How many coulombs are required to deposit 50 g of aluminium when the electrode reaction is `Al^(+3)+ 3e^(-) rarr Al(s)?`

Options

A. 536111 C
B. `536.111` C
C. 96500 C
D. 38600 C

Answer

To determine how many coulombs are required to deposit 50 g of aluminum using the given electrode reaction \( \text{Al}^{3+} + 3e^- \rightarrow \text{Al}(s) \), we can follow these steps:

### Step 1: Calculate the number of moles of aluminum
First, we need to find out how many moles of aluminum are present in 50 g. The molar mass of aluminum (Al) is approximately 27 g/mol.

\[ \text{Number of moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{50 \, \text{g}}{27 \, \text{g/mol}} \approx 1.85 \, \text{mol} \]

### Step 2: Determine the number of moles of electrons required
From the electrode reaction, we see that to deposit 1 mole of aluminum, 3 moles of electrons are required. Therefore, for 1.85 moles of aluminum:

\[ \text{Moles of electrons} = 1.85 \, \text{mol Al} \times 3 \, \text{mol e}^- / \text{mol Al} = 5.55 \, \text{mol e}^- \]

### Step 3: Convert moles of electrons to coulombs
To find the total charge in coulombs, we use Faraday's constant, which is approximately \( 96500 \, \text{C/mol} \). The total charge (Q) can be calculated as follows:

\[ Q = \text{moles of electrons} \times \text{Faraday's constant} = 5.55 \, \text{mol e}^- \times 96500 \, \text{C/mol} \approx 535725 \, \text{C} \]

### Final Answer
Thus, the total charge required to deposit 50 g of aluminum is approximately **535725 coulombs**.

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