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

Explanation

To find the mole ratio of Al, Cu, and Na deposited at the 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 reactions at the cathodes

1. **For Al2O3**:
The reduction reaction can be represented as:
\[ Al^{3+} + 3e^- \rightarrow Al \]
This means that 1 mole of Al is deposited for every 3 moles of electrons (or 1 Faraday).

2. **For CuSO4**:
The reduction reaction is:
\[ Cu^{2+} + 2e^- \rightarrow Cu \]
This means that 1 mole of Cu is deposited for every 2 moles of electrons (or 2 Faradays).

3. **For NaCl**:
The reduction reaction is:
\[ Na^+ + e^- \rightarrow Na \]
This means that 1 mole of Na is deposited for every 1 mole of electrons (or 1 Faraday).

### Step 2: Calculate the moles of each metal deposited

- **For Al**:
Since 1 Faraday corresponds to 3 moles of electrons, the moles of Al deposited will be:
\[ \text{Moles of Al} = \frac{1 \text{ Faraday}}{3 \text{ moles of e}^-} = \frac{1}{3} \text{ moles} \]

- **For Cu**:
Since 1 Faraday corresponds to 2 moles of electrons, the moles of Cu deposited will be:
\[ \text{Moles of Cu} = \frac{1 \text{ Faraday}}{2 \text{ moles of e}^-} = \frac{1}{2} \text{ moles} \]

- **For Na**:
Since 1 Faraday corresponds to 1 mole of electrons, the moles of Na deposited will be:
\[ \text{Moles of Na} = \frac{1 \text{ Faraday}}{1 \text{ mole of e}^-} = 1 \text{ mole} \]

### Step 3: Find the mole ratio

Now, we can express the mole ratio of Al, Cu, and Na:
- Moles of Al = \( \frac{1}{3} \)
- Moles of Cu = \( \frac{1}{2} \)
- Moles of Na = \( 1 \)

To express these in a simple ratio, we can find a common denominator. The least common multiple of the denominators (3, 2, 1) is 6.

- 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 the respective cathodes is **2:3:6**.

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