In a hydrogen oxyge fuel cell, electricity is produced. In this process `H_2`(g) is oxided at anode and `O_2`(g) reduced at cathode
Given: Cathode `O_2(g)+2H_2O(l)+4e^(-)to4OH^(-)(aq)`
Anode `H_2(g)+2OH^(-)(aq)to2H_2O(l)+2e^(-)`
4.48 litre `H_2` at 1atm and 273 k oxidised in 9650 sec.
The mass of water produced is :

Options

A. 7.2g
B. 3.6g
C. 1.8g
D. 0.9g

Answer

To solve the problem of determining the mass of water produced in a hydrogen-oxygen fuel cell, we can follow these steps:

### Step 1: Understand the Reaction
In the hydrogen-oxygen fuel cell, hydrogen gas (H₂) is oxidized at the anode and oxygen gas (O₂) is reduced at the cathode. The relevant half-reactions are:
- **Anode Reaction**: \( H_2(g) + 2OH^-(aq) \rightarrow 2H_2O(l) + 2e^- \)
- **Cathode Reaction**: \( O_2(g) + 2H_2O(l) + 4e^- \rightarrow 4OH^-(aq) \)

From the anode reaction, we can see that 1 mole of hydrogen produces 2 moles of water.

### Step 2: Calculate Moles of Hydrogen
We are given that 4.48 liters of hydrogen gas is oxidized. At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters.

To find the number of moles of hydrogen:
\[ \text{Moles of } H_2 = \frac{\text{Volume of } H_2}{\text{Molar Volume at STP}} = \frac{4.48 \, \text{L}}{22.4 \, \text{L/mol}} = 0.2 \, \text{mol} \]

### Step 3: Calculate Moles of Water Produced
From the anode reaction, we know that 1 mole of H₂ produces 2 moles of H₂O. Therefore, the moles of water produced from 0.2 moles of hydrogen is:
\[ \text{Moles of } H_2O = 0.2 \, \text{mol} \times 2 = 0.4 \, \text{mol} \]

### Step 4: Calculate Mass of Water Produced
The molar mass of water (H₂O) is approximately 18 g/mol. To find the mass of water produced:
\[ \text{Mass of } H_2O = \text{Moles of } H_2O \times \text{Molar Mass of } H_2O = 0.4 \, \text{mol} \times 18 \, \text{g/mol} = 7.2 \, \text{g} \]

### Conclusion
The mass of water produced from the oxidation of 4.48 liters of hydrogen gas is **7.2 grams**.

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