The `EMF` of the cell `:`
`Pt, H_(2)(1atm)|H^(o+)(aq)||AgCl|Ag` is `0.27` and `0.26V` at `25^(@)C`, and `35^(@)C` respectively. The heat of the reaction occuring inside the cell at `25^(@)C` is `………………….. kJ K^(-1)`

Explanation

To find the heat of the reaction occurring inside the cell at 25°C, we can use the following steps:

### Step 1: Understand the relationship between EMF, temperature, and heat of reaction
The heat of reaction (ΔH) can be calculated using the formula:
\[ \Delta H = -nF E_{cell} - T \left( \frac{dE}{dT} \right) \]
where:
- \( n \) = number of moles of electrons transferred (for the given cell, \( n = 1 \))
- \( F \) = Faraday's constant (\( 96500 \, C/mol \))
- \( E_{cell} \) = EMF of the cell at 25°C
- \( T \) = temperature in Kelvin
- \( \frac{dE}{dT} \) = change in EMF with respect to temperature

### Step 2: Calculate \( \frac{dE}{dT} \)
Given:
- \( E_{cell} \) at 25°C = 0.27 V
- \( E_{cell} \) at 35°C = 0.26 V

To find \( \frac{dE}{dT} \):
\[ \frac{dE}{dT} = \frac{E_{cell}(35°C) - E_{cell}(25°C)}{35°C - 25°C} = \frac{0.26 \, V - 0.27 \, V}{10 \, °C} = \frac{-0.01 \, V}{10 \, °C} = -0.001 \, V/°C \]

### Step 3: Convert temperature to Kelvin
Convert 25°C to Kelvin:
\[ T = 25 + 273 = 298 \, K \]

### Step 4: Substitute values into the ΔH formula
Now, substituting the values into the ΔH formula:
\[ \Delta H = -nF E_{cell} - T \left( \frac{dE}{dT} \right) \]
Substituting the known values:
\[ \Delta H = -1 \times 96500 \, C/mol \times 0.27 \, V - 298 \, K \times (-0.001 \, V/°C) \]

### Step 5: Calculate ΔH
Calculating the first term:
\[ -1 \times 96500 \times 0.27 = -26055 \, J \]
Calculating the second term:
\[ 298 \times (-0.001) = -0.298 \, J \]
Now, combining both terms:
\[ \Delta H = -26055 - (-0.298) = -26055 + 0.298 = -26054.702 \, J \]

### Step 6: Convert to kJ/K
Convert Joules to kilojoules:
\[ \Delta H = -26.054702 \, kJ \]
Thus, rounding to three decimal places:
\[ \Delta H \approx -26.055 \, kJ/K \]

### Final Answer
The heat of the reaction occurring inside the cell at 25°C is approximately:
\[ \Delta H = -26.055 \, kJ/K \]