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 current produced is (in amp):

Options

A. 1A
B. 2A
C. 4A
D. 8A

Answer

To find the current produced in a hydrogen-oxygen fuel cell, we can follow these steps:

### Step 1: Calculate the number of moles of hydrogen gas (H₂)
Given that 4.48 liters of hydrogen gas is oxidized at standard temperature and pressure (STP), we can use the molar volume of a gas at STP, which is 22.4 liters per mole.

\[ \text{Number of moles of } H_2 = \frac{\text{Volume of } H_2}{\text{Molar volume at STP}} = \frac{4.48 \, \text{L}}{22.4 \, \text{L/mol}} = 0.2 \, \text{mol} \]

### Step 2: Determine the number of electrons transferred
From the given anode reaction:
\[ H_2(g) + 2OH^-(aq) \rightarrow 2H_2O(l) + 2e^- \]
We see that 1 mole of H₂ produces 2 moles of electrons (n-factor = 2).

Thus, for 0.2 moles of H₂:
\[ \text{Number of electrons} = 0.2 \, \text{mol} \times 2 \, \text{mol e}^- = 0.4 \, \text{mol e}^- \]

### Step 3: Convert moles of electrons to charge
Using Faraday's constant, which is approximately 96500 C/mol, we can calculate the total charge (Q) transferred:
\[ Q = \text{Number of moles of electrons} \times \text{Faraday's constant} = 0.4 \, \text{mol} \times 96500 \, \text{C/mol} = 38600 \, \text{C} \]

### Step 4: Calculate the current
Current (I) can be calculated using the formula:
\[ I = \frac{Q}{t} \]
where \(t\) is the time in seconds. Given that \(t = 9650 \, \text{s}\):
\[ I = \frac{38600 \, \text{C}}{9650 \, \text{s}} \approx 4 \, \text{A} \]

### Final Answer
The current produced is approximately **4 Amperes**.

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