In the following reactions, what weight of substance would be liberated if `1F` of electricity were passed through the cell `:`
`i. Cu^(2+)+2e^(-) rarr Cu` `ii. Al^(3+)+3e^(-) rarr Al`
`iii. 2Cl^(c-) rarr Cl_(2)+2e^(-)`

Explanation

To solve the problem, we need to determine the weight of the substance liberated when 1 Faraday (1F) of electricity is passed through each of the given electrochemical reactions.

### Step-by-Step Solution:

**i. Reaction: Cu²⁺ + 2e⁻ → Cu**

1. **Identify the number of electrons involved:** In this reaction, 2 electrons are required to reduce 1 mole of Cu²⁺ to Cu.
2. **Determine moles of Cu deposited for 1F:** Since 1 Faraday corresponds to 1 mole of electrons, the moles of Cu deposited for 1F is:
\[ \text{Moles of Cu} = \frac{1F}{2e^-} = \frac{1}{2} \text{ moles of Cu} \]
3. **Calculate the mass of Cu:** The atomic mass of copper (Cu) is 63.5 g/mol. Therefore, the mass of Cu deposited is:
\[ \text{Mass of Cu} = \left(\frac{1}{2} \text{ moles}\right) \times 63.5 \text{ g/mol} = 31.75 \text{ g} \]

**ii. Reaction: Al³⁺ + 3e⁻ → Al**

1. **Identify the number of electrons involved:** In this reaction, 3 electrons are required to reduce 1 mole of Al³⁺ to Al.
2. **Determine moles of Al deposited for 1F:** The moles of Al deposited for 1F is:
\[ \text{Moles of Al} = \frac{1F}{3e^-} = \frac{1}{3} \text{ moles of Al} \]
3. **Calculate the mass of Al:** The atomic mass of aluminum (Al) is 27 g/mol. Therefore, the mass of Al deposited is:
\[ \text{Mass of Al} = \left(\frac{1}{3} \text{ moles}\right) \times 27 \text{ g/mol} = 9 \text{ g} \]

**iii. Reaction: 2Cl⁻ → Cl₂ + 2e⁻**

1. **Identify the number of electrons involved:** In this reaction, 2 electrons are required to produce 1 mole of Cl₂ from 2 moles of Cl⁻.
2. **Determine moles of Cl₂ produced for 1F:** The moles of Cl₂ produced for 1F is:
\[ \text{Moles of Cl₂} = \frac{1F}{2e^-} = \frac{1}{2} \text{ moles of Cl₂} \]
3. **Calculate the mass of Cl₂:** The molecular mass of chlorine (Cl₂) is 71 g/mol (since Cl has an atomic mass of 35.5 g). Therefore, the mass of Cl₂ produced is:
\[ \text{Mass of Cl₂} = \left(\frac{1}{2} \text{ moles}\right) \times 71 \text{ g/mol} = 35.5 \text{ g} \]

### Final Results:
- For Cu: 31.75 g
- For Al: 9 g
- For Cl₂: 35.5 g