QJEEAdv18Chf6a5: JEE Advanced 2018 Chemistry - Class 12 JEEAdv18Chf6a5 Solution: Consider an electrochemical cell : The value of for the cell…

At $300 \mathrm{~K}$ , following electrochemical cell operates:

$\mathrm{A}(s)\left|\mathrm{A}^{n+}(a q, 2 \mathrm{M})\right| \mid \mathrm{B}^{2 n+}(a q,(\mathrm{M}) \mid \mathrm{B}(s)$

The reactions at :

(i) Anode :

$\begin{gathered}\mathrm{A}_{(s)} \longrightarrow \mathrm{A}^{n+}(a q)+\mathrm{n} e^{-} \\2 \mathrm{M}\end{gathered}$ ...........(i)

(ii) Cathode :

$\begin{aligned}& \mathrm{B}^{2 n+}(a q)+2 n e^{-} \longrightarrow \mathrm{B}(s) \\& 1 \mathrm{M}\end{aligned}$

Multiplying equation (i) by 2,

$2 \mathrm{~A}(s) \longrightarrow 2 \mathrm{~A}^{n+}(a q)+2 n e^{-}$

The net electrochemical cell is written as

$2 \mathrm{~A}(s)+\mathrm{B}^{2 n+}(a q) \longrightarrow \mathrm{B}(s)+2 \mathrm{~A}^{n+}(a q)$

Given :

Enthalpy change $\left(\Delta \mathrm{H}^{\circ}\right)$ for cell reaction $=2 \times$ Gibbs free energy than for cell reaction

$\Delta \mathrm{H}^{\circ}=2 \times \Delta \mathrm{G}^{\circ}$

According to the Nernst equation :

$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{2.303 \mathrm{RT}}{n \mathrm{~F}} \log \frac{[\text { Product }]^x}{[\text { Reactant }]^y}$

$\begin{aligned}\mathrm{E}_{\text {cell }} & =0 ; \text { hence, } \\\\\mathrm{E}_{\text {cell }}^{\circ} & =\frac{2.303 \mathrm{RT}}{n \mathrm{~F}} \log \frac{[\text { Product }]^x}{[\text { Reactant }]^y} \\\\\mathrm{E}_{\text {cell }}^{\circ} & =\frac{8.3 \mathrm{JK}^{-1} \times 300 \mathrm{~K}}{2 n \mathrm{~F}} \ln \frac{\left[\mathrm{A}^n\right]}{\left[\mathrm{B}^{2 n}\right]} \\\\\mathrm{E}_{\text {cell }}^{\circ} & =\frac{8.3 \times 300 \mathrm{~J}}{2 n \mathrm{~F}} \ln \frac{(2)^2}{1} \\\\\mathrm{E}_{\text {cell }}^{\circ} & =\frac{8.3 \times 300}{2 n \mathrm{~F}} \ln 4\end{aligned}$

For a spontaneous reaction,

$\begin{aligned}\Delta \mathrm{G}^{\circ} & =\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ} \\\\\Delta \mathrm{S}^{\circ} & =\frac{\Delta \mathrm{G}^{\circ}}{\mathrm{T}}=\frac{-2 n \mathrm{FE}_{\text {cell }}^{\mathrm{o}}}{\mathrm{T}} \\\\& =\frac{-2 n \mathrm{~F} \times 8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}}{2 \times n \mathrm{~F}} \times \frac{0.7}{300 \mathrm{~K}} \\\\& =-8.3-0.7 \times 2 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \\\\& =-11.62 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\end{aligned}$

The change in entropy $\left(\Delta S^{\circ}\right)$ per mol of $B$ is $-11.62 \mathrm{~J} \mathrm{~K} \mathrm{~mol}^{-1}$