To solve the problem, we will use Faraday's laws of electrolysis. According to Faraday's first law, the mass of a substance deposited during electrolysis is directly proportional to the quantity of charge (in coulombs) passed through the electrolyte.
Given Data:
- Current (I) = 45 A
- Time (t) = 1 hour 45 minutes = 1 × 60 + 45 = 105 minutes = 105 × 60 seconds = 6300 seconds
- Atomic mass of gold (Au) = 197 g/mol
- Faraday's constant (F) = 96500 C
a. Calculate the mass of gold deposited
-
Calculate the total charge (Q) passed through the solution: \[ Q = I \times t = 45 , \text{A} \times 6300 , \text{s} = 283500 , \text{C} \]
-
Next, we need to find the number of moles of electrons (n) used in the reaction. We know one mole of electrons is equivalent to one Faraday charge: \[ n = \frac{Q}{F} = \frac{283500 , \text{C}}{96500 , \text{C/mol}} \approx 2.94 , \text{mol of electrons} \]
-
Gold (Au) typically requires 3 electrons to deposit one mole of gold (i.e., \( \text{Au}^{3+} + 3e^- \rightarrow \text{Au} \)). Therefore, the moles of gold deposited (moles of Au) can be calculated as: \[ \text{Moles of } \text{Au} = \frac{n}{3} = \frac{2.94}{3} \approx 0.98 , \text{mol} \]
-
Finally, the mass of gold deposited can be calculated using its molar mass: \[ \text{Mass of Au} = \text{moles of Au} \times \text{molar mass of Au} = 0.98 , \text{mol} \times 197 , \text{g/mol} \approx 193.06 , \text{g} \]
b. The number of moles of gold deposited
From the previous calculations: \[ \text{Moles of Au} \approx 0.98 , \text{mol} \]
c. If the same current is used, find the time taken for 5.5 g of gold
-
First, calculate the moles of gold that correspond to 5.5 g: \[ \text{Moles of Au} = \frac{5.5 , \text{g}}{197 , \text{g/mol}} \approx 0.0279 , \text{mol} \]
-
The moles of electrons required to deposit this amount of gold: \[ \text{Moles of electrons} = \text{Moles of Au} \times 3 = 0.0279 , \text{mol} \times 3 \approx 0.0837 , \text{mol} \]
-
Calculate the total charge required: \[ Q = n \times F = 0.0837 , \text{mol} \times 96500 , \text{C/mol} \approx 8065.05 , \text{C} \]
-
Finally, calculate the time taken using the current: \[ t = \frac{Q}{I} = \frac{8065.05 , \text{C}}{45 , \text{A}} \approx 179.22 , \text{s} \approx 2.99 , \text{minutes} \]
Summary:
a. The mass of gold deposited: 193.06 g
b. The number of moles of gold deposited: 0.98 mol
c. Time taken to deposit 5.5 g of gold: about 2.99 minutes