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=10.8g, Cu=12.7g, Fe=1.11g`
B. `Ag=540.8g, Cu=367.5g, Fe=325g`
C. `Ag=108g, Cu=63.5g, Fe=56g`
D. `Ag=540g, Cu=158.8g, Fe=93.3g`

Answer

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 will follow these steps:

### Step 1: Identify the reactions at the cathode
The reduction reactions at the cathode for each electrolyte are as follows:
1. For AgNO₃:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]
2. For CuSO₄:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]
3. For FeCl₃:
\[ \text{Fe}^{3+} + 3e^- \rightarrow \text{Fe} \]

### Step 2: Calculate the moles of each metal deposited
Using Faraday's laws of electrolysis, we can calculate the moles of each metal deposited based on the charge passed:

- **For Ag**:
- Each mole of Ag requires 1 Faraday (1 mole of electrons).
- Therefore, the moles of Ag deposited:
\[ \text{Moles of Ag} = \text{Charge (in Faradays)} = 5 \text{ moles} \]

- **For Cu**:
- Each mole of Cu requires 2 Faradays (2 moles of electrons).
- Therefore, the moles of Cu deposited:
\[ \text{Moles of Cu} = \frac{5 \text{ Faradays}}{2} = 2.5 \text{ moles} \]

- **For Fe**:
- Each mole of Fe requires 3 Faradays (3 moles of electrons).
- Therefore, the moles of Fe deposited:
\[ \text{Moles of Fe} = \frac{5 \text{ Faradays}}{3} \approx 1.67 \text{ moles} \]

### Step 3: Calculate the mass of each metal deposited
Now, we will convert the moles of each metal to grams using their molar masses:

- **For Ag**:
- Molar mass of Ag = 108 g/mol
- Mass of Ag deposited:
\[ \text{Mass of Ag} = 5 \text{ moles} \times 108 \text{ g/mol} = 540 \text{ g} \]

- **For Cu**:
- Molar mass of Cu = 63.5 g/mol
- Mass of Cu deposited:
\[ \text{Mass of Cu} = 2.5 \text{ moles} \times 63.5 \text{ g/mol} = 158.75 \text{ g} \approx 158.8 \text{ g} \]

- **For Fe**:
- Molar mass of Fe = 56 g/mol
- Mass of Fe deposited:
\[ \text{Mass of Fe} = 1.67 \text{ moles} \times 56 \text{ g/mol} \approx 93.3 \text{ g} \]

### Final Results
Thus, the grams of each metal liberated at the cathode are:
- Ag: 540 g
- Cu: 158.8 g
- Fe: 93.3 g