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 aluminium using the electrode reaction \( \text{Al}^{3+} + 3e^- \rightarrow \text{Al}(s) \), we will follow these steps:

### Step 1: Determine the molar mass of aluminium
The molar mass of aluminium (Al) is approximately 27 g/mol.

### Step 2: Calculate the number of moles of aluminium in 50 g
To find the number of moles of aluminium in 50 g, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{50 \, \text{g}}{27 \, \text{g/mol}} \approx 1.85 \, \text{mol} \]

### Step 3: Determine the number of electrons required for the deposition
From the electrode reaction, we see that 1 mole of aluminium requires 3 moles of electrons (3 Faraday) for deposition:
\[ \text{Electrons required for 1 mole of Al} = 3 \, \text{mol of e}^- \]

### Step 4: Calculate the total number of moles of electrons for 1.85 moles of aluminium
To find the total moles of electrons required for 1.85 moles of aluminium:
\[ \text{Total moles of e}^- = 1.85 \, \text{mol Al} \times 3 \, \text{mol e}^- / \text{mol Al} = 5.55 \, \text{mol e}^- \]

### Step 5: Convert moles of electrons to coulombs
Using Faraday's constant, which is approximately \( 96500 \, \text{C/mol} \), we can convert moles of electrons to coulombs:
\[ \text{Total charge (C)} = \text{Total moles of e}^- \times \text{Faraday's constant} = 5.55 \, \text{mol e}^- \times 96500 \, \text{C/mol} \approx 53557.5 \, \text{C} \]

### Step 6: Round the answer
Rounding to the nearest whole number, we find that approximately 53558 coulombs are required to deposit 50 g of aluminium.

### Final Answer
Therefore, the total charge required to deposit 50 g of aluminium is approximately **53558 coulombs**.

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