Consider the following cell reaction Cd$$(s)$$ + Hg$$_2$$SO$$_4$$(s) + $$\frac{9}{5}$$H$$_2$$O$$(l)$$ $$\rightleftharpoons$$ CdSO$$_4$$.$$\frac{9}{5}$$H$$_2$$O$$(s)$$ + 2Hg$$(l)$$.
The value of E$$^0_{cell}$$ is 4.315 V at 25°C. If $$\Delta H°$$ = -825.2 kJ mol$$^{-1}$$, the standard entropy change $$\Delta S°$$ in J K$$^{-1}$$ is _________. (Nearest integer)
[Given : Faraday constant = 96487 C mol$$^{-1}$$]

First, we recall the relation between the standard cell potential and the standard Gibbs free-energy change.

For a cell reaction in which $$n$$ electrons are transferred, the Nernst-derived thermodynamic formula is stated as

$$\Delta G^{\circ} = -\,nF\,E^{\circ}_{\text{cell}}.$$

In the given reaction, solid cadmium $$\mathrm{Cd(s)}$$ is oxidised to $$\mathrm{Cd^{2+}}$$ while the mercurous ion $$\mathrm{Hg_2^{2+}}$$ is reduced to $$\mathrm{2Hg(l)}$$. The balanced equation shows that two electrons move from cadmium to mercury, so we have

$$n = 2.$$

The Faraday constant is given as $$F = 96487 \text{ C mol}^{-1}$$ and the standard cell potential is $$E^{\circ}_{\text{cell}} = 4.315 \text{ V}$$. Substituting these values gives

$$\Delta G^{\circ} = -\,(2)(96487\ \text{C mol}^{-1})(4.315\ \text{V}).$$

Because $$1 \text{ V} = 1 \text{ J C}^{-1}$$, the product of coulombs and volts directly yields joules:

$$\Delta G^{\circ} = -\,2 \times 96487 \times 4.315\ \text{J mol}^{-1}.$$

We now carry out the multiplication step by step.

First,

$$96487 \times 4.315 = 96487 \times (4 + 0.315) = 96487 \times 4 + 96487 \times 0.315 = 385948 + 30393.405 = 416341.405\ \text{J mol}^{-1}.$$

Next, multiply by 2:

$$2 \times 416341.405 = 832682.81\ \text{J mol}^{-1}.$$

Therefore,

$$\Delta G^{\circ} = -\,832682.81\ \text{J mol}^{-1} \approx -\,8.327 \times 10^{5}\ \text{J mol}^{-1} = -\,832.7\ \text{kJ mol}^{-1}.$$

Now we connect the thermodynamic quantities $$\Delta H^{\circ},\; \Delta G^{\circ}$$ and $$\Delta S^{\circ}$$ with the basic relation from the definition of Gibbs free energy:

$$\boxed{\;\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\;}.$$

Rearranging for the entropy change, we obtain

$$\Delta S^{\circ} = \frac{\Delta H^{\circ} - \Delta G^{\circ}}{T}.$$

The data give the standard enthalpy change as $$\Delta H^{\circ} = -\,825.2\ \text{kJ mol}^{-1}$$ and the temperature as $$T = 25^{\circ}\text{C} = 298\ \text{K}$$.

Substituting, while keeping kilojoules consistent, we write

$$\Delta S^{\circ} = \frac{-825.2\ \text{kJ mol}^{-1} - \bigl(-832.6828\ \text{kJ mol}^{-1}\bigr)}{298\ \text{K}}.$$

The numerator simplifies:

$$-825.2 - (-832.6828) = -825.2 + 832.6828 = 7.4828\ \text{kJ mol}^{-1}.$$

Now divide by the absolute temperature:

$$\Delta S^{\circ} = \frac{7.4828\ \text{kJ mol}^{-1}}{298\ \text{K}} = 0.0251\ \text{kJ K}^{-1}\text{ mol}^{-1}.$$

Converting to joules (1 kJ = 1000 J):

$$\Delta S^{\circ} = 0.0251 \times 1000 = 25.1\ \text{J K}^{-1}\text{ mol}^{-1}.$$

Rounding to the nearest integer yields

$$\boxed{25\ \text{J K}^{-1}}.$$

Hence, the correct answer is Option 25.