Three moles of electrons are passed through three solutions in succession containing `AgNO_3`, `CuSO_4` and `AuCL_3` respectively the molar ratio of amounts of cations reduced at cathode will be

Options

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

Answer

To solve the problem, we need to determine the molar ratio of the amounts of cations reduced at the cathode when 3 moles of electrons are passed through solutions of AgNO3, CuSO4, and AuCl3.

### Step-by-Step Solution:

1. **Identify the Reduction Reactions:**
- For **AgNO3**: The reduction reaction is:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]
Here, 1 mole of Ag is produced from 1 mole of electrons.

- For **CuSO4**: The reduction reaction is:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]
Here, 1 mole of Cu is produced from 2 moles of electrons.

- For **AuCl3**: The reduction reaction is:
\[ \text{Au}^{3+} + 3e^- \rightarrow \text{Au} \]
Here, 1 mole of Au is produced from 3 moles of electrons.

2. **Calculate Moles of Cations Reduced:**
- For **AgNO3**:
- Moles of Ag reduced = Moles of electrons passed = 3 moles
- Ratio: 3 moles of Ag

- For **CuSO4**:
- Moles of Cu reduced = \( \frac{3 \text{ moles of electrons}}{2 \text{ moles of electrons per Cu}} = 1.5 \text{ moles of Cu} \)
- Ratio: 1.5 moles of Cu

- For **AuCl3**:
- Moles of Au reduced = \( \frac{3 \text{ moles of electrons}}{3 \text{ moles of electrons per Au}} = 1 \text{ mole of Au} \)
- Ratio: 1 mole of Au

3. **Establish the Molar Ratio:**
- Now we have the moles of each cation reduced:
- Ag: 3 moles
- Cu: 1.5 moles
- Au: 1 mole
- To express this as a ratio, we can write:
\[ \text{Ratio} = 3 : 1.5 : 1 \]
- To convert this into whole numbers, we can multiply each term by 2:
\[ 3 \times 2 : 1.5 \times 2 : 1 \times 2 = 6 : 3 : 2 \]

4. **Final Molar Ratio:**
- The final molar ratio of the amounts of cations reduced at the cathode is:
\[ \text{Ag} : \text{Cu} : \text{Au} = 6 : 3 : 2 \]

### Conclusion:
The molar ratio of the amounts of cations reduced at the cathode is \( 6 : 3 : 2 \).