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 follow these steps:

### Step 1: Understand the electrochemical reactions
- **Silver ion (`Ag^(+)`)**: The reduction reaction is:
\[ Ag^+ + e^- \rightarrow Ag \]
This indicates that 1 mole of `Ag^(+)` requires 1 mole of electrons (1 Faraday) to deposit 1 mole of solid silver.

- **Copper ion (`Cu^(2+)`)**: The reduction reaction is:
\[ Cu^{2+} + 2e^- \rightarrow Cu \]
This indicates that 1 mole of `Cu^(2+)` requires 2 moles of electrons (2 Faradays) to deposit 1 mole of solid copper.

- **Iron ion (`Fe^(3+)`)**: The reduction reaction is:
\[ Fe^{3+} + 3e^- \rightarrow Fe \]
This indicates that 1 mole of `Fe^(3+)` requires 3 moles of electrons (3 Faradays) to deposit 1 mole of solid iron.

### Step 2: Calculate the moles deposited for the same quantity of electricity
Assuming we pass 1 Faraday of electricity through each solution:

- For `Ag^(+)`:
- 1 Faraday deposits 1 mole of `Ag`.

- For `Cu^(2+)`:
- 1 Faraday deposits \(\frac{1}{2}\) mole of `Cu` (since it requires 2 Faradays for 1 mole).

- For `Fe^(3+)`:
- 1 Faraday deposits \(\frac{1}{3}\) mole of `Fe` (since it requires 3 Faradays for 1 mole).

### Step 3: Write the ratio of moles deposited
Now we can express the moles deposited as:
- Moles of `Ag` = 1
- Moles of `Cu` = \(\frac{1}{2}\)
- Moles of `Fe` = \(\frac{1}{3}\)

To find the ratio, we can express it as:
\[ 1 : \frac{1}{2} : \frac{1}{3} \]

### Step 4: Simplify the ratio
To simplify this ratio, we can multiply each term by 6 (the least common multiple of the denominators 1, 2, and 3):
\[ 1 \times 6 : \frac{1}{2} \times 6 : \frac{1}{3} \times 6 = 6 : 3 : 2 \]

Thus, the final ratio of moles of `Ag^(+)`, `Cu^(2+)`, and `Fe^(3+)` deposited is:
\[ 6 : 3 : 2 \]

### Conclusion
The ratio of moles of `Ag^(+)`, `Cu^(2+)`, and `Fe^(3+)` ions deposited by the passage of the same quantity of electricity is **6:3:2**.