Electrochemistry - Result Question 29
28. Consider an electrochemical cell : $A(s)\left|A^{n+}(a q, 2 M) | B^{2 n+}(a q, 1 M)\right| B(s)$. The value of $\Delta H^{\ominus}$ for the cell reaction is twice of $\Delta G^{\ominus}$ at $300 K$. If the emf of the cell is zero, the $\Delta S^{\ominus}$ (in $J K^{-1} mol^{-1}$ ) of the cell reaction per mole of B formed at $300 K$ is
(Given : $\ln (2)=0.7, R$ (universal gas constant) $=8.3 J K^{-1}$ $mol^{-1} . H, S$ and $G$ are enthalpy, entropy and Gibbs energy, respectively.)
(2018 Adv.)
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Answer:
Correct Answer: 28. $ (-11.62 \mathrm{~J} \mathrm{K} \mathrm{mol^{-1}}) $
Solution:
- Given,
$ A(s)\left|A^{n+}(a q, 2 M) | B^{2 n+}(a q, 1 M)\right| B(s) $
So, reactions at respective electrode will be
Anode $ A(s) \longrightarrow A^{n+}+n e^{-} \times 2 $
Cathode $ B^{2 n+}+2 n e^{-} \longrightarrow B(s)$
Overall reaction
$ 2 A(s)+B^{2 n+}(a q) \longrightarrow 2 A^{n+}(a q)+B(s) $
Further,
$\Delta H^{\circ}=2 \Delta G^{\circ}$ and $E _{\text {cell }}=0$ is also given
Now by using the Nernst equation
$ E_ {\text {cell }}=E_ {\text {cell }}^{\circ}-\frac{R T}{n F} \ln \frac{\text { [Product] }}{\text { [Reactant] }} $
After putting the values
$0 =E_{\text {cell }}^{\circ}-\frac{R T}{2 n F} \ln \frac{\left[A^{n+}\right]^2}{\left[B^{2 n+}\right]} $
or $E^{\circ} =\frac{R T}{2 n F} \ln \frac{[2]^2}{[1]}=\frac{R T}{2 n F} \ln 4 \quad $…. (i)
Further from the formula,
$ \Delta G^{\circ}=-n F E^{\circ} \Rightarrow \Delta G^{\circ}=-2 n F E^{\circ} $
Now putting the value of $E^{\circ}$ from eq. (i)
$ \Delta G^{\circ}=-2 n F \times \frac{R T}{2 n F} \ln 4 \quad$ ….(ii)
$ \Delta G^{\circ}=-R T \ln 4 $
Finally, using the formula
$ \begin{aligned} & \Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} \\ & \left.\Delta G^{\circ}=2 \Delta G^{\circ}-T \Delta S^{\circ} \quad \text { (as } \Delta H^{\circ}=2 \Delta G^{\circ}, \text { given }\right) \\ & \Delta G^{\circ}=T \Delta S^{\circ} \end{aligned} $
or
$ \Delta S^{\circ}=\frac{\Delta G^{\circ}}{T}=\frac{-R T \ln 4}{T} $
(from eq. (ii), $\Delta G^{\circ}=-R T \ln 4$ )
$ =-R \ln 4=-8.3 \times 2 \times 0.7 $
(as all values given)
$ =-11.62 \mathrm{~J} \mathrm{K} \mathrm{mol^{-1}} $
BETA
