Suppose that gold is being plated on to another metal in an electrolytic cell. The half - cell reaction producing the `Au(s)` is `AuCl_(4)^(-)+3e^(-)+Au(s)+4Cl^(-)`. If a 0.30 A current runs for 15.00 minute, what mass of `Au(s)` will be plated, assume all the electrons are used in the reduction of `AuCl_(4)^(-)` ? the Faraday constant is 96485 C/mol and molar mass of Au is 197.

Options

A. 0.184 g
B. 0.551 g
C. 1.84 g
D. 0.613 g

Answer

To solve the problem of how much gold (Au) will be plated in an electrolytic cell when a current of 0.30 A runs for 15.00 minutes, we can follow these steps:

### Step 1: Calculate the total charge (Q) passed through the cell
The total charge can be calculated using the formula:
\[ Q = I \times t \]
where:
- \( I \) is the current in amperes (A)
- \( t \) is the time in seconds (s)

Given:
- \( I = 0.30 \, A \)
- \( t = 15.00 \, \text{minutes} = 15.00 \times 60 \, \text{seconds} = 900 \, \text{s} \)

Now, substituting the values:
\[ Q = 0.30 \, A \times 900 \, s = 270 \, C \]

### Step 2: Calculate the number of moles of electrons (n)
Using Faraday's constant, we can find the number of moles of electrons:
\[ n = \frac{Q}{F} \]
where:
- \( F \) is Faraday's constant, \( F = 96485 \, C/mol \)

Substituting the values:
\[ n = \frac{270 \, C}{96485 \, C/mol} \approx 0.00280 \, mol \]

### Step 3: Determine the number of moles of gold plated
From the half-cell reaction:
\[ AuCl_4^{-} + 3e^{-} \rightarrow Au(s) + 4Cl^{-} \]
We see that 3 moles of electrons are required to plate 1 mole of gold. Therefore, the moles of gold plated can be calculated as:
\[ \text{Moles of } Au = \frac{n}{3} = \frac{0.00280 \, mol}{3} \approx 0.000933 \, mol \]

### Step 4: Calculate the mass of gold plated
Using the molar mass of gold (Au), which is 197 g/mol, we can find the mass:
\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \]
Substituting the values:
\[ \text{Mass} = 0.000933 \, mol \times 197 \, g/mol \approx 0.1837 \, g \]

### Final Result
The mass of gold plated is approximately:
\[ \text{Mass of } Au \approx 0.184 \, g \]