What volume of hydrogen and oxygen would be obtained at `27^@C` and 740 mm of Hg by passing a current of 25 amperes through acidulated water for 24 hours?

Answer

To solve the problem, we will follow these steps:

### Step 1: Convert time to seconds
Given time is 24 hours. We need to convert this into seconds.
\[ \text{Time in seconds} = 24 \text{ hours} \times 3600 \text{ seconds/hour} = 86400 \text{ seconds} \]

### Step 2: Calculate the total charge (Q)
Using the formula \( Q = I \times T \), where \( I \) is the current in amperes and \( T \) is the time in seconds:
\[ Q = 25 \text{ A} \times 86400 \text{ s} = 2160000 \text{ C} \]

### Step 3: Calculate moles of electrons
Using Faraday's constant, which is approximately \( 96500 \text{ C/mol} \):
\[ \text{Moles of electrons} = \frac{Q}{F} = \frac{2160000 \text{ C}}{96500 \text{ C/mol}} \approx 22.38 \text{ moles} \]

### Step 4: Determine moles of hydrogen (H2) produced
From the electrolysis of water, the stoichiometry shows that 2 moles of electrons produce 1 mole of H2:
\[ \text{Moles of } H2 = \frac{22.38 \text{ moles of electrons}}{2} \approx 11.19 \text{ moles of } H2 \]

### Step 5: Determine moles of oxygen (O2) produced
From the electrolysis of water, 4 moles of electrons produce 1 mole of O2:
\[ \text{Moles of } O2 = \frac{22.38 \text{ moles of electrons}}{4} \approx 5.595 \text{ moles of } O2 \]

### Step 6: Use the ideal gas law to find volumes
The ideal gas law is given by \( PV = nRT \). We need to calculate the volume for both gases.

#### For Hydrogen (H2):
- \( n = 11.19 \text{ moles} \)
- \( R = 0.0821 \text{ L atm/(K mol)} \)
- \( T = 27^\circ C = 300 \text{ K} \)
- \( P = 740 \text{ mmHg} = \frac{740}{760} \text{ atm} \approx 0.9737 \text{ atm} \)

Using the ideal gas law:
\[ V_{H2} = \frac{nRT}{P} = \frac{11.19 \times 0.0821 \times 300}{0.9737} \approx 283.06 \text{ L} \]

#### For Oxygen (O2):
- \( n = 5.595 \text{ moles} \)

Using the ideal gas law:
\[ V_{O2} = \frac{nRT}{P} = \frac{5.595 \times 0.0821 \times 300}{0.9737} \approx 141.53 \text{ L} \]

### Final Answer:
- Volume of Hydrogen (H2) = 283.06 L
- Volume of Oxygen (O2) = 141.53 L

---