Entropy Changes in Reaction Dynamics and Electrochemical Cells

To determine the correct option(s) concerning entropy ($\Delta S$), we evaluate each statement step-by-step:

Analysis of Option (A):

The given chemical reaction is: $\text{M}(s) + 2\text{H}^+(aq) \rightarrow \text{H}_2(g) + \text{M}^{2+}(aq)$

• The oxidation half-reaction: $\text{M}(s) \rightarrow \text{M}^{2+}(aq) + 2e^-$
• The reduction half-reaction: $2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g)$

Thus, the number of moles of electrons transferred per mole of reaction is $n = 2$.

The relationship between the entropy change of the cell reaction ($\Delta S$) and the temperature coefficient of cell potential ($\frac{dE_{\text{cell}}}{dT}$) is given by: $\Delta S = n F \left(\frac{dE_{\text{cell}}}{dT}\right)$

Given that $\frac{dE_{\text{cell}}}{dT} = \frac{R}{F}$, we substitute $n = 2$: $\Delta S = 2 \times F \times \left(\frac{R}{F}\right) = 2R$

Since the option states that the entropy change of the reaction is $R$ (instead of $2R$), Option (A) is incorrect.

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Analysis of Option (B):

The given cell representation is: $\text{Pt}(s) \mid \text{H}_2(g, 1\text{ bar}) \mid \text{H}^+(aq, 0.01\text{ M}) \parallel \text{H}^+(aq, 0.1\text{ M}) \mid \text{H}_2(g, 1\text{ bar}) \mid \text{Pt}(s)$

• Anode reaction: $\text{H}_2(g, 1\text{ bar}) \rightarrow 2\text{H}^+(aq, 0.01\text{ M}) + 2e^-$
• Cathode reaction: $2\text{H}^+(aq, 0.1\text{ M}) + 2e^- \rightarrow \text{H}_2(g, 1\text{ bar})$
• Overall cell reaction: $\text{H}^+(aq, 0.1\text{ M}) \rightarrow \text{H}^+(aq, 0.01\text{ M})$

For an ideal concentration cell, there is no net chemical change involving chemical bond formation or breaking, so the enthalpy change of the process is zero ($\Delta H = 0$).

The Gibbs free energy change ($\Delta G$) for spontaneous dilution is negative: $\Delta G = -nFE_{\text{cell}} = RT \ln\left(\frac{0.01}{0.1}\right) = -RT \ln(10) < 0$

Using the fundamental thermodynamic relation: $\Delta G = \Delta H - T\Delta S$ Since $\Delta H = 0$: $\Delta G = -T\Delta S \implies \Delta S = -\frac{\Delta G}{T} > 0$

Since the process is spontaneous ($\Delta G < 0$) purely due to a positive entropy change ($\Delta S > 0$) with $\Delta H \approx 0$, it is an entropy-driven process. Thus, Option (B) is correct.

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Analysis of Option (C):

Racemization is the conversion of an optically active compound (a single pure enantiomer) into a racemic mixture containing equal amounts ($50:50$) of $d$- and $l$-enantiomers.

The entropy of mixing two enantiomers in equal proportions is given by: $\Delta S_{\text{mix}} = -R \left( x_d \ln x_d + x_l \ln x_l \right)$ Where $x_d = x_l = 0.5$: $\Delta S_{\text{mix}} = -R \left( 0.5 \ln 0.5 + 0.5 \ln 0.5 \right) = R \ln 2 > 0$

Since disorder/randomness increases when a pure state converts to a mixture, $\Delta S > 0$. Thus, Option (C) is correct.

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Analysis of Option (D):

Consider the ligand exchange reaction: $[\text{Ni}(\text{H}_2\text{O})_6]^{2+} + 3\text{ en} \rightarrow [\text{Ni}(\text{en})_3]^{2+} + 6\text{H}_2\text{O}$

• On the reactant side, there are $1 + 3 = 4$ species.
• On the product side, there are $1 + 6 = 7$ species.

Since one bidentate ethylenediamine ($\text{en}$) ligand replaces two unidentate $\text{H}_2\text{O}$ ligands, the total number of independent molecules in solution increases from 4 to 7. This increase in the number of free particles leads to a significant increase in microstates (randomness), resulting in: $\Delta S > 0$

This favorable entropy change is the primary driving force for the chelate effect. Thus, Option (D) is correct.

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Conclusion:

The correct options are B, C, and D.