Verified Solution • 2025–26 Session moderate

QJEEAdv22Ch7235: JEE Advanced 2022 Chemistry - Class 11 JEEAdv22Ch7235 Solution: The correct option(s) about entropy (S) is(are) gas constant,…

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

[R=[\mathrm{R}= gas constant, F=\mathrm{F}= Faraday constant, T=\mathrm{T}= Temperature ]]

Multiple Choice Options

  • A For the reaction, M(s)+2H+(aq)→H2(g)+M2+(aq)\mathrm{M}(s)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{H}_{2}(g)+\mathrm{M}^{2+}(a q) , if dEcelldT=RF\frac{d E_\text{cell}}{d T}=\frac{R}{F} , then the entropy change of the reaction is R\mathrm{R} (assume that entropy and internal energy changes are temperature independent).
  • B The cell reaction, Pt⁡(s)∣H2(g,1\operatorname{Pt}(s) \mid \mathrm{H}_{2}(g, 1 bar )∣H+(aq,0.01M)∥H+(aq,0.1M)∣H2(g,1bar)∣Pt⁡(s))\left|\mathrm{H}^{+}(a q, 0.01 \mathrm{M}) \| \mathrm{H}^{+}(a q, 0.1 \mathrm{M})\right| \mathrm{H}_{2}(g, 1 \mathrm{bar}) \mid \operatorname{Pt}(s) , is an entropy driven process.
  • C For racemization of an optically active compound, ΔS>0\Delta \mathrm{S}>0 .
  • D ΔS>0\Delta \mathrm{S}>0 , for [Ni(H2O)6]2++3\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}+3 en →[Ni(en)3]2++6H2O\rightarrow\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}+6 \mathrm{H}_{2} \mathrm{O} (where en == ethylenediamine).

Step-by-Step Verified Solution

(A) Given, Temperature coefficient dEcell d T=RF\frac{d \mathrm{E}_{\text {cell }}}{d \mathrm{~T}}=\frac{\mathrm{R}}{\mathrm{F}}

M(s)+2H+(aq)→H2( g)+M2+(aq)\mathrm{M}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{H}_2(\mathrm{~g})+\mathrm{M}^{2+}(\mathrm{aq})

Here, number of electron (n)=2(n)=2 transferred

Since,

ΔSamp;=(dEcell d T)n FΔSamp;=RF×2×FΔSamp;=2R\begin{aligned}\Delta S & =\left(\frac{d \mathrm{E}_{\text {cell }}}{d \mathrm{~T}}\right) n \mathrm{~F} \\\\\Delta S & =\frac{\mathrm{R}}{\mathrm{F}} \times 2 \times \mathrm{F} \\\\\Delta \mathrm{S} & =2 \mathrm{R}\end{aligned}

Option (A) is incorrect.

(B)

amp;Ecell =−2.303RTFlog⁡0.010.1=2.303RTFamp;dEcell dT=2.303RFamp;∴ΔS=nFdEdTgt;0\begin{aligned}& \mathrm{E}_{\text {cell }}=\frac{-2.303 \mathrm{RT}}{\mathrm{F}} \log \frac{0.01}{0.1}=\frac{2.303 \mathrm{RT}}{\mathrm{F}} \\\\& \frac{\mathrm{dE}_{\text {cell }}}{\mathrm{dT}}=\frac{2.303 \mathrm{R}}{\mathrm{F}} \\\\& \therefore \Delta \mathrm{S}=\mathrm{nF} \frac{\mathrm{dE}}{\mathrm{dT}}>0\end{aligned}

It is an entropy driven process.

(C) It is correct

During racemisation of optically active compound, disorder increases and hence, entropy increases.

(D) For [Ni(H2O)6]2++3en→[Ni(en)3]+3+6H2O\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}+3 \mathrm{en} \rightarrow\left[\mathrm{Ni}(\mathrm{en})_3\right]^{+3}+6 \mathrm{H}_2 \mathrm{O} ,

Entropy increases when bidentate ligands replace monodentate ligands due to increase in the number of molecules on the product side.

Hence, (B, C, D) are correct.

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