Electrochemistry Result Question 9
9. If the standard electrode potential for a cell is 2 V at 300 K , the equilibrium constant $(K)$ for the reaction,
$ \mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \rightleftharpoons \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s) $
at 300 K is approximately
$ \left(R=8 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \mathrm{~F}=96000 \mathrm{C} \mathrm{mol}^{-1}\right) $
(a) $e^{-160}$
(b) $e^{160}$
(c) $e^{-80}$
(d) $e^{320}$
(2019 Main, 9 Jan II)
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Answer:
Correct Answer: 9. ( b )
Solution:
- The relationship between standard electrode potential $\left(E^{\circ}\right)$ and equilibrium constant $(K)$ of the cell reaction,
$ \mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \rightleftharpoons \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s) $
can be expressed as,
$ E^{\circ}=\frac{R T}{n F} \ln K \Rightarrow K=e^{n F E^{\circ} / R T} $
Given, $n=2, F=96000 \mathrm{Cmol}^{-1}$
$ \begin{aligned} E^{\circ} & =2 \mathrm{~V}, R=8 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\ T & =300 \mathrm{~K} \\ \therefore \quad K & =e^{\frac{2 \times 96000 \times 2}{8 \times 300}}=e^{160} \end{aligned} $
BETA
