Thermodynamics and Thermochemistry - Result Question 10
10. The standard electrode potential $E^{\Theta}$ and its temperature coefficient $\left(\frac{d E^{\Theta}}{d T}\right)$ for a cell are $2 V$ and $-5 \times 10^{-4} VK^{-1}$ at $300 K$ respectively. The cell reaction is
$Zn(s)+Cu^{2+}(a q) \rightarrow Zn^{2+}(a q)+Cu(s)$
The standard reaction enthalpy $\left(\Delta _r H^{\Theta}\right)$ at $300 K$ in $kJ $ $mol^{-1}$ is, $\quad[Use, R=8 $ $JK^{-1}$ $ mol^{-1}$ and $F=96,000$ $ C$ $ mol^{-1}$ ]
(2019 Main, 12 Jan I)
(a) $-412.8$
(b) $-384.0$
(c) $206.4$
(d) $192.0$
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Answer:
Correct Answer: 10. (a)
Solution:
Given,
$E^{\circ} =2 V,\left(\frac{d E^{\circ}}{d T}\right)=-5 \times 10^{-4} VK^{-1} $
$T =300 K, R=8 J K^{-1} mol^{-1} $
$F =96000$ $ C$ $ mol^{-1}$
According to Gibbs-Helmholtz equation,
$ \hspace{20mm} \Delta G=\Delta H-T \Delta S \hspace{29mm}…(i) $
$ \text { Also, } \hspace{10mm} \Delta G=-n F E^{\circ} \text { cell }\hspace{29mm}…(ii) $
On substituting the given values in equation (ii), we get
$ \Delta G=-2 \times 96000 $ $C $ $mol^{-1} \times 2 V $
$[\because n=2$ for the given reaction $]$
$ =-4 \times 96000 $ $J$ $ mol^{-1} $
$ =-384000 $ $J $ $mol^{-1} $
Now, $\quad \Delta S=n F\left(\frac{d E^{\circ}}{d T}\right)$ or
$\Delta S =2 \times 96000$ $ C$ $ mol^{-1} \times\left(-5 \times 10^{-4} VK^{-1}\right) $
$=-96 $ $JK^{-1}$ $ mol^{-1}$
Thus, on substituting the values of $\Delta G$ and $\Delta S$ in Eq. (i), we get $-384000 $ $J$ $ mol^{-1}$
$=\Delta H-300 K \times(-96 $ $JK^{-1} $ $mol^{-1}) $
$\Delta H =-384000-28800 $ $J $ $mol^{-1} $
$=-412800$ $ J$ $ mol^{-1} $
$=-412.800 $ $kJ $ $mol^{-1}$
BETA
