One mole of electron passes through each of the solution of `AgNO_3, CuSO_4 and AlCl_3`when Ag, Cu and Al are deposited at cathode. The molar ratio of Ag,Cu and Al deposited are

Options

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

Answer

To find the molar ratio of Ag, Cu, and Al deposited at the cathode when one mole of electrons passes through each of the solutions of AgNO₃, CuSO₄, and AlCl₃, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the reactions**:
- For AgNO₃: The reduction reaction at the cathode is:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]
- For CuSO₄: The reduction reaction at the cathode is:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]
- For AlCl₃: The reduction reaction at the cathode is:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]

2. **Determine the n-factor for each ion**:
- For Ag⁺, the n-factor is 1 (1 electron is needed to deposit 1 mole of Ag).
- For Cu²⁺, the n-factor is 2 (2 electrons are needed to deposit 1 mole of Cu).
- For Al³⁺, the n-factor is 3 (3 electrons are needed to deposit 1 mole of Al).

3. **Calculate the number of moles deposited for each metal**:
- For Ag:
\[ \text{Moles of Ag} = \frac{1 \text{ equivalent}}{n-factor} = \frac{1}{1} = 1 \text{ mole} \]
- For Cu:
\[ \text{Moles of Cu} = \frac{1 \text{ equivalent}}{n-factor} = \frac{1}{2} = 0.5 \text{ moles} \]
- For Al:
\[ \text{Moles of Al} = \frac{1 \text{ equivalent}}{n-factor} = \frac{1}{3} \approx 0.33 \text{ moles} \]

4. **Convert to a common ratio**:
- To find the molar ratio, we can express all quantities in terms of a common denominator. The least common multiple of 1, 0.5, and 0.33 is 6.
- Therefore, we multiply each by 6:
- Ag: \(1 \times 6 = 6\)
- Cu: \(0.5 \times 6 = 3\)
- Al: \(0.33 \times 6 \approx 2\)

5. **Final molar ratio**:
- The molar ratio of Ag:Cu:Al is \(6:3:2\).

### Conclusion:
The molar ratio of Ag, Cu, and Al deposited at the cathode is **6:3:2**.