Hydrazine can be used in fuel cell
`N_(2)H_(4 (aq)) + O_(2(g)) rarr N_(2(g)) + 2H_(2)O_((l))`
If `DeltaG^(@)` for this reaction is -600kJ, what will be the `E^(@)` for the cell?

Options

A. 1.25V
B. 1.50V
C. 1.57V
D. 1.75V

Answer

To solve the problem, we need to find the standard cell potential \( E^\circ \) for the given reaction using the relationship between Gibbs free energy change \( \Delta G^\circ \) and the cell potential \( E^\circ \).

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction given is:
\[ N_2H_4(aq) + O_2(g) \rightarrow N_2(g) + 2 H_2O(l) \]

2. **Given Data**:
We know that:
\[ \Delta G^\circ = -600 \, \text{kJ} \]
We need to convert this value into joules for consistency in units:
\[ \Delta G^\circ = -600 \times 10^3 \, \text{J} = -600000 \, \text{J} \]

3. **Determine the Number of Electrons Transferred (n)**:
In the reaction, we need to determine how many electrons are involved. The oxidation states of nitrogen in hydrazine \( N_2H_4 \) and nitrogen gas \( N_2 \) indicate that there is a change in oxidation states. Each nitrogen in \( N_2H_4 \) goes from -2 to 0, which means:
- 2 nitrogen atoms change from -2 to 0, contributing to a total change of 4 electrons (2 electrons per nitrogen atom).
Therefore, \( n = 4 \).

4. **Use the Relationship Between \( \Delta G^\circ \) and \( E^\circ \)**:
The relationship is given by:
\[ \Delta G^\circ = -nFE^\circ \]
Where:
- \( F \) is Faraday's constant, approximately \( 96485 \, \text{C/mol} \).

5. **Rearranging the Equation to Solve for \( E^\circ \)**:
Rearranging gives:
\[ E^\circ = -\frac{\Delta G^\circ}{nF} \]

6. **Substituting the Values**:
Substituting the known values into the equation:
\[ E^\circ = -\frac{-600000 \, \text{J}}{4 \times 96485 \, \text{C/mol}} \]

7. **Calculating \( E^\circ \)**:
\[ E^\circ = \frac{600000}{4 \times 96485} \]
\[ E^\circ = \frac{600000}{385940} \approx 1.55 \, \text{V} \]

8. **Final Result**:
Rounding to two decimal places, we find:
\[ E^\circ \approx 1.55 \, \text{V} \]

### Conclusion:
Thus, the standard cell potential \( E^\circ \) for the reaction is approximately \( 1.55 \, \text{V} \).