How long will it take for a uniform current of 6.00 A to deposit 78 g of gold from a solution of `AuCl_(4)^(-)`? What mass of chlorine gas will be formed simultaneously at anode of the cell?
(Atomic mass of Au = 197)

Options

A. t = 3010 sec, w = 35.50 g
B. t = 20306 sec, w = 45.54 g
C. t = 19500 sec, w = 54.5 g
D. t = 19139.16 sec, w = 42.24 g

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of gold deposited
We know that the mass of gold deposited is 78 g. The molar mass of gold (Au) is 197 g/mol.

\[ \text{Number of moles of Au} = \frac{\text{mass}}{\text{molar mass}} = \frac{78 \, \text{g}}{197 \, \text{g/mol}} \approx 0.396 \, \text{mol} \]

### Step 2: Determine the number of moles of electrons required
From the electrochemical reaction, we know that to deposit 1 mole of gold, 3 moles of electrons are required. Therefore, for 0.396 moles of gold:

\[ \text{Number of moles of electrons} = 0.396 \, \text{mol} \times 3 = 1.188 \, \text{mol} \]

### Step 3: Calculate the total charge (Q) required
Using Faraday's constant (F = 96500 C/mol), we can calculate the total charge required to deposit the gold:

\[ Q = \text{Number of moles of electrons} \times F = 1.188 \, \text{mol} \times 96500 \, \text{C/mol} \approx 114,000 \, \text{C} \]

### Step 4: Calculate the time (T) required using the current (I)
We know that current (I) is defined as charge (Q) per unit time (T):

\[ I = \frac{Q}{T} \implies T = \frac{Q}{I} \]

Substituting the values we have:

\[ T = \frac{114,000 \, \text{C}}{6 \, \text{A}} = 19,000 \, \text{s} \]

### Step 5: Calculate the mass of chlorine gas produced
The reaction at the anode involves the oxidation of chloride ions (Cl⁻) to form chlorine gas (Cl₂). The balanced reaction is:

\[ 2 \, \text{Cl}^- \rightarrow \text{Cl}_2 + 2 \, e^- \]

From this, we see that 2 moles of electrons produce 1 mole of Cl₂. Therefore, the number of moles of Cl₂ produced from 1.188 moles of electrons is:

\[ \text{Number of moles of Cl}_2 = \frac{1.188 \, \text{mol}}{2} = 0.594 \, \text{mol} \]

### Step 6: Calculate the mass of chlorine gas produced
The molar mass of chlorine gas (Cl₂) is approximately 71 g/mol. Thus, the mass of chlorine gas produced is:

\[ \text{Mass of Cl}_2 = \text{Number of moles} \times \text{Molar mass} = 0.594 \, \text{mol} \times 71 \, \text{g/mol} \approx 42.14 \, \text{g} \]

### Final Answers:
- Time taken to deposit 78 g of gold: **19,000 seconds**
- Mass of chlorine gas produced: **42.14 g**