The standard electrode potential

Question:

The standard electrode potential $\mathrm{E}^{0}$ and its temperature coefficient $\left(\frac{\mathrm{dE}^{0}}{\mathrm{dT}}\right)$ for a cell are $2 \mathrm{~V}$ and $-5 \times 10^{-4} \mathrm{VK}^{-1}$ at 300 $K$ respectively. The cell reaction is: $\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}$ The standard reaction enthalpy $\left(\Delta_{\mathrm{r}} \mathrm{H}^{0}\right)$ at $300 \mathrm{~K}$ in $\mathrm{kJ} \mathrm{mol}^{-1}$ is, $\left[\right.$ Use $\mathrm{R}=8 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ and $\mathrm{F}=96,000 \mathrm{C} \mathrm{mol}^{-1}$ ]

  1.  $-412.8$

  2. $-384.0$

  3.  $192.0$

  4. $206.4$


Correct Option: 1

Solution:

$\Delta \mathrm{G}^{\circ}=\Delta_{r} \mathrm{H}^{\circ}-T \Delta \mathrm{S}^{\circ}$

$\Delta_{r} \mathrm{H}^{\circ}=\Delta \mathrm{G}^{\circ}+T \Delta \mathrm{S}^{\circ}$

$\Delta_{r} \mathrm{H}^{\circ}=-n \mathrm{FE}^{\circ}+\mathrm{T} n \mathrm{~F} \frac{\mathrm{dE}^{\circ}}{\mathrm{dT}}$

$\Delta_{r} \mathrm{H}^{\circ}=-n \mathrm{FE}^{\circ}+n \mathrm{FT} \frac{\mathrm{dE}^{\circ}}{\mathrm{dT}}$

Cell reaction: $\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$

$\Delta_{r} \mathrm{H}^{\circ}=-n \mathrm{~F}\left(\mathrm{E}^{\circ}=\frac{\mathrm{TdE}^{\circ}}{\mathrm{dT}}\right)$

$\Delta_{r} \mathrm{H}^{\circ}=-2 \times 96000\left(2-300 \times-5 \times 10^{-4}\right)$

$\Delta_{r} \mathrm{H}^{\circ}=-2 \times 96000\left(2+300 \times 5 \times 10^{-4}\right)$

$=-2 \times 96000(2+0.15)$

$=-412.8 \times 10^{3} \mathrm{~J} / \mathrm{mol} ;=-412.8 \mathrm{~kJ} / \mathrm{mol}$

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