Two different electrolytic cells filled with molten `Cu(NO_3)_(2)` and molten `Al(NO_3)_(3)` respectively are connected in series. When electricity is passed 2.7 g Al is deposited on electrode. Calculate the weight of Cu deposited on cathode.
`[Cu = 63.5 , Al = 27.0 g mol^(-1)]`

Options

A. 190.5 g
B. 9.525 g
C. 63.5 g
D. 31.75 g

Answer

To solve the problem, we will use Faraday's law of electrolysis, which states that the amount of substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Weight of aluminum deposited (W_Al) = 2.7 g
- Atomic mass of copper (Cu) = 63.5 g/mol
- Atomic mass of aluminum (Al) = 27.0 g/mol

2. **Calculate the Equivalent Weights:**
- For Copper (Cu):
- The n-factor for Cu in Cu(NO₃)₂ is 2 (since Cu is in +2 oxidation state).
- Equivalent weight of Cu = Atomic mass of Cu / n-factor = 63.5 g/mol / 2 = 31.75 g/equiv.

- For Aluminum (Al):
- The n-factor for Al in Al(NO₃)₃ is 3 (since Al is in +3 oxidation state).
- Equivalent weight of Al = Atomic mass of Al / n-factor = 27.0 g/mol / 3 = 9.0 g/equiv.

3. **Set Up the Ratio Using Faraday's Law:**
- According to Faraday's law, the ratio of the weights of the substances deposited is equal to the ratio of their equivalent weights:
\[ \frac{W_{Cu}}{W_{Al}} = \frac{E_{Cu}}{E_{Al}} \]
Where:
- \( W_{Cu} \) = weight of copper deposited
- \( W_{Al} \) = weight of aluminum deposited
- \( E_{Cu} \) = equivalent weight of copper
- \( E_{Al} \) = equivalent weight of aluminum

4. **Substitute the Known Values:**
\[ \frac{W_{Cu}}{2.7 \, \text{g}} = \frac{31.75 \, \text{g/equiv.}}{9.0 \, \text{g/equiv.}} \]

5. **Cross-Multiply and Solve for \( W_{Cu} \):**
\[ W_{Cu} = 2.7 \, \text{g} \times \frac{31.75}{9.0} \]
\[ W_{Cu} = 2.7 \, \text{g} \times 3.5278 \approx 9.525 \, \text{g} \]

6. **Final Answer:**
The weight of copper deposited on the cathode is approximately **9.525 g**.