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 \( Fe_2(SO_4)_3 \) and acidified \( FeCl_3 \) when the same charge is passed, we can follow these steps:

### Step 1: Identify the oxidation states of iron in both electrolytes.
- In \( Fe_2(SO_4)_3 \), iron is in the +3 oxidation state (as \( Fe^{3+} \)).
- In \( FeSO_4 \), iron is in the +2 oxidation state (as \( Fe^{2+} \)).

### Step 2: Write the reduction half-reactions for both electrolytes.
- For \( FeSO_4 \):
\[ Fe^{2+} + 2e^- \rightarrow Fe \]
- For \( FeCl_3 \):
\[ Fe^{3+} + 3e^- \rightarrow Fe \]

### Step 3: Determine the number of electrons required for deposition.
- The reaction for \( Fe^{2+} \) requires 2 electrons to deposit 1 mole of iron.
- The reaction for \( Fe^{3+} \) requires 3 electrons to deposit 1 mole of iron.

### Step 4: Set up a common basis to compare the two reactions.
To compare the amount of iron deposited from both reactions when the same charge is passed, we need to find a common multiple of the electrons involved:
- The least common multiple of 2 and 3 is 6.

### Step 5: Calculate the moles of iron deposited from each electrolyte.
- For \( Fe^{2+} \):
\[ 6 \text{ electrons} \rightarrow \frac{6}{2} = 3 \text{ moles of } Fe \]
- For \( Fe^{3+} \):
\[ 6 \text{ electrons} \rightarrow \frac{6}{3} = 2 \text{ moles of } Fe \]

### Step 6: Determine the ratio of ions deposited at the cathode.
The ratio of moles of iron deposited from \( Fe^{2+} \) to \( Fe^{3+} \) is:
\[ \text{Ratio} = \frac{3 \text{ moles of } Fe^{2+}}{2 \text{ moles of } Fe^{3+}} = \frac{3}{2} \]

### Final Answer:
The ratio of ions deposited at the cathode from \( FeSO_4 \) to \( FeCl_3 \) is \( 3:2 \).

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