Calculate the number of coulombs required to deposit 5.4 g of Al when the electrode reaction is
`Al^(3+)+3e^(-) to Al`
(Given , atomic mass of Al =27 g `mol^(-1), F=96500 C mol^(-1)`)

Explanation

To calculate the number of coulombs required to deposit 5.4 g of Al using the given electrode reaction, we can follow these steps:

### Step 1: Identify the reaction and the number of electrons involved
The electrode reaction is:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]
This indicates that 3 moles of electrons are required to deposit 1 mole of aluminum.

### Step 2: Calculate the number of moles of Al to be deposited
Using the atomic mass of aluminum (Al = 27 g/mol), we can calculate the number of moles of aluminum in 5.4 g:
\[ \text{Number of moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{5.4 \text{ g}}{27 \text{ g/mol}} = 0.2 \text{ mol} \]

### Step 3: Calculate the number of moles of electrons required
Since 1 mole of Al requires 3 moles of electrons, the number of moles of electrons required to deposit 0.2 moles of Al is:
\[ \text{Moles of electrons} = 0.2 \text{ mol Al} \times 3 \text{ mol e}^- \text{ per mol Al} = 0.6 \text{ mol e}^- \]

### Step 4: Convert moles of electrons to coulombs
Using Faraday's constant (F = 96500 C/mol), we can calculate the total charge (in coulombs) required to deposit the aluminum:
\[ \text{Charge (Q)} = \text{moles of electrons} \times F = 0.6 \text{ mol e}^- \times 96500 \text{ C/mol} = 57900 \text{ C} \]

### Step 5: Final result
Thus, the number of coulombs required to deposit 5.4 g of Al is:
\[ Q = 57900 \text{ C} \quad \text{or} \quad Q = 5.79 \times 10^4 \text{ C} \]

### Summary
The total charge required to deposit 5.4 g of aluminum is \( 57900 \text{ C} \) or \( 5.79 \times 10^4 \text{ C} \).

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