The correct option(s) about entropy ( $S$ ) is(are)\ [ $R=$ gas constant, $F=$ Faraday constant,\

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The correct option(s) about entropy ( $S$ ) is(are)\ [ $R=$ gas constant, $F=$ Faraday constant,\ $T=$ Temperature]\

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(b, c, d): $\Delta G=\Delta H-T \Delta S$\ $\Delta S=\frac{d(\Delta G)}{d t}$ at constant pressure\ So, $\Delta S=n F \frac{d E}{d t}$\ $\frac{d E}{d t}=\frac{\Delta S}{n F}$\ $\frac{d E}{d t}=\frac{R}{F}$ (Given)\ On comparing equation (i) and (ii), we get\ $\frac{\Delta S}{n F}=\frac{R}{F} \Rightarrow \Delta S=n R$\ $M_{(s)}+2 \mathrm{H}^{+}{ }_{(a q)} \longrightarrow \mathrm{H}_{2(g)}+M^{2+}{ }_{(a q)}$\ $n=2 ; \Delta S=2 R$\ Option (a) is incorrect.\ $E_{	ext {cell }}=E_{	ext {cell }}^{\circ}-\frac{0.059}{n} \log \frac{[P]}{[R]}$\ $\frac{1}{2} \mathrm{H}_{2} \longrightarrow \underset{(0.01 \mathrm{M})}{\mathrm{H}^{+}}+e^{-}$\ $\underset{(0.1 \mathrm{M})}{\mathrm{H}^{+}}+e^{-} \longrightarrow \frac{1}{2} \mathrm{H}_{2}$ $\xrightarrow{\frac{1}{2} \mathrm{H}_{2(g)}+\underset{(0.1 \mathrm{M})}{\mathrm{H}_{(a q)}^{+}} \longrightarrow \underset{(0.01 \mathrm{M})}{ }} \mathrm{H}_{(a)}^{+}+\frac{1}{2} \mathrm{H}_{2(g)}$\ Thus, $E_{	ext {cell }}^{\circ}$ for the reaction is zero.\ $E_{	ext {cell }}=0-\frac{0.059}{1} \log \frac{\left[\mathrm{H}^{+}ight]_{2}}{\left[\mathrm{H}^{+}ight]_{1}}$\ $=-0.059 \log \left(\frac{0.01}{0.1}ight)=-0.059 \log \left(\frac{1}{10}ight)$ $$ =-0.059 	imes(-1)=0.059 \mathrm{~V} $$ So, $E_{	ext {cell }}$ is positive, $\Delta G$ is negative.\ $	herefore \Delta S=n F \frac{d E}{d t}=+\mathrm{ve}$\ $	herefore$ Option (b) is correct.\ $	herefore \quad \Delta H=0$\ ( $	herefore$ Same type of bonds break) Hence, if $\Delta S=+v e$, only then $\Delta G$ is negative to make the process spontaneous.\ Option (c) is correct.\ Option (d) is correct.\