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. Amount of iron deposited in `FeSO_(4) and Fe_(2)(SO_(4))_(3)` is equal
B. Amount of iron deposite in `Fe(NO_(3))_(3) " is " (2)/(3)` of the amount of iron deposited in `FeSO_(4)`
C. Amount of iron deposited in `Fe_(2)(SO_(4))_(3) and Fe(NO_(3))_(3)` are equal
D. Same gas willl evolve in all three cases at anode

Answer

To solve the problem, we need to analyze the electrolysis of the three different electrolytic cells containing FeSO4, Fe2(SO4)3, and Fe(NO3)3. We will use Faraday's laws of electrolysis to determine the amount of iron deposited in each case.

### Step-by-Step Solution:

1. **Identify the Equivalent Weights**:
- For FeSO4: The iron ion is Fe²⁺. The equivalent weight of iron (Fe) in FeSO4 is calculated as:
\[ \text{Equivalent weight} = \frac{\text{Molar mass of Fe}}{2} = \frac{m}{2} \]
- For Fe2(SO4)3: The iron ion is Fe³⁺. The equivalent weight of iron in Fe2(SO4)3 is:
\[ \text{Equivalent weight} = \frac{\text{Molar mass of Fe}}{3} = \frac{m}{3} \]
- For Fe(NO3)3: The iron ion is also Fe³⁺. The equivalent weight is the same as for Fe2(SO4)3:
\[ \text{Equivalent weight} = \frac{m}{3} \]

2. **Apply Faraday's Second Law of Electrolysis**:
- According to Faraday's second law, the amount of substance deposited (or liberated) at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte and inversely proportional to the equivalent weight of the substance.
- If the same quantity of electricity (Q) is passed through all three cells, the amount of iron deposited can be expressed as:
\[ \text{Weight of Fe deposited} \propto \frac{Q}{\text{Equivalent weight}} \]

3. **Calculate the Ratios of Iron Deposited**:
- For FeSO4:
\[ \text{Weight of Fe} \propto \frac{Q}{m/2} = \frac{2Q}{m} \]
- For Fe2(SO4)3:
\[ \text{Weight of Fe} \propto \frac{Q}{m/3} = \frac{3Q}{m} \]
- For Fe(NO3)3:
\[ \text{Weight of Fe} \propto \frac{Q}{m/3} = \frac{3Q}{m} \]

4. **Establish the Relationships**:
- The ratio of the weights of iron deposited in FeSO4, Fe2(SO4)3, and Fe(NO3)3 can be summarized as:
\[ \text{Weight of Fe in FeSO4} : \text{Weight of Fe in Fe2(SO4)3} : \text{Weight of Fe in Fe(NO3)3} = \frac{2Q}{m} : \frac{3Q}{m} : \frac{3Q}{m} \]
- This simplifies to:
\[ 2 : 3 : 3 \]

5. **Analyze Each Option**:
- **Option A**: Amount of iron deposited in FeSO4 and Fe2(SO4)3 is equal. **(False)**
- **Option B**: Amount of iron deposited in Fe(NO3)3 is \( \frac{2}{3} \) of the amount in FeSO4. **(True)**
- **Option C**: Amount of iron deposited in Fe2(SO4)3 and Fe(NO3)3 are equal. **(True)**
- **Option D**: Same gas will evolve in all three cases at anode. **(False)**

### Conclusion:
The correct options are B and C.