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

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The cell reaction, $A g | A g^{+} (0.1 M) | | A g^{+} (0.2 M) | A g$ , is an entropy driven process.

$Δ S > 0$ for micellization.

For racemization of an optically active compound, $Δ S < 0$ .

For the reaction, $2 A l (s) + 6 H^{+} (a q) \overset{}{\to} 2 A l^{3 +} (a q) + 3 H_{2} (g)$ , if $\frac{d E_{c e l l}}{d T} = \frac{2 R}{3 F}$ , then the entropy change of the reaction is 4R.

answer is A, B, D.

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Detailed Solution

A) $Δ S = n F \frac{d E}{d t} = 6 F \times \frac{2 R}{3 F} = 4 R$
B) The cell is spontaneous and there is decrease in concentration on moving from reactant side to product side and hence $Δ S > 0$
$\begin{array}{l} A g^{+} (0.2 M) \overset{}{\to} A g^{+} (0.1 M) \\ E = - \frac{0.06}{1} \log \frac{0.1}{0.2} > 0 \end{array}$
C) Racemization leads to increase in entropy as randomness increases.
D) $Δ S > 0$ for micellization due to release of solvated water molecules from the hydrophilc head of surfactant molecules.