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, $\quad\left[Use, R=8 JK^{-1} mol^{-1}\right.$ and $F=96,000 C mol^{-1}$ ]
(a) -412.8
(b) -384.0
(c) 206.4
(d) 192.0
(2019 Main, 12 Jan I)
Thermodynamics and Thermochemistry
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Solution:
- Given,
$$ \begin{aligned} E^{\circ} & =2 V,\left(\frac{d E^{\circ}}{d T}\right)=-5 \times 10^{-4} VK^{-1} \ T & =300 K, R=8 JK^{-1} mol^{-1} \ F & =96000 Cmol^{-1} \end{aligned} $$
According to Gibbs-Helmholtz equation,
$$ \Delta G=\Delta H-T \Delta S $$
$$ \text { Also, } \quad \Delta G=-n F E^{\circ} \text { cell } $$
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
$$ \begin{aligned} \Delta S & =2 \times 96000 C mol^{-1} \times\left(-5 \times 10^{-4} VK^{-1}\right) \ & =-96 JK^{-1} mol^{-1} \end{aligned} $$
Thus, on substituting the values of $\Delta G$ and $\Delta S$ in Eq. (i), we get $-384000 J mol^{-1}$
$$ \begin{aligned} & =\Delta H-300 K \times\left(-96 JK^{-1} mol^{-1}\right) \ \Delta H & =-384000-28800 Jmol^{-1} \ & =-412800 J mol^{-1} \ & =-412.800 kJ mol^{-1} \end{aligned} $$
