Aluminium can be prepared by:

Options

A. Electrolytic reduction of aluminia in presence of Cryolite & fluorspar.
B. Reduction of `AlCl_3(s)` by potassium amalgam.
C. Reduction of aqueous solution of `Al_(2)(SO_(4))_(3) (aq)` by zinc metal
D. Electrolysis of aqueous solution of its salt .

Answer

To determine how aluminum can be prepared, we will analyze the given methods step by step.

### Step 1: Electrolytic Reduction of Alumina
The primary method for the preparation of aluminum is through the **electrolytic reduction of alumina (Al2O3)** in the presence of **cryolite (Na3AlF6)** and fluorspar. This method is known as the **Hall-Héroult process**.

- **Reaction**: In this process, alumina is dissolved in molten cryolite, which lowers the melting point and increases conductivity. During electrolysis, aluminum ions (Al³⁺) are reduced at the cathode:

\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]

- **Conclusion**: This method is a widely accepted and efficient way to extract aluminum.

### Step 2: Reduction of AlCl3 by Potassium Amalgam
Another method involves the **reduction of aluminum chloride (AlCl3)** using **potassium amalgam**.

- **Reaction**: When a very dilute solution of potassium amalgam (about 1.5% K) is used, aluminum can be extracted from AlCl3:

\[ 3\text{K} + \text{AlCl}_3 \rightarrow \text{Al} + 3\text{KCl} \]

- **Conclusion**: This reaction is valid and can successfully produce aluminum.

### Step 3: Reduction of Aqueous Solution of Al2(SO4)3 by Zinc
Now, let's consider the reduction of an aqueous solution of aluminum sulfate (Al2(SO4)3) using zinc metal.

- **Analysis**: Since aluminum is higher than zinc in the electrochemical series, zinc cannot displace aluminum from its salt. Therefore, no reaction occurs:

\[ \text{No Reaction} \]

- **Conclusion**: This method is not applicable for the preparation of aluminum.

### Step 4: Electrolysis of Aqueous Solution of Aluminum Salt
Finally, we examine the electrolysis of an aqueous solution of aluminum salt.

- **Analysis**: In this case, both Al³⁺ ions and H⁺ ions are present in the solution. However, the reduction potential of hydrogen is higher than that of aluminum, meaning hydrogen will be preferentially discharged at the cathode:

\[ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2(g) \]

- **Conclusion**: This method will not yield aluminum, as hydrogen gas is evolved instead.

### Final Conclusion
From the analysis above, the methods that can prepare aluminum are:
1. Electrolytic reduction of alumina in the presence of cryolite and fluorspar.
2. Reduction of AlCl3 by potassium amalgam.

Thus, the correct answers are **Option 1 and Option 2**.

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