To solve the problem, we will use Faraday's laws of electrolysis.
Given:
- Current (I) = 45 A
- Time (t) = 1 hour 45 minutes = 105 minutes = \( 105 \times 60 \) seconds = 6300 seconds
- Atomic mass of gold (Au) = 193 g/mol
- Faraday's constant (F) = 96500 C
a. The mass of gold deposited
First, we'll calculate the total charge (Q) that has passed through the cell using the formula: \[ Q = I \times t \] Substituting the values: \[ Q = 45 , \text{A} \times 6300 , \text{s} = 283500 , \text{C} \]
Next, we need to determine the number of moles of electrons (n) used to deposit the gold. Since the reduction of gold ions (Au\(^3+\)) to gold (Au) requires 3 moles of electrons per mole of gold: \[ n = \frac{Q}{F} = \frac{283500 , \text{C}}{96500 , \text{C/mol}} \approx 2.94 , \text{mol of electrons} \] To find the moles of gold deposited: \[ \text{Moles of Au} = \frac{n}{3} = \frac{2.94}{3} \approx 0.98 , \text{mol} \]
Finally, we can calculate the mass of gold (m) deposited using the formula: \[ m = \text{moles} \times \text{molar mass} = 0.98 , \text{mol} \times 193 , \text{g/mol} \approx 189.34 , \text{g} \]
b. The number of moles of gold deposited
As calculated above, the moles of gold deposited is: \[ \text{Moles of Au} \approx 0.98 , \text{mol} \]
c. Time taken for 5.5 g of gold
Now we want to calculate the time taken to deposit 5.5 g of gold. First, we find the number of moles of gold from 5.5 g: \[ \text{Moles of Au} = \frac{5.5 , \text{g}}{193 , \text{g/mol}} \approx 0.0285 , \text{mol} \] For this amount of gold, the number of moles of electrons required is: \[ n = 3 \times \text{Moles of Au} = 3 \times 0.0285 \approx 0.0855 , \text{mol} \]
Next, we calculate the total charge required to deposit this amount of gold: \[ Q = n \times F = 0.0855 , \text{mol} \times 96500 , \text{C/mol} \approx 8258 , \text{C} \]
Now we can use the current to find the time required: \[ t = \frac{Q}{I} = \frac{8258 , \text{C}}{45 , \text{A}} \approx 183.51 , \text{s} \approx 3.06 , \text{minutes} \]
Final Answers:
a. The mass of gold deposited = 189.34 g
b. The number of moles of gold deposited = 0.98 mol
c. Time taken for 5.5 g of gold = 3.06 minutes