At `25^(@)C, DeltaH_(f)(H_(2)O,l) =- 56700 J//mol` and energy of ionization of `H_(2)O(l) = 19050J//mol`. What will be the reversible `EMF` at `25^(@)C` of the cell,
`Pt|H_(2)(g) (1atm) |H^(+) || OH^(-) |O_(2)(g) (1atm)|Pt`, if at `26^(@)C` the emf increase by `0.001158V`.

Explanation

To solve the problem, we need to calculate the reversible EMF (E_cell) of the given electrochemical cell at 25°C using the provided thermodynamic data. Here are the steps to arrive at the solution:

### Step 1: Write the reactions and their enthalpy changes
1. **Enthalpy of formation of water (H2O)**:
\[ \text{Reaction 1: } \text{H}_2 + \frac{1}{2} \text{O}_2 \rightarrow \text{H}_2\text{O} \quad \Delta H_1 = -56700 \, \text{J/mol} \]

2. **Ionization of water**:
\[ \text{Reaction 2: } 2 \text{H}_2\text{O} \rightarrow 2 \text{H}^+ + 2 \text{OH}^- \quad \Delta H_2 = 2 \times 1950 = 3900 \, \text{J/mol} \]

### Step 2: Write the overall cell reaction
The overall cell reaction can be derived from the anode and cathode reactions:
- **Anode (oxidation)**:
\[ \text{H}_2 \rightarrow 2 \text{H}^+ + 2 \text{e}^- \]
- **Cathode (reduction)**:
\[ 2 \text{e}^- + 2 \text{H}_2\text{O} + \frac{1}{2} \text{O}_2 \rightarrow 2 \text{OH}^- \]

Combining these gives:
\[ \text{H}_2 + 2 \text{H}_2\text{O} + \frac{1}{2} \text{O}_2 \rightarrow 2 \text{H}^+ + 2 \text{OH}^- \]

### Step 3: Relate Gibbs free energy to enthalpy and entropy
Using the relation:
\[ \Delta G = \Delta H - T \Delta S \]
For the overall reaction, we can express:
\[ \Delta G_3 = \Delta H_1 + \Delta H_2 - T(\Delta S_1 + \Delta S_2) \]

### Step 4: Calculate the change in Gibbs free energy
Using the relation for cell potential:
\[ \Delta G = -nFE_{cell} \]
We can relate the Gibbs free energy change to the EMF of the cell:
\[ -nFE_{cell} = \Delta H_1 + \Delta H_2 - T \Delta S_{net} \]

### Step 5: Substitute and rearrange for E_cell
Rearranging gives:
\[ E_{cell} = -\frac{\Delta H_1 + \Delta H_2}{nF} + T \frac{dE}{dT} \]

### Step 6: Substitute values
- \(n = 2\) (number of electrons)
- \(F = 96500 \, \text{C/mol}\)
- \(T = 298 \, \text{K}\)
- \(\Delta H_1 = -56700 \, \text{J/mol}\)
- \(\Delta H_2 = 3900 \, \text{J/mol}\)
- \(\frac{dE}{dT} = 0.001158 \, \text{V/K}\)

Substituting these values into the equation:
\[ E_{cell} = -\frac{-56700 + 3900}{2 \times 96500} + 298 \times 0.001158 \]

### Step 7: Calculate E_cell
Calculating the first term:
\[ E_{cell} = \frac{52800}{193000} + 0.345 \]
\[ E_{cell} = 0.273 + 0.345 = 0.618 \, \text{V} \]

### Final Answer
The reversible EMF at 25°C of the cell is approximately:
\[ E_{cell} \approx 0.618 \, \text{V} \]