Three faradays of electricity are passed through molten `Al_2O_3` aqueous solution of `CuSO_4` and molten `NaCl` taken in deffernt electrolytic cells. The amout of `Al,Cu` and Na deposited at the cathodes will be in the ration of .

Options

A. ` 1` mole : 15 ,mole `1` moles
B. 1 mole : 2 moles :3 moles
C. 1 mole : 2 moles : 3 moles
D. 1.5 moles : 2 moles : 3 moles

Explanation

To solve the problem of determining the ratio of the amounts of aluminum (Al), copper (Cu), and sodium (Na) deposited at the cathodes when 3 faradays of electricity are passed through molten Al2O3, aqueous CuSO4, and molten NaCl, we can follow these steps:

### Step 1: Determine the deposition of Aluminum from Al2O3
- The dissociation of Al2O3 in electrolysis gives:
\[ Al_2O_3 \rightarrow 2Al^{3+} + 3O^{2-} \]
- The cathode reaction for aluminum is:
\[ 2Al^{3+} + 6e^- \rightarrow 2Al \]
- This means that to deposit 2 moles of Al, 6 moles of electrons (or 6 faradays) are required.
- Therefore, for 3 faradays:
\[ \text{Moles of Al deposited} = \frac{3 \text{ faradays}}{6 \text{ faradays/mole}} = 0.5 \text{ moles of Al} \]

### Step 2: Determine the deposition of Copper from CuSO4
- The dissociation of CuSO4 in electrolysis gives:
\[ CuSO_4 \rightarrow Cu^{2+} + SO_4^{2-} \]
- The cathode reaction for copper is:
\[ Cu^{2+} + 2e^- \rightarrow Cu \]
- This means that to deposit 1 mole of Cu, 2 faradays are required.
- Therefore, for 3 faradays:
\[ \text{Moles of Cu deposited} = \frac{3 \text{ faradays}}{2 \text{ faradays/mole}} = 1.5 \text{ moles of Cu} \]

### Step 3: Determine the deposition of Sodium from NaCl
- The dissociation of NaCl in electrolysis gives:
\[ NaCl \rightarrow Na^{+} + Cl^{-} \]
- The cathode reaction for sodium is:
\[ Na^{+} + e^- \rightarrow Na \]
- This means that to deposit 1 mole of Na, 1 faraday is required.
- Therefore, for 3 faradays:
\[ \text{Moles of Na deposited} = 3 \text{ moles of Na} \]

### Step 4: Calculate the ratio of Al, Cu, and Na deposited
- We have:
- Al: 0.5 moles
- Cu: 1.5 moles
- Na: 3 moles
- The ratio of Al:Cu:Na can be written as:
\[ \text{Ratio} = 0.5 : 1.5 : 3 \]
- To simplify this ratio, we can multiply each term by 2:
\[ \text{Ratio} = 1 : 3 : 6 \]

### Final Answer
The amounts of Al, Cu, and Na deposited at the cathodes will be in the ratio of **1:3:6**.

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