The correct option(s) about entropy (S) is(are)

[R = gas constant, F = Faraday constant, T = Temperature]

The entropy change of an electro‐chemical reaction is connected with the temperature-coefficient of its e.m.f. by the relation
$$\Delta S = nF\,\frac{dE}{dT}$$
where $$n$$ = number of electrons transferred, $$F$$ = Faraday constant.

Option A

The reaction is $$\text{M(s)} + 2H^{+}(aq) \rightarrow H_2(g) + M^{2+}(aq)$$. Here $$n = 2$$. Given $$\dfrac{dE_{\text{cell}}}{dT}=\dfrac{R}{F}$$, therefore
$$\Delta S = nF\frac{dE}{dT}=2F\left(\frac{R}{F}\right)=2R$$. The statement claims $$\Delta S = R$$, hence it is incorrect.

Option B

Cell: Pt | $$H_2(1\,\text{bar})$$ | $$H^{+}(0.01\,\text{M})$$ || $$H^{+}(0.1\,\text{M})$$ | $$H_2(1\,\text{bar})$$ | Pt. For a concentration cell the e.m.f. is
$$E = \frac{2.303\,RT}{nF}\,\log\!\left(\frac{[H^{+}]_{\text{cathode}}}{[H^{+}]_{\text{anode}}}\right).$$ Taking the left-hand compartment (0.01 M) as anode and the right-hand (0.1 M) as cathode,
$$E = \frac{2.303\,RT}{2F}\,\log\!\left(\frac{0.10}{0.010}\right) = \frac{2.303\,RT}{2F}\,(1).$$ Thus
$$\frac{dE}{dT}= \frac{2.303\,R}{2F}\,\log\!\left(\frac{0.10}{0.010}\right)=\frac{2.303\,R}{2F}.$$ Hence
$$\Delta S = nF\frac{dE}{dT}=2F\left(\frac{2.303\,R}{2F}\right)=2.303\,R\;(\gt 0).$$ Because $$\Delta S$$ is positive, the process is entropy-driven. Option B is correct.

Option C

A single enantiomer is converted into an equimolar (racemic) mixture of two enantiomers. The number of distinguishable species doubles, so disorder increases and
$$\Delta S \gt 0.$$ Therefore Option C is correct.

Option D

Reaction: $$[\text{Ni(H}_2\text{O})_6]^{2+} + 3\,\text{en} \rightarrow [\text{Ni(en)}_3]^{2+} + 6\,\text{H}_2\text{O}.$$ Number of moles before reaction: $$1 + 3 = 4.$$ Number of moles after reaction: $$1 + 6 = 7.$$ The increase in the number of particles and the chelate effect both raise the randomness of the system, hence $$\Delta S \gt 0.$$ Option D is correct.

Final answer: Option B (entropy-driven concentration cell), Option C (racemization, $$\Delta S \gt 0$$), Option D (chelation, $$\Delta S \gt 0$$).