If 54 g of silver is deposited during an electrolysis reaction, how much aluminium will be deposited by the same amount of electric current?

Options

A. 2.7 g
B. 4.5 g
C. 27 g
D. 5.4 g

Explanation

To solve the problem of how much aluminum will be deposited when 54 g of silver is deposited during electrolysis, we can follow these steps:

### Step 1: Determine the moles of silver deposited
First, we need to calculate the number of moles of silver (Ag) deposited. The molar mass of silver is approximately 108 g/mol.

\[ \text{Moles of Ag} = \frac{\text{mass of Ag}}{\text{molar mass of Ag}} = \frac{54 \, \text{g}}{108 \, \text{g/mol}} = 0.5 \, \text{mol} \]

### Step 2: Determine the charge required for silver deposition
In electrolysis, the deposition of silver involves the reduction of Ag⁺ ions. The reaction can be represented as:

\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]

This means that 1 mole of silver requires 1 mole of electrons (or 1 Faraday) for deposition. Therefore, for 0.5 moles of silver, the charge required is:

\[ \text{Charge for 0.5 mol of Ag} = 0.5 \, \text{mol} \times 1 \, \text{Faraday} = 0.5 \, \text{Faraday} \]

### Step 3: Determine the charge required for aluminum deposition
The deposition of aluminum involves the reduction of Al³⁺ ions. The reaction can be represented as:

\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \]

This means that 1 mole of aluminum requires 3 moles of electrons (or 3 Faradays) for deposition. The molar mass of aluminum is approximately 27 g/mol.

### Step 4: Calculate how much aluminum can be deposited with the same charge
We have already established that we have 0.5 Faraday of charge available. To find out how much aluminum can be deposited with this charge, we can set up a proportion based on the charges required:

\[ \text{For 3 Faraday, 27 g of Al is deposited} \]
\[ \text{For 0.5 Faraday, let } x \text{ g of Al be deposited} \]

Using the formula:

\[ \frac{27 \, \text{g}}{3 \, \text{Faraday}} = \frac{x \, \text{g}}{0.5 \, \text{Faraday}} \]

Cross-multiplying gives:

\[ 27 \times 0.5 = 3 \times x \]

\[ 13.5 = 3x \]

\[ x = \frac{13.5}{3} = 4.5 \, \text{g} \]

### Conclusion
Thus, when 54 g of silver is deposited during electrolysis, 4.5 g of aluminum will also be deposited by the same amount of electric current.

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