If `3F` of electricity is passed through the solutions of `AgNO_3, CuSO_4` and `AuCL_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 Faradays of electricity are passed through the solutions of AgNO3, CuSO4, and AuCl3, we will follow these steps:

### Step 1: Identify the reactions at the cathode
1. **AgNO3 dissociation**:
\[ \text{AgNO}_3 \rightarrow \text{Ag}^+ + \text{NO}_3^- \]
At the cathode, the reduction reaction is:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]

2. **CuSO4 dissociation**:
\[ \text{CuSO}_4 \rightarrow \text{Cu}^{2+} + \text{SO}_4^{2-} \]
At the cathode, the reduction reaction is:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

3. **AuCl3 dissociation**:
\[ \text{AuCl}_3 \rightarrow \text{Au}^{3+} + 3\text{Cl}^- \]
At the cathode, the reduction reaction is:
\[ \text{Au}^{3+} + 3e^- \rightarrow \text{Au} \]

### Step 2: Calculate the moles of each metal deposited
1. **For Ag**:
- 1 mole of Ag requires 1 Faraday (1 mole of electrons).
- Therefore, 3 Faradays will deposit:
\[ \text{Moles of Ag} = 3 \text{ Faradays} \times 1 \text{ mole of Ag/Faraday} = 3 \text{ moles of Ag} \]

2. **For Cu**:
- 1 mole of Cu requires 2 Faradays (2 moles of electrons).
- Therefore, 3 Faradays will deposit:
\[ \text{Moles of Cu} = \frac{3 \text{ Faradays}}{2 \text{ Faradays/mole of Cu}} = 1.5 \text{ moles of Cu} \]

3. **For Au**:
- 1 mole of Au requires 3 Faradays (3 moles of electrons).
- Therefore, 3 Faradays will deposit:
\[ \text{Moles of Au} = \frac{3 \text{ Faradays}}{3 \text{ Faradays/mole of Au}} = 1 \text{ mole of Au} \]

### Step 3: Establish the molar ratio
Now we have the following moles deposited:
- Ag: 3 moles
- Cu: 1.5 moles
- Au: 1 mole

To express these in a simple ratio, we can multiply each by 2 to eliminate the decimal:
- Ag: \(3 \times 2 = 6\)
- Cu: \(1.5 \times 2 = 3\)
- Au: \(1 \times 2 = 2\)

Thus, the 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**.

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