The current needed to reduce 26.6 g of nitrobenzene to aniline in acidic medium, is

Options

A. 0.4 F
B. 0.6 F
C. 0.8 F
D. 1.2 F

Answer

To determine the current needed to reduce 26.6 g of nitrobenzene (C6H5NO2) to aniline (C6H5NH2) in acidic medium, we will follow these steps:

### Step 1: Write the balanced reduction reaction
The reduction of nitrobenzene to aniline can be represented as:
\[ \text{C}_6\text{H}_5\text{NO}_2 + 6 \text{H}^+ + 6 \text{e}^- \rightarrow \text{C}_6\text{H}_5\text{NH}_2 + 2 \text{H}_2\text{O} \]

### Step 2: Calculate the molar mass of nitrobenzene
The molar mass of nitrobenzene (C6H5NO2) is calculated as follows:
- Carbon (C): 12 g/mol × 6 = 72 g/mol
- Hydrogen (H): 1 g/mol × 5 = 5 g/mol
- Nitrogen (N): 14 g/mol × 1 = 14 g/mol
- Oxygen (O): 16 g/mol × 2 = 32 g/mol

Adding these together:
\[ 72 + 5 + 14 + 32 = 123 \text{ g/mol} \]

### Step 3: Determine the number of moles of nitrobenzene in 26.6 g
To find the number of moles of nitrobenzene in 26.6 g, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
\[ \text{Number of moles} = \frac{26.6 \text{ g}}{123 \text{ g/mol}} \approx 0.216 \text{ moles} \]

### Step 4: Calculate the number of electrons required
From the balanced equation, we see that 1 mole of nitrobenzene requires 6 moles of electrons for reduction. Therefore, for 0.216 moles of nitrobenzene:
\[ \text{Electrons required} = 0.216 \text{ moles} \times 6 \text{ moles of e}^- = 1.296 \text{ moles of e}^- \]

### Step 5: Convert moles of electrons to Faraday
1 mole of electrons corresponds to 96500 Coulombs (Faraday constant). Thus, the total charge (in Faraday) needed is:
\[ \text{Charge (in Faraday)} = \frac{1.296 \text{ moles}}{1} \approx 1.296 \text{ Faraday} \]

### Step 6: Conclusion
The current needed to reduce 26.6 g of nitrobenzene to aniline in acidic medium is approximately 1.296 Faraday.