The correct option(s) about entropy (S) is(ar | vedclass

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(A) The relationship between entropy change and cell potential is given by $\Delta S = nF \left( \frac{dE_{cell}}{dT} \right)$.For option $(A)$: Given

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$B, C, D$

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(A) The relationship between entropy change and cell potential is given by $\Delta S = nF \left( \frac{dE_{cell}}{dT} ight)$.For option $(A)$: Given $\frac{dE_{cell}}{dT} = \frac{R}{F}$ and for the reaction $M_{(s)} + 2H^{+}_{(aq)} ightarrow H_{2(g)} + M^{2+}_{(aq)}$, $n = 2$. Thus, $\Delta S = 2 	imes F 	imes \frac{R}{F} = 2R$. Since $2R 
eq R$, $(A)$ is incorrect.For option $(B)$: This is a concentration cell. $\Delta H = 0$ for concentration cells. Since the reaction is spontaneous $(E_{cell} > 0)$, $\Delta G  0$. Thus, it is an entropy-driven process. $(B)$ is correct.For option $(C)$: Racemization involves the conversion of an optically active compound into a racemic mixture. This increases the disorder of the system, so $\Delta S > 0$. $(C)$ is correct.For option $(D)$: The reaction $[Ni(H_2O)_6]^{2+} + 3en ightarrow [Ni(en)_3]^{2+} + 6H_2O$ involves the replacement of $6$ monodentate ligands with $3$ bidentate ligands, increasing the number of free molecules in the solution ($4$ particles on the left to $7$ particles on the right). This increases entropy, so $\Delta S > 0$. $(D)$ is correct.

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