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Step-by-Step Solution
Step 1: Identify the Given Data
• Standard electromotive force, E_{cell}^0 = 4.315 \text{ V} at 25^\circ \text{C} (= 298 K).
• Change in enthalpy, \Delta H^\circ = -825.2 \text{ kJ mol}^{-1} (which is -825.2 \times 10^3 \text{ J mol}^{-1} ).
• Faraday constant, F = 96487 \text{ C mol}^{-1} .
• Temperature, T = 298 \text{ K} .
• Number of moles of electrons, n = 2 (since 2 moles of Hg are formed from Hg_2^{2+} , indicating 2 electrons in the half-cell process).
Step 2: Write the Relevant Thermodynamic Relationship
From standard thermodynamics and electrochemistry, the Gibbs energy change at standard conditions is related to cell potential by:
\Delta G^\circ = -\,n F E^\circ
Also, the relationship between Gibbs energy, enthalpy, and entropy is given by:
\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ
Combining both, we get:
-n F E^\circ = \Delta H^\circ - T \Delta S^\circ
Rearranging for \Delta S^\circ :
\Delta S^\circ = \frac{\Delta H^\circ + n F E^\circ}{T}
Step 3: Substitute the Numerical Values
Substitute \Delta H^\circ = -825.2 \times 10^3 \text{ J mol}^{-1} , n = 2 , F = 96487 \text{ C mol}^{-1} , E^\circ = 4.315 \text{ V} , and T = 298 \text{ K} into the formula:
\Delta S^\circ
= \frac{\bigl(-825.2 \times 10^3\bigr) + \bigl(2 \times 96487 \times 4.315\bigr)}{298}
Step 4: Perform the Calculations
Compute each term:
1) \Delta H^\circ = -825.2 \times 10^3\, \text{J mol}^{-1}
2) n F E^\circ = 2 \times 96487 \times 4.315
= 2 \times 96487 \times 4.315 \text{ J mol}^{-1}
\approx 832.682 \times 10^3 \text{ J mol}^{-1}.
Summation of the terms in the numerator:
-825.2 \times 10^3 + 832.682 \times 10^3
\approx 7.482 \times 10^3 \text{ J mol}^{-1}.
Divide by T = 298 \text{ K} :
\Delta S^\circ
= \frac{7.482 \times 10^3}{298}
\approx 25.11 \text{ J K}^{-1}\text{mol}^{-1}.
Step 5: Final Answer (Nearest Integer)
Rounding to the nearest integer:
\Delta S^\circ \approx 25 \text{ J K}^{-1}\text{mol}^{-1}.
