How long does it take to deposit 100 g of Al from an electrolytic cell containing `Al_(2)O_(3)` using a current of 125 ampere ?

Options

A. 1.54 h
B. 1.42 h
C. 1.32 h
D. 2.15 h

Explanation

To solve the problem of how long it takes to deposit 100 g of aluminum (Al) from an electrolytic cell containing Al₂O₃ using a current of 125 amperes, we can follow these steps:

### Step 1: Determine the moles of aluminum to be deposited
First, we need to calculate the number of moles of aluminum in 100 g. The molar mass of aluminum (Al) is approximately 27 g/mol.

\[ \text{Moles of Al} = \frac{\text{mass}}{\text{molar mass}} = \frac{100 \, \text{g}}{27 \, \text{g/mol}} \approx 3.70 \, \text{moles} \]

### Step 2: Calculate the total charge required
From electrochemistry, we know that to reduce one mole of Al³⁺ to Al, we need 3 moles of electrons (3 Faradays). Therefore, for 3.70 moles of Al, the total charge (Q) required can be calculated as follows:

\[ Q = \text{moles of Al} \times 3 \times F \]

Where \( F \) (Faraday's constant) is approximately 96500 C/mol.

\[ Q = 3.70 \, \text{moles} \times 3 \times 96500 \, \text{C/mol} \approx 107,0000 \, \text{C} \]

### Step 3: Use the relationship between charge, current, and time
According to the formula from Faraday's laws of electrolysis:

\[ Q = I \times t \]

Where \( I \) is the current in amperes and \( t \) is the time in seconds. Rearranging the formula to find time gives us:

\[ t = \frac{Q}{I} \]

Substituting the values we have:

\[ t = \frac{1070000 \, \text{C}}{125 \, \text{A}} \approx 8560 \, \text{seconds} \]

### Step 4: Convert time from seconds to hours
To convert seconds into hours, we divide by 3600 (the number of seconds in an hour):

\[ t \approx \frac{8560 \, \text{s}}{3600 \, \text{s/h}} \approx 2.38 \, \text{hours} \]

### Conclusion
It takes approximately **2.38 hours** to deposit 100 g of aluminum from the electrolytic cell.

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