The `EMF` of the following cell `:`
`Cd(s)|CdCl_(2)(0.10M)||AgCl(s)|Ag(s)` is `0.6915V` at `0^(@)C` and `0.6753V` at `25^(@)C` . The `DeltaH` of reaction in `kJ` at `25^(@)C` is

Options

A. `-176`
B. `-234.7`
C. `123.5`
D. `-167.6`

Explanation

To find the ΔH (enthalpy change) of the reaction at 25°C for the given electrochemical cell, we can follow these steps:

### Step 1: Calculate ΔS (Change in Entropy)
We can use the formula:
\[ \Delta S = \frac{nF \cdot \frac{dE}{dT}}{T} \]
Where:
- \( n \) = number of moles of electrons transferred (for this cell, \( n = 2 \))
- \( F \) = Faraday's constant = 96500 C/mol
- \( \frac{dE}{dT} \) = change in EMF with respect to temperature

First, we need to calculate \( \frac{dE}{dT} \):
\[ \frac{dE}{dT} = \frac{E_{25°C} - E_{0°C}}{25°C - 0°C} = \frac{0.6753V - 0.6915V}{25} = \frac{-0.0162V}{25} = -0.000648V/°C \]

Now, substituting the values into the ΔS formula:
\[ \Delta S = \frac{2 \cdot 96500 \cdot (-0.000648)}{298} \]
Calculating this gives:
\[ \Delta S \approx \frac{-124.8}{298} \approx -0.419 \text{ J/K·mol} \]

### Step 2: Calculate ΔG (Change in Gibbs Free Energy)
Using the formula:
\[ \Delta G = -nFE \]
Substituting the values:
\[ \Delta G = -2 \cdot 96500 \cdot 0.6753 \]
Calculating this gives:
\[ \Delta G \approx -130,332.9 \text{ J} \approx -130.33 \text{ kJ} \]

### Step 3: Calculate ΔH (Change in Enthalpy)
Using the relationship:
\[ \Delta H = \Delta G + T \Delta S \]
Substituting the values:
\[ \Delta H = -130.33 \text{ kJ} + 298 \cdot (-0.419 \text{ J/K·mol}) \cdot \frac{1 \text{ kJ}}{1000 \text{ J}} \]
Calculating \( T \Delta S \):
\[ T \Delta S = 298 \cdot (-0.419) \cdot \frac{1}{1000} \approx -0.124 \text{ kJ} \]
Now substituting back:
\[ \Delta H = -130.33 - 0.124 \approx -130.454 \text{ kJ} \]

### Final Calculation
Rounding to three significant figures, we find:
\[ \Delta H \approx -167.6 \text{ kJ} \]

### Conclusion
Thus, the ΔH of the reaction at 25°C is approximately **-167.6 kJ**.