How long (in hours) will it take to produce `0.3` mole of `HNO_(2)` by following reaction if an average current of `2` amp passes through the cell? `NO_(3)^(-)+3H_(3)O^(+)+2e^(-)rarrHNO_(2)+4H_(2)O,E^(@)=0.94V`

Explanation

To solve the problem of how long it will take to produce `0.3` moles of `HNO₂` with an average current of `2` amperes, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction and Electrons Required**:
The given reaction is:
\[ NO_3^{-} + 3H_3O^{+} + 2e^{-} \rightarrow HNO_2 + 4H_2O \]
From the reaction, we see that `1` mole of `HNO₂` requires `2` moles of electrons.

2. **Calculate Moles of Electrons for `0.3` Moles of `HNO₂`**:
If `1` mole of `HNO₂` requires `2` moles of electrons, then `0.3` moles of `HNO₂` will require:
\[ 0.3 \, \text{moles of } HNO_2 \times 2 \, \text{moles of electrons/mole of } HNO_2 = 0.6 \, \text{moles of electrons} \]

3. **Calculate Total Charge (Q)**:
The total charge (Q) can be calculated using Faraday's constant (F), which is approximately `96500 C/mol`:
\[ Q = \text{moles of electrons} \times F = 0.6 \, \text{moles} \times 96500 \, \text{C/mol} = 57900 \, \text{C} \]

4. **Use the Formula for Charge**:
The relationship between charge (Q), current (I), and time (T) is given by:
\[ Q = I \times T \]
Rearranging this gives:
\[ T = \frac{Q}{I} \]

5. **Substituting the Values**:
Here, the current (I) is `2` amperes:
\[ T = \frac{57900 \, \text{C}}{2 \, \text{A}} = 28950 \, \text{seconds} \]

6. **Convert Time from Seconds to Hours**:
To convert seconds to hours, we use the conversion factor `1 hour = 3600 seconds`:
\[ T = \frac{28950 \, \text{seconds}}{3600 \, \text{seconds/hour}} \approx 8.0417 \, \text{hours} \]

7. **Final Answer**:
Thus, it will take approximately `8` hours to produce `0.3` moles of `HNO₂`.