On passing 1 Faraday charge through the electrolytic cells containing `Ag^(+),Ni^(2+),Cr^(3+)` ions solution, the mass of deposited Ag is (At mass = 108 u ), Ni (At mass = 59u) and Cr (At mass = 52U):

Options

A. `{:(,"Ag","Ni","cr"),((1),108 " gm",29.5 " gm",17.3" gm"):}`
B. `{:(,"Ag","Ni","cr"),((2),108 " gm",59 " gm",52" gm"):}`
C. `{:(,"Ag","Ni","cr"),((3),108 " gm",108 " gm",108" gm"):}`
D. `{:(,"Ag","Ni","cr"),((4),108 " gm",117.5 " gm",166" gm"):}`

Answer

To solve the problem of calculating the mass of deposited metals (Ag, Ni, Cr) when 1 Faraday of charge is passed through an electrolytic cell containing their ions, we will use Faraday's laws of electrolysis.

### Step-by-Step Solution:

1. **Understanding Faraday's Law**:
Faraday's first law states that the mass of a substance deposited or liberated during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte. The formula can be expressed as:
\[ W = \frac{Q}{F} \times E \]
where \( W \) is the mass of the substance deposited, \( Q \) is the total charge passed (in Coulombs), \( F \) is Faraday's constant (approximately 96500 C/mol), and \( E \) is the equivalent weight of the substance.

2. **Calculating Equivalent Weight**:
The equivalent weight \( E \) can be calculated using the formula:
\[ E = \frac{\text{Molar Mass}}{n} \]
where \( n \) is the number of electrons transferred in the half-reaction.

3. **For Silver (Ag)**:
- Molar mass of Ag = 108 u
- \( n \) for Ag = 1 (Ag⁺ + e⁻ → Ag)
\[ E_{Ag} = \frac{108}{1} = 108 \text{ g/equiv} \]

4. **For Nickel (Ni)**:
- Molar mass of Ni = 59 u
- \( n \) for Ni = 2 (Ni²⁺ + 2e⁻ → Ni)
\[ E_{Ni} = \frac{59}{2} = 29.5 \text{ g/equiv} \]

5. **For Chromium (Cr)**:
- Molar mass of Cr = 52 u
- \( n \) for Cr = 3 (Cr³⁺ + 3e⁻ → Cr)
\[ E_{Cr} = \frac{52}{3} \approx 17.33 \text{ g/equiv} \]

6. **Calculating Mass Deposited**:
Since we are passing 1 Faraday (1 equiv), the mass deposited for each metal is equal to its equivalent weight:
- Mass of Ag deposited = 108 g
- Mass of Ni deposited = 29.5 g
- Mass of Cr deposited = 17.33 g

### Final Results:
- Mass of Ag deposited = 108 g
- Mass of Ni deposited = 29.5 g
- Mass of Cr deposited = 17.33 g