Question

Consider the following cell reaction
\(\text{Cd}_{(s)} + \text{Hg}_2\text{SO}_{4(s)} + \frac{9}{5} \text{H}_2\text{O}_{(l)} \rightleftharpoons \text{CdSO}_4 \cdot \frac{9}{5} \text{H}_2\text{O}_{(s)} + 2\text{Hg}_{(l)}\)
The value of \(E_{cell}^0\) is \(4.315\text{ V}\) at \(25^\circ\text{C}\). If \(\Delta H^0 = -825.2\text{ kJ mol}^{-1}\), the standard entropy change \(\Delta S^0\) in J K\(^{-1}\) is \(\dots\dots\dots\). (Nearest integer)
[Given : Faraday constant = \(96487\text{ C mol}^{-1}\)]

Hint:

Ensure all energy units are consistent (either all in J or all in kJ) before performing addition or subtraction in thermodynamic equations.

Answer 1

Correct Answer: 25

Step 1: Understanding the Concept:
We use the relationship between the standard cell potential and the standard Gibbs free energy change, and then apply the thermodynamic relation involving enthalpy and entropy.
Step 2: Key Formula or Approach:
1. \(\Delta G^0 = -nFE_{cell}^0\)
2. \(\Delta G^0 = \Delta H^0 - T\Delta S^0\)
Step 3: Detailed Explanation:
1. Determine \(n\): In the reaction, \(\text{Cd}\) is oxidised to \(\text{Cd}^{2+}\) in \(\text{CdSO}_4\), and \(\text{Hg}_2^{2+}\) in \(\text{Hg}_2\text{SO}_4\) is reduced to \(2\text{Hg}\). Thus, \(n = 2\).
2. Calculate \(\Delta G^0\):
\[ \Delta G^0 = -2 \times 96487 \text{ C/mol} \times 4.315\text{ V} \]
\[ \Delta G^0 = -832682.8\text{ J/mol} = -832.68\text{ kJ/mol} \]
3. Calculate \(\Delta S^0\):
Given \(\Delta H^0 = -825.2\text{ kJ/mol} = -825200\text{ J/mol}\).
Temperature \(T = 25^\circ\text{C} = 298\text{ K}\).
\[ \Delta G^0 = \Delta H^0 - T\Delta S^0 \]
\[ -832682.8 = -825200 - 298 \times \Delta S^0 \]
\[ 298 \times \Delta S^0 = -825200 + 832682.8 \]
\[ 298 \times \Delta S^0 = 7482.8 \]
\[ \Delta S^0 = \frac{7482.8}{298} \approx 25.11\text{ J K}^{-1}\text{ mol}^{-1} \]
Step 4: Final Answer:
The standard entropy change to the nearest integer is 25.