The quantity of charge required to obtain one mole of aluminium from `Al_(2)0_(3)` is

Options

A. 1 F
B. 6 F
C. 3 F
D. 2 F

Explanation

To determine the quantity of charge required to obtain one mole of aluminum from Al2O3, we can follow these steps:

### Step 1: Write the Reduction Reaction
The first step is to write the reduction half-reaction for the conversion of aluminum oxide (Al2O3) to aluminum (Al). The balanced half-reaction is:

\[ \text{Al}_2\text{O}_3 + 6 \text{e}^- \rightarrow 2 \text{Al} + 3 \text{O}^{2-} \]

### Step 2: Determine the Number of Electrons Required
From the balanced half-reaction, we can see that to produce 2 moles of aluminum, 6 electrons are required. Therefore, to obtain 1 mole of aluminum, we need:

\[ \frac{6 \text{ e}^-}{2} = 3 \text{ e}^- \]

### Step 3: Calculate the Charge Required
The charge (Q) required can be calculated using Faraday's law of electrolysis, which states that the charge required to transfer n moles of electrons is given by:

\[ Q = n \cdot F \]

Where:
- \( n \) is the number of moles of electrons (in this case, 3 moles for 1 mole of aluminum).
- \( F \) is Faraday's constant, approximately \( 96500 \, \text{C/mol} \).

Substituting the values:

\[ Q = 3 \, \text{mol} \times 96500 \, \text{C/mol} = 289500 \, \text{C} \]

### Step 4: Conclusion
Thus, the quantity of charge required to obtain one mole of aluminum from Al2O3 is 289500 C.

### Final Answer
The quantity of charge required to obtain one mole of aluminum from Al2O3 is **289500 C**.

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