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 find the standard cell potential \( E^\circ \) for the given reaction, we can use the relationship between Gibbs free energy change \( \Delta G^\circ \) and cell potential \( E^\circ \):

\[ \Delta G^\circ = -nFE^\circ \]

Where:
- \( \Delta G^\circ \) is the standard Gibbs free energy change (in joules),
- \( n \) is the number of moles of electrons transferred in the reaction,
- \( F \) is Faraday's constant (approximately \( 96500 \, \text{C/mol} \)),
- \( E^\circ \) is the standard cell potential (in volts).

### Step 1: Convert \( \Delta G^\circ \) to Joules
Given \( \Delta G^\circ = -600 \, \text{kJ} \), we convert this to joules:

\[ \Delta G^\circ = -600 \times 10^3 \, \text{J} = -600000 \, \text{J} \]

### Step 2: Determine the number of electrons transferred (\( n \))
In the reaction:

\[ \text{N}_2\text{H}_4 (aq) + \text{O}_2 (g) \rightarrow \text{N}_2 (g) + 2 \text{H}_2\text{O} (l) \]

We need to balance the half-reactions to find the total number of electrons transferred. The oxidation of hydrazine involves the loss of electrons. For this reaction, the number of electrons transferred (\( n \)) is found to be 4.

### Step 3: Rearranging the equation to solve for \( E^\circ \)
Using the equation from Step 1:

\[ -600000 = -nFE^\circ \]

Substituting \( n = 4 \) and \( F = 96500 \):

\[ -600000 = -4 \times 96500 \times E^\circ \]

### Step 4: Solve for \( E^\circ \)
Rearranging the equation gives:

\[ E^\circ = \frac{600000}{4 \times 96500} \]

Calculating the denominator:

\[ 4 \times 96500 = 386000 \]

Now substituting back:

\[ E^\circ = \frac{600000}{386000} \approx 1.55 \, \text{V} \]

### Step 5: Rounding to the nearest option
The calculated value of \( E^\circ \) is approximately \( 1.55 \, \text{V} \). The closest option available is \( 1.57 \, \text{V} \).

### Final Answer
Thus, the standard cell potential \( E^\circ \) for the reaction is approximately \( 1.57 \, \text{V} \).

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