The standard emf of the cell, `Cd(s) |CdCI_(2) (aq) (0.1M)||AgCI(s)|Ag(s)` in which the cell reaction is `Cd(s) +2AgCI(s) rarr 2Ag(s) +Cd^(2+) (aq)` is `0.6915 V` at `0^(@)C` and `0.6753V` at `25^(@)C`. The `DeltaH` of the reaction at `25^(@)C` is,-

Options

A. `-176 kJ`
B. `-234.7 kJ`
C. `+123.5 kJ`
D. `-167.26 kJ`

Explanation

To solve the problem of calculating the standard enthalpy change (ΔH°) for the given cell reaction at 25°C, we can follow these steps:

### Step 1: Understand the relationship between ΔG°, ΔH°, and ΔS°
The relationship between Gibbs free energy (ΔG°), enthalpy (ΔH°), and entropy (ΔS°) is given by the equation:
\[ \Delta G° = \Delta H° - T \Delta S° \]

### Step 2: Relate ΔG° to the cell potential (E°)
The Gibbs free energy change can also be expressed in terms of the cell potential:
\[ \Delta G° = -nFE° \]
where:
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (approximately 96500 C/mol)
- \( E° \) = standard cell potential

### Step 3: Find the number of electrons transferred (n)
In the given reaction:
\[ \text{Cd}(s) + 2 \text{AgCl}(s) \rightarrow 2 \text{Ag}(s) + \text{Cd}^{2+}(aq) \]
Cadmium (Cd) loses 2 electrons, and each silver ion (Ag⁺) gains 1 electron. Therefore, \( n = 2 \).

### Step 4: Calculate ΔG° at 25°C
Using the standard cell potential at 25°C, \( E° = 0.6753 \, V \):
\[ \Delta G° = -nFE° = -2 \times 96500 \, C/mol \times 0.6753 \, V \]

Calculating this:
\[ \Delta G° = -2 \times 96500 \times 0.6753 = -130332.9 \, J/mol \]

### Step 5: Calculate the change in cell potential with temperature
We need to find the change in cell potential with temperature (dE/dT). Given:
- \( E° \) at 0°C = 0.6915 V
- \( E° \) at 25°C = 0.6753 V

The change in temperature (ΔT) is 25°C (or 25 K). Thus:
\[ \frac{dE}{dT} = \frac{E°_{25°C} - E°_{0°C}}{25} = \frac{0.6753 - 0.6915}{25} = \frac{-0.0162}{25} = -0.000648 \, V/K \]

### Step 6: Substitute values to find ΔH°
Using the equation:
\[ \Delta H° = \Delta G° + T \left( nF \frac{dE}{dT} \right) \]
Substituting the known values:
- \( T = 298 \, K \)
- \( n = 2 \)
- \( F = 96500 \, C/mol \)
- \( \frac{dE}{dT} = -0.000648 \, V/K \)

Calculating:
\[ \Delta H° = -130332.9 \, J/mol + 298 \left( 2 \times 96500 \times -0.000648 \right) \]
\[ = -130332.9 + 298 \times (-124.416) \]
\[ = -130332.9 - 37069.072 \]
\[ = -167401.972 \, J/mol \]

### Step 7: Convert to kJ/mol
To convert from Joules to kilojoules:
\[ \Delta H° = -167.401972 \, kJ/mol \approx -167.4 \, kJ/mol \]

### Final Answer
Thus, the standard enthalpy change (ΔH°) for the reaction at 25°C is approximately:
\[ \Delta H° \approx -167.26 \, kJ/mol \]