Moles of Al} = \frac{1 \text{ mole of electrons}}{3 \text{ electrons/mole of Al}} \approx 0.333 \text{ moles of Al} \]
- The mass of aluminum deposited can be calculated as:
\[ \text{Mass} = 0.333 \text{ moles} \times 27 \text{ g/mole} \approx 9 \text{ g} \]
- **Conclusion**: 9 grams of aluminum is deposited, so this statement is correct.

### Final Summary:
- The correct statements are:
- 0.5 moles of magnesium is deposited (True).
- 5.6 liters of O2 is produced (True).
- 9 grams of aluminum is deposited (True).
- The statement regarding 1 mole of copper being deposited is incorrect.