At 298K the standard free energy of formation of `H_(2)O(l)` is `-256.5 kJ..mol`, while that of its ionisation to `H^(+)` & `OH^(-)` is `80 kJ//mol`. What will be emf at 298 K of the cell.
`H_(2)(g,1 bar) |H^(+)(1M) || O_(2)(g,1bar)`
Fill your answer by multiplying it with 10.

Explanation

To solve the problem, we need to calculate the EMF (electromotive force) of the given electrochemical cell at 298 K. We will follow these steps:

### Step 1: Write the Cell Reaction
The cell reaction can be represented as:
\[ \text{H}_2(g, 1 \text{ bar}) + \frac{1}{2} \text{O}_2(g, 1 \text{ bar}) \rightarrow 2 \text{H}^+(1 \text{ M}) + 2 \text{OH}^- \]

### Step 2: Calculate the Standard Gibbs Free Energy Change (ΔG°)
The standard Gibbs free energy change for the cell reaction can be calculated using the standard free energy of formation of water and the standard free energy of ionization.

Given:
- Standard free energy of formation of \( \text{H}_2\text{O}(l) = -256.5 \, \text{kJ/mol} \)
- Standard free energy of ionization of \( \text{H}_2\text{O}(l) = 80 \, \text{kJ/mol} \)

Since the ionization produces 2 moles of \( \text{H}^+ \) and \( \text{OH}^- \), we need to multiply the ionization energy by 2:
\[ \Delta G^\circ = \Delta G^\circ_{\text{formation}} + 2 \times \Delta G^\circ_{\text{ionization}} \]
\[ \Delta G^\circ = -256.5 \, \text{kJ/mol} + 2 \times 80 \, \text{kJ/mol} \]
\[ \Delta G^\circ = -256.5 \, \text{kJ/mol} + 160 \, \text{kJ/mol} = -96.5 \, \text{kJ/mol} \]

### Step 3: Convert ΔG° to Joules
To use the value in calculations, we convert it from kJ to J:
\[ \Delta G^\circ = -96.5 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = -96500 \, \text{J/mol} \]

### Step 4: Calculate the Number of Electrons Transferred (n)
In the reaction, 2 moles of \( \text{H}^+ \) are produced from 1 mole of \( \text{H}_2 \), which means 2 electrons are transferred. Thus, \( n = 2 \).

### Step 5: Use the Nernst Equation to Find EMF (E°cell)
The relationship between Gibbs free energy and EMF is given by:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]
Where:
- \( F \) is Faraday's constant, approximately \( 96500 \, \text{C/mol} \).

Rearranging the equation to solve for \( E^\circ_{\text{cell}} \):
\[ E^\circ_{\text{cell}} = -\frac{\Delta G^\circ}{nF} \]
Substituting the values:
\[ E^\circ_{\text{cell}} = -\frac{-96500 \, \text{J/mol}}{2 \times 96500 \, \text{C/mol}} = \frac{96500}{193000} \, \text{V} \]
\[ E^\circ_{\text{cell}} = 0.5 \, \text{V} \]

### Step 6: Final Answer Adjustment
The question asks to multiply the answer by 10:
\[ \text{Final Answer} = 0.5 \times 10 = 5 \, \text{V} \]

### Conclusion
The EMF of the cell at 298 K is **5 V**.