If same quantity of electricity is passed through three electrolytic cells containing `FeSO_(4),Fe_(2)(SO_(4))_(3)`, and `Fe(NO_(3))_(3)`, then

Options

A. The amount of iron deposited in `FeSO_(4)` and `Fe_(2)(SO_(4))_(3)` are equal.
B. The amount of iron deposited in `FeSO_(4)` is `1.5` times of the amount of iron deposited in `Fe(NO_(3))_(3)`
C. The amount of iron deposited in `Fe_(2)(SO_(4))_(3)` and `Fe(NO_(3))_(3)` are equal.
D. The same amount of gas is evolved in all three cases of the anode.

Explanation

To solve the problem, we need to analyze the electrochemical reactions occurring in the three electrolytic cells containing `FeSO4`, `Fe2(SO4)3`, and `Fe(NO3)3` when the same quantity of electricity is passed through them.

### Step-by-Step Solution:

1. **Identify the Electrolytes and Their Ions**:
- In the first cell with `FeSO4`, the ions present are `Fe^2+` and `SO4^2-`.
- In the second cell with `Fe2(SO4)3`, the ions present are `Fe^3+` and `SO4^2-`.
- In the third cell with `Fe(NO3)3`, the ions present are `Fe^3+` and `NO3^-`.

2. **Determine the Reduction Reactions**:
- For `FeSO4`:
\[ Fe^{2+} + 2e^- \rightarrow Fe(s) \]
This means 2 moles of electrons are required to deposit 1 mole of iron.
- For `Fe2(SO4)3` and `Fe(NO3)3`:
\[ Fe^{3+} + 3e^- \rightarrow Fe(s) \]
This means 3 moles of electrons are required to deposit 1 mole of iron.

3. **Calculate the Amount of Iron Deposited**:
- Let the quantity of electricity passed be `Q` coulombs.
- The relationship between charge (Q), moles of electrons (n), and Faraday's constant (F) is given by:
\[ n = \frac{Q}{F} \]
- For `FeSO4`, the amount of iron deposited (moles) is:
\[ \text{Moles of Fe} = \frac{Q}{2F} \]
- For `Fe2(SO4)3` and `Fe(NO3)3`, the amount of iron deposited (moles) is:
\[ \text{Moles of Fe} = \frac{Q}{3F} \]

4. **Compare the Amounts of Iron Deposited**:
- From the above calculations:
- In `FeSO4`, the amount of iron deposited is proportional to \(\frac{1}{2}\).
- In `Fe2(SO4)3` and `Fe(NO3)3`, the amount of iron deposited is proportional to \(\frac{1}{3}\).
- Therefore, the ratio of iron deposited in `FeSO4` to `Fe(NO3)3` is:
\[ \frac{\frac{Q}{2F}}{\frac{Q}{3F}} = \frac{3}{2} \quad \text{(or 1.5 times)} \]

5. **Anode Reactions**:
- In all three cells, the anode reaction involves the oxidation of hydroxide ions:
\[ 4OH^- \rightarrow O2 + 2H2O + 4e^- \]
- Since the same quantity of electricity is passed through all cells, the amount of oxygen gas evolved will be the same.

6. **Conclusion**:
- The correct statements based on the analysis are:
- The amount of iron deposited in `FeSO4` is not equal to that in `Fe2(SO4)3` or `Fe(NO3)3`.
- The amount of iron deposited in `FeSO4` is 1.5 times that in `Fe(NO3)3`.
- The amount of iron deposited in `Fe2(SO4)3` and `Fe(NO3)3` is equal.
- The same amount of gas is evolved in all three cases.

### Final Answer:
- The correct statements are B, C, and D.