Thermodynamics and Thermochemistry 1 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 $2 \mathrm{~V}$ and $-5 \times 10^{-4} \mathrm{VK}^{-1}$ at $300 \mathrm{~K}$ respectively. The cell reaction is
$\mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \rightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s)$
The standard reaction enthalpy $\left(\Delta_{r} H^{\Theta}\right)$ at $300 \mathrm{~K} \mathrm{in} \mathrm{kJ} \mathrm{mol}^{-1}$ is, $\quad\left[\mathrm{Use}, R=8 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right.$ and $F=96,000 \mathrm{C} \mathrm{mol}^{-1}$ ]
(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,
$$ \begin{aligned} E^{\circ} & =2 \mathrm{~V},\left(\frac{d E^{\circ}}{d T}\right)=-5 \times 10^{-4} \mathrm{VK}^{-1} \ T & =300 \mathrm{~K}, R=8 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \ F & =96000 \mathrm{Cmol}^{-1} \end{aligned} $$
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
$$ \Delta G=-2 \times 96000 \mathrm{C} \mathrm{mol}^{-1} \times 2 \mathrm{~V} $$
$[\because n=2$ for the given reaction $]$
$$ =-4 \times 96000 \mathrm{~J} \mathrm{~mol}^{-1} $$
$$ =-384000 \mathrm{~J} \mathrm{~mol}^{-1} $$
Now, $\quad \Delta S=n F\left(\frac{d E^{\circ}}{d T}\right)$ or
$$ \begin{aligned} \Delta S & =2 \times 96000 \mathrm{C} \mathrm{mol}^{-1} \times\left(-5 \times 10^{-4} \mathrm{VK}^{-1}\right) \ & =-96 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \end{aligned} $$
Thus, on substituting the values of $\Delta G$ and $\Delta S$ in Eq. (i), we get $-384000 \mathrm{~J} \mathrm{~mol}^{-1}$
$$ \begin{aligned} & =\Delta H-300 \mathrm{~K} \times\left(-96 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right) \ \Delta H & =-384000-28800 \mathrm{Jmol}^{-1} \ & =-412800 \mathrm{~J} \mathrm{~mol}^{-1} \ & =-412.800 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$
