Two elcetrolytic cells are containing acidified `Fe_(2)(SO_(4))_(3)` and the other containing acidified
The ratio of ions deposited at cathode on two cells on passing same charge is respectively:

Options

A. (2:3)
B. (1:2)
C. (2:1)
D. (3:2)

Explanation

To solve the problem of finding the ratio of ions deposited at the cathode in two electrolytic cells containing acidified \( \text{Fe}_2(\text{SO}_4)_3 \) and acidified \( \text{FeCl}_3 \) when the same charge is passed, we can follow these steps:

### Step 1: Identify the ions in each solution
- In the first cell containing \( \text{Fe}_2(\text{SO}_4)_3 \), the ions present are \( \text{Fe}^{3+} \) and \( \text{SO}_4^{2-} \).
- In the second cell containing \( \text{FeCl}_3 \), the ions present are \( \text{Fe}^{3+} \) and \( \text{Cl}^- \).

### Step 2: Determine the reduction reactions at the cathode
- For \( \text{Fe}_2(\text{SO}_4)_3 \):
\[ \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \quad \text{(1 mole of Fe requires 3 moles of electrons)} \]
- For \( \text{FeCl}_3 \):
\[ \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \quad \text{(1 mole of Fe requires 3 moles of electrons)} \]

### Step 3: Calculate the amount of metal deposited using Faraday's laws of electrolysis
According to Faraday's laws, the mass of a substance deposited at an electrode is directly proportional to the quantity of electricity passed through the cell. The formula is given by:
\[ m = \frac{Q}{F} \times \frac{M}{n} \]
where:
- \( m \) = mass of the deposited metal
- \( Q \) = total electric charge passed
- \( F \) = Faraday's constant (approximately 96500 C/mol)
- \( M \) = molar mass of the metal
- \( n \) = number of electrons transferred per mole of metal

### Step 4: Find the ratio of deposits
- For \( \text{Fe}_2(\text{SO}_4)_3 \):
- 1 mole of \( \text{Fe} \) requires 3 moles of electrons.

- For \( \text{FeCl}_3 \):
- Similarly, 1 mole of \( \text{Fe} \) also requires 3 moles of electrons.

Since both reactions require the same number of electrons (3) to deposit 1 mole of iron, the ratio of the amount of iron deposited from both solutions when the same charge is passed is:
\[ \text{Ratio} = \frac{\text{Amount of Fe from } \text{Fe}_2(\text{SO}_4)_3}{\text{Amount of Fe from } \text{FeCl}_3} = 1:1 \]

### Conclusion
The ratio of ions deposited at the cathode in the two cells when the same charge is passed is \( 1:1 \).

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