Assume that impure copper contains only iron, silver and a gold as impurities. After passage of `140A`, for `482.5` sec, of the mass of the anode decreased by `22.260g` and the cathode increased in mass by `22.011g`. Estimate the % iron and % copper originally present.

Options

A. `Cu =98.88%, Fe = 0.85%`
B. `Cu =0.85%, Fe = 98.88%`
C. `Cu =90%, Fe = 10%`
D. `Cu =60%, Fe = 40%`

Explanation

To solve the problem, we need to determine the percentage of copper and iron in the impure copper sample based on the given data. Here's a step-by-step breakdown of the solution:

### Step 1: Calculate the total charge passed
We know that charge (Q) can be calculated using the formula:
\[ Q = I \times t \]
where:
- \( I = 140 \, \text{A} \) (current)
- \( t = 482.5 \, \text{s} \) (time)

Calculating the charge:
\[ Q = 140 \, \text{A} \times 482.5 \, \text{s} = 67550 \, \text{C} \]

### Step 2: Convert charge to Faraday (F)
We know that 1 Faraday (F) is approximately 96500 C. Therefore, we can convert the charge to Faraday:
\[ \text{Charge in Faraday} = \frac{Q}{96500} = \frac{67550}{96500} \approx 0.7 \, F \]

### Step 3: Determine the mass of copper deposited at the cathode
The mass of copper deposited at the cathode is given as:
\[ \text{Mass of copper deposited} = 22.011 \, \text{g} \]

### Step 4: Calculate the moles of copper deposited
Using the molar mass of copper (63.5 g/mol) and knowing that the reduction of copper involves 2 moles of electrons:
\[ \text{Moles of copper} = \frac{\text{mass}}{\text{molar mass}} = \frac{22.011 \, \text{g}}{63.5 \, \text{g/mol}} \approx 0.346 \, \text{mol} \]

### Step 5: Calculate the Faraday used for copper
Using the formula:
\[ \text{Faraday} = \text{moles of copper} \times \text{number of electrons} \]
\[ \text{Faraday for copper} = 0.346 \, \text{mol} \times 2 \approx 0.692 \, F \]

### Step 6: Calculate the Faraday used for iron
The total Faraday used is 0.7 F, so the Faraday used for iron is:
\[ \text{Faraday for iron} = 0.7 \, F - 0.692 \, F \approx 0.008 \, F \]

### Step 7: Calculate the mass of iron oxidized
Using the molar mass of iron (56 g/mol) and knowing that oxidation of iron involves 2 moles of electrons:
\[ \text{Weight of iron} = \text{Faraday for iron} \times \text{molar mass of iron} \times \frac{1}{2} \]
\[ \text{Weight of iron} = 0.008 \, F \times 56 \, \text{g/mol} \times \frac{1}{2} \approx 0.224 \, \text{g} \]

### Step 8: Calculate the percentage of iron
To find the percentage of iron in the original sample:
\[ \text{Percentage of iron} = \left( \frac{\text{Weight of iron}}{\text{Total mass of anode loss}} \right) \times 100 \]
\[ \text{Percentage of iron} = \left( \frac{0.224 \, \text{g}}{22.260 \, \text{g}} \right) \times 100 \approx 1.01\% \]

### Step 9: Calculate the percentage of copper
The percentage of copper can be calculated as:
\[ \text{Percentage of copper} = 100\% - \text{Percentage of iron} \]
\[ \text{Percentage of copper} = 100\% - 1.01\% \approx 98.99\% \]

### Final Results:
- Percentage of copper: **98.99%**
- Percentage of iron: **1.01%**