State True or False
The same quantity of electricity is passed through `Al_(2)(SO_(4))_(3)` and `AgNO_(3)` solution with platinum electrodes. If `n` number of `Al` atoms are deposited on the cathode, `3n` number of `Ag` atoms will be deposited on the cathode.

Explanation

To solve the question, we need to analyze the electrochemical reactions occurring when the same quantity of electricity is passed through `Al_(2)(SO_(4))_(3)` and `AgNO_(3)` solutions.

### Step-by-Step Solution:

1. **Identify the Reduction Reactions**:
- For aluminum sulfate (`Al_(2)(SO_(4))_(3)`), the reduction reaction at the cathode is:
\[ Al^{3+} + 3e^- \rightarrow Al \]
- For silver nitrate (`AgNO_(3)`), the reduction reaction at the cathode is:
\[ Ag^{+} + e^- \rightarrow Ag \]

2. **Determine the Number of Electrons Required**:
- From the aluminum reaction, it takes 3 moles of electrons to deposit 1 mole of aluminum.
- From the silver reaction, it takes 1 mole of electron to deposit 1 mole of silver.

3. **Calculate the Moles of Metal Deposited**:
- If we pass a certain quantity of electricity (let's say `Q`), we can express this in terms of Faraday's laws of electrolysis:
- For aluminum, if `Q` is the total charge passed, the number of moles of aluminum deposited can be calculated as:
\[ \text{Moles of } Al = \frac{Q}{3F} \]
- For silver, the number of moles of silver deposited is:
\[ \text{Moles of } Ag = \frac{Q}{F} \]

4. **Establish the Relationship**:
- If `n` moles of aluminum are deposited, then:
\[ n = \frac{Q}{3F} \implies Q = 3nF \]
- Substituting `Q` in the silver equation:
\[ \text{Moles of } Ag = \frac{3nF}{F} = 3n \]

5. **Conclusion**:
- Therefore, if `n` aluminum atoms are deposited, `3n` silver atoms will be deposited. This confirms the statement in the question.

### Final Answer:
The statement is **True**.