Three faraday of electricity is passed through molten solutions of `AgNO_3` , `NiSO_4` and `CrCl_3` kept in three vessels using inert electrodes. The ratio in mol in which the metals Ag, Ni and Cr will be deposited is-

Options

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

Answer

To solve the problem of determining the ratio in which metals Ag, Ni, and Cr will be deposited when 3 Faraday of electricity is passed through their molten solutions, we can follow these steps:

### Step 1: Understand the electrochemical reactions
1. **Silver (Ag)**: The half-reaction for the reduction of silver ions is:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]
This indicates that 1 mole of Ag is deposited for every 1 Faraday of charge.

2. **Nickel (Ni)**: The half-reaction for the reduction of nickel ions is:
\[ \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni} \]
This indicates that 2 moles of electrons (2 Faraday) are required to deposit 1 mole of Ni.

3. **Chromium (Cr)**: The half-reaction for the reduction of chromium ions is:
\[ \text{Cr}^{3+} + 3e^- \rightarrow \text{Cr} \]
This indicates that 3 moles of electrons (3 Faraday) are required to deposit 1 mole of Cr.

### Step 2: Calculate the amount of each metal deposited using 3 Faraday
1. **For Silver (Ag)**:
- Since 1 Faraday deposits 1 mole of Ag, then 3 Faraday will deposit:
\[ 3 \text{ Faraday} \rightarrow 3 \text{ moles of Ag} \]

2. **For Nickel (Ni)**:
- Since 2 Faraday deposits 1 mole of Ni, then 3 Faraday will deposit:
\[ 3 \text{ Faraday} \rightarrow \frac{3}{2} \text{ moles of Ni} = 1.5 \text{ moles of Ni} \]

3. **For Chromium (Cr)**:
- Since 3 Faraday deposits 1 mole of Cr, then:
\[ 3 \text{ Faraday} \rightarrow 1 \text{ mole of Cr} \]

### Step 3: Establish the ratio of deposited metals
Now we have:
- Moles of Ag = 3
- Moles of Ni = 1.5
- Moles of Cr = 1

To express this in a simple ratio, we can multiply each term by 2 to eliminate the fraction:
- Moles of Ag = 3 × 2 = 6
- Moles of Ni = 1.5 × 2 = 3
- Moles of Cr = 1 × 2 = 2

Thus, the ratio of Ag : Ni : Cr is:
\[ 6 : 3 : 2 \]

### Final Answer
The ratio in which the metals Ag, Ni, and Cr will be deposited is **6 : 3 : 2**.

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