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 determining 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 can follow these steps:

### Step 1: Understand the Electrochemical Reactions
1. **For AgNO₃**: The reaction is:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]
Here, 1 mole of Ag is deposited per 1 Faraday of charge.

2. **For CuSO₄**: The reaction is:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]
Here, 1 mole of Cu is deposited per 2 Faradays of charge.

3. **For FeCl₃**: The reaction is:
\[ \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \]
Here, 1 mole of Fe is deposited per 3 Faradays of charge.

### Step 2: Calculate the Amount of Each Metal Deposited
1. **Calculate for Silver (Ag)**:
- Charge required for 1 mole of Ag = 1 Faraday.
- Therefore, for 5 Faradays:
\[ \text{Moles of Ag} = 5 \, \text{Faradays} \times \frac{1 \, \text{mole Ag}}{1 \, \text{Faraday}} = 5 \, \text{moles Ag} \]
- Molar mass of Ag = 108 g/mol.
- Mass of Ag deposited:
\[ \text{Mass of Ag} = 5 \, \text{moles} \times 108 \, \text{g/mol} = 540 \, \text{grams} \]

2. **Calculate for Copper (Cu)**:
- Charge required for 1 mole of Cu = 2 Faradays.
- Therefore, for 5 Faradays:
\[ \text{Moles of Cu} = 5 \, \text{Faradays} \times \frac{1 \, \text{mole Cu}}{2 \, \text{Faradays}} = 2.5 \, \text{moles Cu} \]
- Molar mass of Cu = 63.5 g/mol.
- Mass of Cu deposited:
\[ \text{Mass of Cu} = 2.5 \, \text{moles} \times 63.5 \, \text{g/mol} = 158.75 \, \text{grams} \]

3. **Calculate for Iron (Fe)**:
- Charge required for 1 mole of Fe = 3 Faradays.
- Therefore, for 5 Faradays:
\[ \text{Moles of Fe} = 5 \, \text{Faradays} \times \frac{1 \, \text{mole Fe}}{3 \, \text{Faradays}} = \frac{5}{3} \, \text{moles Fe} \approx 1.67 \, \text{moles Fe} \]
- Molar mass of Fe = 56 g/mol.
- Mass of Fe deposited:
\[ \text{Mass of Fe} = \frac{5}{3} \, \text{moles} \times 56 \, \text{g/mol} \approx 93.33 \, \text{grams} \]

### Final Results
- Mass of Ag deposited = 540 grams
- Mass of Cu deposited = 158.75 grams
- Mass of Fe deposited = 93.33 grams

### Summary
The grams of each metal liberated at the cathode are:
- **Ag**: 540 grams
- **Cu**: 158.75 grams
- **Fe**: 93.33 grams