1 mol of electrons pass through each of the solution of `AgNO_(3), CuSO_(4) and AlCl_(3) ` when Ag, Cu and Al are deposited. Molar ratio of their deposition will be:

Options

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

Answer

To solve the problem of finding the molar ratio of deposition of Ag, Cu, and Al when 1 mole of electrons passes through their respective solutions, we can follow these steps:

### Step 1: Write the half-reactions for each metal ion
- For silver (Ag):
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]
This shows that 1 mole of Ag^+ requires 1 mole of electrons to deposit 1 mole of Ag.

- For copper (Cu):
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]
This indicates that 1 mole of Cu^2+ requires 2 moles of electrons to deposit 1 mole of Cu.

- For aluminum (Al):
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]
This shows that 1 mole of Al^3+ requires 3 moles of electrons to deposit 1 mole of Al.

### Step 2: Determine the amount of metal deposited per mole of electrons
- For Ag: 1 mole of Ag is deposited with 1 mole of electrons.
- For Cu: With 1 mole of electrons, only \( \frac{1}{2} \) mole of Cu is deposited (since it requires 2 moles of electrons for 1 mole of Cu).
- For Al: With 1 mole of electrons, only \( \frac{1}{3} \) mole of Al is deposited (since it requires 3 moles of electrons for 1 mole of Al).

### Step 3: Establish the molar ratios
Now we can summarize the deposition amounts:
- Ag: 1 mole
- Cu: \( \frac{1}{2} \) mole
- Al: \( \frac{1}{3} \) mole

To find a common ratio, we can express these amounts with a common denominator. The least common multiple of 1, 2, and 3 is 6.

- Ag: \( 1 \times 6 = 6 \)
- Cu: \( \frac{1}{2} \times 6 = 3 \)
- Al: \( \frac{1}{3} \times 6 = 2 \)

Thus, the molar ratio of their deposition is:
\[ \text{Ag} : \text{Cu} : \text{Al} = 6 : 3 : 2 \]

### Final Answer
The molar ratio of their deposition will be \( 6 : 3 : 2 \).

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