We check each statement using thermodynamic relations.

1. Option A

Given reaction: $\mathrm{M}(s)+2\mathrm{H}^+(aq) \rightarrow \mathrm{H}_2(g)+\mathrm{M}^{2+}(aq)$

For an electrochemical reaction, $\Delta G = -nFE_{\text{cell}}$ and $\left(\frac{\partial \Delta G}{\partial T}\right)_P = -\Delta S$

So, $-\Delta S = -nF\left(\frac{dE_{\text{cell}}}{dT}\right)$ which gives $\Delta S = nF\left(\frac{dE_{\text{cell}}}{dT}\right)$

Here, for the reaction written, $n=2$ electrons. Given $\frac{dE_{\text{cell}}}{dT} = \frac{R}{F}$ Therefore, $\Delta S = 2F\cdot \frac{R}{F}=2R$

So entropy change is $2R$, not $R$.

Hence, A is incorrect.

2. Option B

Cell: $\mathrm{Pt}(s)|\mathrm{H}_2(g,1\,bar)|\mathrm{H}^+(aq,0.01M)\|\mathrm{H}^+(aq,0.1M)|\mathrm{H}_2(g,1\,bar)|\mathrm{Pt}(s)$

This is a concentration cell with hydrogen electrodes.

Overall reaction is effectively transfer of $\mathrm{H}^+$ from higher concentration to lower concentration, i.e. dilution/mixing process.

For such a concentration cell:

• $E^\circ_{\text{cell}}=0$
• hence $\Delta H \approx 0$ for ideal dilute solutions
• but because concentrations differ, there is a free energy decrease due to mixing: $\Delta G < 0$

Thus, $\Delta G = \Delta H - T\Delta S$ With $\Delta H \approx 0$ and $\Delta G<0$, we must have $-T\Delta S<0 \Rightarrow \Delta S>0$

So the process is entropy driven.

Hence, B is correct.

3. Option C

Racemization of an optically active compound converts one pure enantiomer into an equimolar mixture of two enantiomers.

Initially: one form only. Finally: mixture of two distinguishable enantiomers.

Mixing increases randomness, so entropy increases. Therefore, $\Delta S > 0$

Hence, C is correct.

4. Option D

Reaction: $[\mathrm{Ni}(\mathrm{H}_2\mathrm{O})_6]^{2+}+3\,\mathrm{en}\rightarrow [\mathrm{Ni}(\mathrm{en})_3]^{2+}+6\mathrm{H}_2\mathrm{O}$

Count species:

• Reactants: $1+3=4$ particles
• Products: $1+6=7$ particles

Here a chelating ligand (en) replaces monodentate water ligands, releasing more free molecules into solution. This is the chelate effect, which has a favorable entropy contribution.

Since the number of independently moving particles increases, entropy increases: $\Delta S>0$

Hence, D is correct.

Final conclusion

• A: Incorrect
• B: Correct
• C: Correct
• D: Correct

Therefore the correct options are: $\boxed{\text{B, C, D}}$

This matches the stored correct answer.