What will be the ratio of moles each of `Ag^(+), Cu^(+2), Fe^(+3)` ions deposited by passage of same quantity of electricity through solutions of their salts:

Options

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

Answer

To determine the ratio of moles of `Ag^(+)`, `Cu^(2+)`, and `Fe^(3+)` ions deposited by the passage of the same quantity of electricity through their salt solutions, we can use Faraday's laws of electrolysis. Here's a step-by-step solution:

### Step 1: Understand the Charges of Ions
- The ions involved are:
- Silver ion: `Ag^(+)` (1+ charge)
- Copper ion: `Cu^(2+)` (2+ charge)
- Iron ion: `Fe^(3+)` (3+ charge)

### Step 2: Determine the Number of Electrons Required for Deposition
- To deposit 1 mole of `Ag`, 1 mole of electrons is required.
- To deposit 1 mole of `Cu`, 2 moles of electrons are required.
- To deposit 1 mole of `Fe`, 3 moles of electrons are required.

### Step 3: Relate Moles of Ions to Charge
- According to Faraday's laws:
- 1 mole of electrons corresponds to 1 Faraday (approximately 96500 Coulombs).
- Therefore, the moles of ions deposited can be expressed as:
- For `Ag`: 1 mole of `Ag` requires 1 Faraday.
- For `Cu`: 1 mole of `Cu` requires 2 Faradays.
- For `Fe`: 1 mole of `Fe` requires 3 Faradays.

### Step 4: Calculate the Ratio of Moles
- If we pass the same quantity of electricity (let's say 1 Faraday), we can calculate how many moles of each ion will be deposited:
- Moles of `Ag` deposited = 1 mole (for 1 Faraday)
- Moles of `Cu` deposited = 1/2 mole (for 1 Faraday, since it requires 2 Faradays for 1 mole)
- Moles of `Fe` deposited = 1/3 mole (for 1 Faraday, since it requires 3 Faradays for 1 mole)

### Step 5: Express the Ratios
- The ratio of moles of `Ag`, `Cu`, and `Fe` deposited can be expressed as:
- `Ag : Cu : Fe = 1 : (1/2) : (1/3)`

### Step 6: Simplify the Ratio
- To simplify the ratio, we can multiply all parts by 6 (the least common multiple of the denominators 1, 2, and 3):
- `Ag : Cu : Fe = 6 : 3 : 2`

### Final Ratio
- Thus, the final ratio of moles of `Ag^(+)`, `Cu^(2+)`, and `Fe^(3+)` deposited is:
- **6 : 3 : 2**