In ` H_2 -O_2` fuel cell, `6.72 L` of hydrogen at `NTP` reacts in `15` minutes, the averge current produced in ampres is .

Options

A. 64.3 amp
B. 643.3 amp
C. 6.43 amp
D. 0.643 amp

Answer

To solve the problem of finding the average current produced in an H2-O2 fuel cell when 6.72 L of hydrogen reacts in 15 minutes, we can follow these steps:

### Step 1: Convert the volume of hydrogen to moles
At Normal Temperature and Pressure (NTP), 22.4 L of gas corresponds to 1 mole. Therefore, we can calculate the number of moles of hydrogen (H2) in 6.72 L.

\[ \text{Moles of } H_2 = \frac{6.72 \text{ L}}{22.4 \text{ L/mol}} = 0.3 \text{ moles} \]

### Step 2: Determine the charge produced
From the reaction in the fuel cell, we know that 1 mole of H2 requires 2 moles of electrons. The charge (Q) can be calculated using Faraday's constant, which is approximately 96500 Coulombs per mole of electrons.

\[ \text{Charge for } 0.3 \text{ moles of } H_2 = 0.3 \text{ moles} \times 2 \text{ moles of electrons/mole of } H_2 \times 96500 \text{ C/mole} \]

Calculating this gives:

\[ Q = 0.3 \times 2 \times 96500 = 57900 \text{ C} \]

### Step 3: Convert time from minutes to seconds
Since the current is measured in Amperes (Coulombs per second), we need to convert the time from minutes to seconds.

\[ \text{Time} = 15 \text{ minutes} \times 60 \text{ seconds/minute} = 900 \text{ seconds} \]

### Step 4: Calculate the average current
The average current (I) can be calculated using the formula:

\[ I = \frac{Q}{t} \]

Substituting the values we found:

\[ I = \frac{57900 \text{ C}}{900 \text{ s}} \approx 64.33 \text{ A} \]

### Final Answer
The average current produced is approximately **64.3 Amperes**.

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