Electrochemistry - Result Question 80
####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.)
Passage Based Questions
Passage
Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately $6.023 \times 10^{23}$ ) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept. A $4.0 M$ aqueous solution of $NaCl$ is prepared and $500 mL$ of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes
(atomic mass : $Na=23, Hg=200,1 F=96500 C$ ).
$(2007,3 \times 4 M=12 M)$
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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
$$ \begin{array}{ll} \text { Anode } & A(s) \longrightarrow A^{n+}+n e^{-} \times 2 \ \text { Cathode } & B^{2 n+}+2 n e^{-} \longrightarrow B(s) \end{array} $$
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
$$ E _{\text {cell }}=E _{\text {cell }}^{o}-\frac{R T}{n F} \ln \frac{[\text { Product }]}{[\text { Reactant }]} $$
After putting the values
or
$$ \begin{aligned} 0 & =E _{\text {cell }}^{\circ}-\frac{R T}{2 n F} \ln \frac{\left[A^{n+}\right]^{2}}{\left[B^{2 n+}\right]} \ E^{\circ} & =\frac{R T}{2 n F} \ln \frac{[2]^{2}}{[1]}=\frac{R T}{2 n F} \ln 4 \end{aligned} $$
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)
$$ \begin{aligned} \Delta G^{\circ} & =-2 n F \times \frac{R T}{2 n F} \ln 4 \ \Delta G^{\circ} & =-R T \ln 4 \end{aligned} $$
Finally, using the formula
$$ \begin{aligned} \Delta G^{\circ} & =\Delta H^{\circ}-T \Delta S^{\circ} \ \Delta G^{\circ} & =2 \Delta G^{\circ}-T \Delta S^{\circ} \quad\left(\text { as } \Delta H^{\circ}=2 \Delta G^{\circ}, \text { given }\right) \ \Delta G^{\circ} & =T \Delta S^{\circ} \ \Delta S^{\circ} & =\frac{\Delta G^{\circ}}{T}=\frac{-R T \ln 4}{T} \ \text { or } \quad \text { (from eq. (ii), } \Delta G^{\circ} & =-R T \ln 4) \ & =-R \ln 4=-8.3 \times 2 \times 0.7 \ \text { (as all values given) } & \ & =-11.62 J / K-mol \end{aligned} $$
