An electric charge of 5 Faradays is passed through three electrolytes `AgNO_3, CuSO_4` and `FeCl_3` solution. The grams of each metal liberted at cathode will be

Options

A. Ag = 10.8 g, Cu = 12.7g, Fe = 1.11g
B. Ag = 540 g, Cu = 367.5 g, Fe = 325 g
C. Ag=108 g , Cu=63.5 g , Fe=56 g
D. Ag=540 g , Cu=158.8 g , Fe = 93.3 g

Explanation

To solve the problem of calculating the grams of each metal liberated at the cathode when an electric charge of 5 Faradays is passed through the electrolytes AgNO₃, CuSO₄, and FeCl₃, we will follow these steps:

### Step 1: Calculate the mass of silver (Ag) liberated from AgNO₃
1. The half-reaction for silver in AgNO₃ is:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]
This indicates that 1 mole of silver is deposited for every 1 Faraday of charge.

2. Given that 5 Faradays of charge is passed:
\[ \text{Moles of Ag} = 5 \text{ Faradays} \times \frac{1 \text{ mole Ag}}{1 \text{ Faraday}} = 5 \text{ moles of Ag} \]

3. The molar mass of silver (Ag) is approximately 108 g/mol. Therefore, the mass of silver liberated is:
\[ \text{Mass of Ag} = 5 \text{ moles} \times 108 \text{ g/mol} = 540 \text{ grams} \]

### Step 2: Calculate the mass of copper (Cu) liberated from CuSO₄
1. The half-reaction for copper in CuSO₄ is:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]
This indicates that 1 mole of copper is deposited for every 2 Faradays of charge.

2. Given that 5 Faradays of charge is passed:
\[ \text{Moles of Cu} = 5 \text{ Faradays} \times \frac{1 \text{ mole Cu}}{2 \text{ Faradays}} = 2.5 \text{ moles of Cu} \]

3. The molar mass of copper (Cu) is approximately 63.5 g/mol. Therefore, the mass of copper liberated is:
\[ \text{Mass of Cu} = 2.5 \text{ moles} \times 63.5 \text{ g/mol} = 158.75 \text{ grams} \approx 158.8 \text{ grams} \]

### Step 3: Calculate the mass of iron (Fe) liberated from FeCl₃
1. The half-reaction for iron in FeCl₃ is:
\[ \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \]
This indicates that 1 mole of iron is deposited for every 3 Faradays of charge.

2. Given that 5 Faradays of charge is passed:
\[ \text{Moles of Fe} = 5 \text{ Faradays} \times \frac{1 \text{ mole Fe}}{3 \text{ Faradays}} = \frac{5}{3} \text{ moles of Fe} \approx 1.67 \text{ moles of Fe} \]

3. The molar mass of iron (Fe) is approximately 56 g/mol. Therefore, the mass of iron liberated is:
\[ \text{Mass of Fe} = \frac{5}{3} \text{ moles} \times 56 \text{ g/mol} \approx 93.33 \text{ grams} \approx 93.3 \text{ grams} \]

### Final Results
- Mass of Ag liberated: **540 grams**
- Mass of Cu liberated: **158.8 grams**
- Mass of Fe liberated: **93.3 grams**

### Summary
The final amounts of metals liberated at the cathode are:
- Silver (Ag): 540 grams
- Copper (Cu): 158.8 grams
- Iron (Fe): 93.3 grams