Question

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

• A. For the reaction, $M(s)+2H^{+}(aq)\to H_2(g)+M^{2+}(aq)$, if $\frac{d{E}_{\text{cell }}}{dT}=\frac{R}{F}$, then the entropy change of the reaction is $R$ (assume that entropy and internal energy changes are temperature independent).
• B. The cell reaction, $Pt(s)\left|H_2(g,1bar)\right|H^{+}(aq,0.01M)\Vert H^{+}(aq,0.1M)\left|H_2(g,1bar)\right|Pt(s)$, is an entropy driven process.
• C. For racemization of an optically active compound, $\Delta S>0$.
• D. $\Delta S>0$, for ${\left[Ni{\left(H_{2}O\right)}_6\right]}^{2+}+3en\to {\left[Ni(en{)}_3\right]}^{2+}+6H_{2}O$ (where en $=$ ethylenediamine).

Answer 1

Answer: (B), (C), (D)

$\Delta G=\Delta H-T\Delta S$
$\Delta G=\Delta H+T{\left(\frac{d\Delta G}{dT}\right)}_p$
$-nF\left(\frac{d{E}_{\text{cell }}}{dT}\right)=-\Delta S$
$\frac{d{E}_{\text{cell }}}{dT}=\frac{\Delta S}{nF}=\frac{R}{F}\text{ (given })$
$\Rightarrow \Delta S=nR$
For the reaction, $M(g)+2H^{\oplus }(aq)⟶H_2(g)+M^{2\oplus }(aq)$
$n=2$
$\Rightarrow \Delta S=2R$
Hence, option (A) is incorrect
For the reaction, $P{t}_{(s)}\mid H_{2(g)},1$ bar $\left|H^{\oplus }{}_{aq}(0.01M)\right|\left|H^{\oplus }(aq,0.1M)\right|H_2(g,1bar)\mid P{t}_{(s)}$
$E_{\text{cell }}=E_{\text{cell }}^{\circ }-\frac{0.0591}{1}\log \frac{0.01}{0.1}=0.0591V$
$E_{cell}$ is positive $\Rightarrow \Delta G<0$ and $\Delta S>0(\Delta H=0$ for concentration cells $)$
Hence, option (B) is correct
Racemization of an optically active compound is a spontaneous process.
Here, $\Delta H=0$ (similar type of bonds are present in enantiomers)
$\Rightarrow \Delta S>0$
Hence, option (C) is correct.
${\left[Ni{\left(H_{2}O\right)}_6\right]}^{2+}+3en\to {\left[Ni(en{)}_3\right]}^{2+}+6H_{2}O$ is a spontaneous process
more stable complex is formed
$\Rightarrow \Delta S>0$