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 ?

Answer

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: Understand the Electrochemical Reaction
Aluminum oxide (Al₂O₃) dissociates in the electrolytic process to yield aluminum ions (Al³⁺) and oxide ions (O²⁻). The reduction reaction for aluminum can be represented as:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]
This indicates that to deposit one mole of aluminum, three moles of electrons are required.

### Step 2: Calculate the Molar Mass of Aluminum
The molar mass of aluminum (Al) is 27 g/mol.

### Step 3: Determine the Number of Moles of Aluminum to be Deposited
To find the number of moles of aluminum to be deposited, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
For 100 g of aluminum:
\[ \text{Number of moles of Al} = \frac{100 \, \text{g}}{27 \, \text{g/mol}} \approx 3.70 \, \text{mol} \]

### Step 4: Calculate the Total Charge Required
Using Faraday's law, the total charge (Q) required to deposit the aluminum can be calculated as:
\[ Q = n \times F \]
where:
- \( n \) is the total number of moles of electrons required,
- \( F \) is Faraday's constant (approximately 96500 C/mol).

Since each mole of aluminum requires 3 moles of electrons:
\[ n = 3 \times \text{Number of moles of Al} = 3 \times 3.70 \approx 11.1 \, \text{mol of electrons} \]

Now, calculate the total charge:
\[ Q = 11.1 \, \text{mol} \times 96500 \, \text{C/mol} \approx 1070000 \, \text{C} \]

### Step 5: Calculate the Time Required
Using the formula:
\[ Q = I \times t \]
we can rearrange it to find time (t):
\[ t = \frac{Q}{I} \]
Substituting the values:
\[ t = \frac{1070000 \, \text{C}}{125 \, \text{A}} \approx 8560 \, \text{s} \]

### Step 6: Convert Time to More Convenient Units
To convert seconds into hours:
\[ t \approx \frac{8560 \, \text{s}}{3600 \, \text{s/h}} \approx 2.38 \, \text{hours} \]

### Final Answer
It takes approximately 8560 seconds (or about 2.38 hours) to deposit 100 g of aluminum from the electrolytic cell.

---