One Faraday of electricity is pa ssed through molten `Al_(2)O_(3)`, aqeusous solution of `CuSO_(4)` and molten NaCl taken in three different electrolytic cells connected in seris. The mole ratio of Al, Cu,Na deposted at the respective cathode is

Options

A. `2:3:6`
B. `6:3:2`
C. `1:2:3`
D. `3:6:2`

Answer

To determine the mole ratio of Al, Cu, and Na deposited at their respective cathodes when one Faraday of electricity is passed through molten Al₂O₃, aqueous CuSO₄, and molten NaCl, we can follow these steps:

### Step 1: Determine the number of moles of electrons required for each metal deposition.
- **Aluminum (Al)**: The reduction half-reaction for aluminum is:
\[ Al^{3+} + 3e^- \rightarrow Al \]
This means that 3 moles of electrons are required to deposit 1 mole of aluminum.

- **Copper (Cu)**: The reduction half-reaction for copper is:
\[ Cu^{2+} + 2e^- \rightarrow Cu \]
This means that 2 moles of electrons are required to deposit 1 mole of copper.

- **Sodium (Na)**: The reduction half-reaction for sodium is:
\[ Na^{+} + e^- \rightarrow Na \]
This means that 1 mole of electron is required to deposit 1 mole of sodium.

### Step 2: Calculate the moles of each metal deposited using 1 Faraday (1 Faraday = 96500 C).
Since 1 Faraday corresponds to 1 mole of electrons (96500 C), we can calculate the moles of each metal deposited:

- For **Al**:
\[ \text{Moles of Al} = \frac{1 \text{ Faraday}}{3 \text{ moles of e}^-} = \frac{1}{3} \text{ moles of Al} \]

- For **Cu**:
\[ \text{Moles of Cu} = \frac{1 \text{ Faraday}}{2 \text{ moles of e}^-} = \frac{1}{2} \text{ moles of Cu} \]

- For **Na**:
\[ \text{Moles of Na} = \frac{1 \text{ Faraday}}{1 \text{ mole of e}^-} = 1 \text{ mole of Na} \]

### Step 3: Establish the mole ratio.
Now we have:
- Moles of Al = \( \frac{1}{3} \)
- Moles of Cu = \( \frac{1}{2} \)
- Moles of Na = \( 1 \)

To find the mole ratio, we can express these in terms of a common denominator. The least common multiple of 3, 2, and 1 is 6. Therefore, we can convert each fraction:

- Moles of Al = \( \frac{1}{3} = \frac{2}{6} \)
- Moles of Cu = \( \frac{1}{2} = \frac{3}{6} \)
- Moles of Na = \( 1 = \frac{6}{6} \)

Thus, the mole ratio of Al : Cu : Na is:
\[ 2 : 3 : 6 \]

### Final Answer:
The mole ratio of Al, Cu, and Na deposited at their respective cathodes is **2 : 3 : 6**.

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