Three Faraday of electricity is passed through aqueous solutions of `AgNO_3, NiSO_4 and CoCl_3` kept in three vessels using inert electrodes. The ratio in mol in which the metals Ag, Ni and Co will deposited is

Options

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

Answer

To solve the problem of finding the ratio in moles of metals deposited (Ag, Ni, and Co) when 3 Faraday of electricity is passed through their respective aqueous solutions, we can follow these steps:

### Step 1: Understand the Reduction Reactions
Each metal ion in the solution will undergo a reduction reaction at the cathode, where they gain electrons to form solid metals.

1. **Silver (Ag)**:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \quad (\text{1 mole of Ag requires 1 Faraday}) \]

2. **Nickel (Ni)**:
\[ \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni} \quad (\text{1 mole of Ni requires 2 Faraday}) \]

3. **Cobalt (Co)**:
\[ \text{Co}^{3+} + 3e^- \rightarrow \text{Co} \quad (\text{1 mole of Co requires 3 Faraday}) \]

### Step 2: Calculate Moles of Each Metal Deposited
Now, we need to determine how many moles of each metal can be deposited using 3 Faraday of electricity.

1. **For Silver (Ag)**:
- 1 mole of Ag requires 1 Faraday.
- Therefore, with 3 Faraday, the moles of Ag deposited:
\[ \text{Moles of Ag} = 3 \text{ Faraday} \times \frac{1 \text{ mole}}{1 \text{ Faraday}} = 3 \text{ moles} \]

2. **For Nickel (Ni)**:
- 1 mole of Ni requires 2 Faraday.
- Therefore, with 3 Faraday, the moles of Ni deposited:
\[ \text{Moles of Ni} = 3 \text{ Faraday} \times \frac{1 \text{ mole}}{2 \text{ Faraday}} = \frac{3}{2} \text{ moles} \]

3. **For Cobalt (Co)**:
- 1 mole of Co requires 3 Faraday.
- Therefore, with 3 Faraday, the moles of Co deposited:
\[ \text{Moles of Co} = 3 \text{ Faraday} \times \frac{1 \text{ mole}}{3 \text{ Faraday}} = 1 \text{ mole} \]

### Step 3: Write the Ratio of Moles
Now that we have the moles of each metal:
- Moles of Ag = 3
- Moles of Ni = \( \frac{3}{2} \)
- Moles of Co = 1

To express these in a simple ratio, we can multiply each term by 2 to eliminate the fraction:
- Moles of Ag = \( 3 \times 2 = 6 \)
- Moles of Ni = \( \frac{3}{2} \times 2 = 3 \)
- Moles of Co = \( 1 \times 2 = 2 \)

Thus, the ratio of moles of Ag:Ni:Co is:
\[ \text{Ag:Ni:Co} = 6:3:2 \]

### Final Answer
The ratio in which the metals Ag, Ni, and Co will be deposited is:
\[ \text{6:3:2} \]