JEE Advance - Chemistry (2018 - Paper 2 Offline - No. 14)
Consider an electrochemical cell :
$$A\left( s \right)\left| {{A^{n + }}\left( {aq,2M} \right)} \right|{B^{2n + }}\left( {aq,1M} \right)\left| {B\left( s \right).} \right.$$
The value of $$\Delta {H^ \circ }$$ for the cell reaction is twice that of $$\Delta {G^ \circ }$$ at $$300$$ $$K.$$ If the $$emf$$ of the cell is zero, the $$\Delta {S^ \circ }$$ (in $$J\,{K^{ - 1}}mo{l^{ - 1}}$$) of the cell reaction per mole of $$B$$ formed at $$300$$ $$K$$ is ___________.
(Given: $$\ln \left( 2 \right) = 0.7,R$$ (universal gas constant) $$ = 8.3J\,{K^{ - 1}}\,mo{l^{ - 1}}.$$ $$H,S$$ and $$G$$ are enthalpy, entropy and Gibbs energy, respectively.)
$$A\left( s \right)\left| {{A^{n + }}\left( {aq,2M} \right)} \right|{B^{2n + }}\left( {aq,1M} \right)\left| {B\left( s \right).} \right.$$
The value of $$\Delta {H^ \circ }$$ for the cell reaction is twice that of $$\Delta {G^ \circ }$$ at $$300$$ $$K.$$ If the $$emf$$ of the cell is zero, the $$\Delta {S^ \circ }$$ (in $$J\,{K^{ - 1}}mo{l^{ - 1}}$$) of the cell reaction per mole of $$B$$ formed at $$300$$ $$K$$ is ___________.
(Given: $$\ln \left( 2 \right) = 0.7,R$$ (universal gas constant) $$ = 8.3J\,{K^{ - 1}}\,mo{l^{ - 1}}.$$ $$H,S$$ and $$G$$ are enthalpy, entropy and Gibbs energy, respectively.)
Answer
-11.62
Explanation
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 $-11.62 \mathrm{~J} \mathrm{~K} \mathrm{~mol}^{-1}$
$$ \mathrm{A}(s)\left|\mathrm{A}^{n+}(a q, 2 \mathrm{M})\right| \mid \mathrm{B}^{2 n+}(a q,(\mathrm{M}) \mid \mathrm{B}(s) $$
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 $-11.62 \mathrm{~J} \mathrm{~K} \mathrm{~mol}^{-1}$
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