In a fuel cell `H_(2)` and `O_(2)` react to produce electricity. In the process `H_(2)` gas is oxidised at the anode and `O_(2)` is reduced at the cathode. If 6.72 litre of `H_(2)` at NTP reacts in 15 minute, what is the average current produced ? If the entire current is used for electro-deposition of Cu from `Cu^(2+)`, how many g of Cu are deposited ?

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of \( H_2 \)

We are given that 6.72 liters of \( H_2 \) gas reacts at NTP (Normal Temperature and Pressure). At NTP, 1 mole of gas occupies 22.4 liters.

\[ \text{Number of moles of } H_2 = \frac{\text{Volume of } H_2}{\text{Molar volume at NTP}} = \frac{6.72 \, \text{liters}}{22.4 \, \text{liters/mole}} = 0.3 \, \text{moles} \]

### Step 2: Determine the total charge (Q) produced

The oxidation of \( H_2 \) at the anode can be represented as:
\[ H_2 \rightarrow 2H^+ + 2e^- \]
This shows that 1 mole of \( H_2 \) produces 2 moles of electrons. Therefore, for 0.3 moles of \( H_2 \):
\[ \text{Moles of electrons} = 0.3 \, \text{moles} \times 2 = 0.6 \, \text{moles of electrons} \]

Using Faraday's constant (96500 C/mole), we can calculate the total charge:
\[ Q = \text{Moles of electrons} \times \text{Faraday's constant} = 0.6 \, \text{moles} \times 96500 \, \text{C/mole} = 57900 \, \text{C} \]

### Step 3: Calculate the average current (I)

The current can be calculated using the formula:
\[ I = \frac{Q}{t} \]
where \( t \) is the time in seconds. Given that the time is 15 minutes:
\[ t = 15 \, \text{minutes} \times 60 \, \text{seconds/minute} = 900 \, \text{seconds} \]
Now substituting the values:
\[ I = \frac{57900 \, \text{C}}{900 \, \text{s}} = 64.33 \, \text{A} \]

### Step 4: Calculate the mass of copper deposited

The reduction of copper ions can be represented as:
\[ Cu^{2+} + 2e^- \rightarrow Cu \]
This shows that 1 mole of \( Cu^{2+} \) requires 2 moles of electrons. The number of moles of \( Cu \) deposited can be calculated from the charge used for deposition:
\[ \text{Moles of } Cu = \frac{Q}{\text{Faraday's constant} \times 2} = \frac{57900 \, \text{C}}{96500 \, \text{C/mole} \times 2} = 0.3 \, \text{moles} \]

Now, we can find the mass of copper deposited using its molar mass (63.5 g/mole):
\[ \text{Mass of } Cu = \text{Moles of } Cu \times \text{Molar mass of } Cu = 0.3 \, \text{moles} \times 63.5 \, \text{g/mole} = 19.05 \, \text{g} \]

### Final Answers
- Average current produced: \( 64.33 \, \text{A} \)
- Mass of copper deposited: \( 19.05 \, \text{g} \)