For hydrogen oxygen fuel cell with reaction
`2H_(2)(g)+O_(2)(g) rarr 2 H_(2)O(l)`
`DeltaG_(f)^(c-)(H_(2)O)=-237.2kJ mol^(-1)`. Hence, `EMF` of the fuel cell is
1) `+2.46 V`
2) `-2.46V`
3) `+1.23 V`
4) `-1.23V`

Options

A. `+2.46 V`
B. `-2.46V`
C. `+1.23 V`
D. `-1.23V`

Explanation

To find the EMF (Electromotive Force) of the hydrogen-oxygen fuel cell for the reaction:

\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \]

we can use the relationship between Gibbs free energy change (\( \Delta G \)) and EMF:

\[ \Delta G = -nFE \]

Where:
- \( \Delta G \) is the Gibbs free energy change (in Joules),
- \( n \) is the number of moles of electrons transferred in the reaction,
- \( F \) is Faraday's constant (approximately \( 96500 \, C/mol \)),
- \( E \) is the EMF of the cell (in Volts).

### Step 1: Identify \( \Delta G \)
We are given:
\[ \Delta G_f^{\circ}(H_2O) = -237.2 \, kJ/mol \]
To convert this to Joules:
\[ \Delta G_f^{\circ}(H_2O) = -237.2 \times 1000 \, J/mol = -237200 \, J/mol \]

### Step 2: Determine \( n \)
In the reaction:
\[ 2H_2 + O_2 \rightarrow 2H_2O \]
Each \( H_2 \) molecule donates 2 electrons to form \( H_2O \). Therefore, for 2 moles of \( H_2 \):
- \( n = 2 \) moles of electrons are transferred.

### Step 3: Use the formula to find \( E \)
Substituting the values into the equation:
\[ \Delta G = -nFE \]
\[ -237200 \, J/mol = -2 \times 96500 \, C/mol \times E \]

### Step 4: Solve for \( E \)
Rearranging the equation to solve for \( E \):
\[ E = \frac{237200}{2 \times 96500} \]
Calculating the denominator:
\[ 2 \times 96500 = 193000 \, C/mol \]
Now substituting back:
\[ E = \frac{237200}{193000} \approx 1.23 \, V \]

### Conclusion
The EMF of the hydrogen-oxygen fuel cell is:
\[ E \approx +1.23 \, V \]

### Final Answer
The correct option is:
3) \( +1.23 \, V \)

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