If `3F` of electrictiy is passed through the solutions of `AgNO_3, CuSO_4` and `Auc=CL_3`, the molar ration of the cations deposited at the cathode is .

Options

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

Explanation

To determine the molar ratio of the cations deposited at the cathode when 3 Faraday of electricity is passed through the solutions of AgNO₃, CuSO₄, and AuCl₃, we will follow these steps:

### Step 1: Analyze the Electrolytes
1. **AgNO₃ dissociation**:
- AgNO₃ → Ag⁺ + NO₃⁻
- At the cathode, Ag⁺ + e⁻ → Ag (s)
- Each mole of Ag requires 1 Faraday (1 mole of electrons) to deposit 1 mole of Ag.

2. **CuSO₄ dissociation**:
- CuSO₄ → Cu²⁺ + SO₄²⁻
- At the cathode, Cu²⁺ + 2e⁻ → Cu (s)
- Each mole of Cu requires 2 Faraday (2 moles of electrons) to deposit 1 mole of Cu.

3. **AuCl₃ dissociation**:
- AuCl₃ → Au³⁺ + 3Cl⁻
- At the cathode, Au³⁺ + 3e⁻ → Au (s)
- Each mole of Au requires 3 Faraday (3 moles of electrons) to deposit 1 mole of Au.

### Step 2: Calculate Moles of Cations Deposited
1. **For Ag**:
- 3 Faraday can deposit:
\[ \text{Moles of Ag} = \frac{3 \text{ Faraday}}{1 \text{ Faraday/mole}} = 3 \text{ moles of Ag} \]

2. **For Cu**:
- 3 Faraday can deposit:
\[ \text{Moles of Cu} = \frac{3 \text{ Faraday}}{2 \text{ Faraday/mole}} = 1.5 \text{ moles of Cu} \]

3. **For Au**:
- 3 Faraday can deposit:
\[ \text{Moles of Au} = \frac{3 \text{ Faraday}}{3 \text{ Faraday/mole}} = 1 \text{ mole of Au} \]

### Step 3: Determine the Molar Ratio
Now, we have the following moles deposited:
- Moles of Ag = 3
- Moles of Cu = 1.5
- Moles of Au = 1

To find the molar ratio of Ag:Cu:Au, we can express it as:
- Ag : Cu : Au = 3 : 1.5 : 1

To simplify this ratio, we can multiply each term by 2 to eliminate the decimal:
- Ag : Cu : Au = 6 : 3 : 2

Thus, the final molar ratio of the cations deposited at the cathode is:
\[ \text{Ag : Cu : Au} = 6 : 3 : 2 \]

### Final Answer
The molar ratio of the cations deposited at the cathode is **6:3:2**.