Electrochemistry Result Question 20
20. $\mathrm{Zn}\left|\mathrm{Zn}^{2+}(a=0.1 \mathrm{M}) | \mathrm{Fe}^{2+}(a=0.01 \mathrm{M})\right| \mathrm{Fe}$.
The emf of the above cell is 0.2905 V . Equilibrium constant for the cell reaction is
(2004, 1M)
(a) $10^{0.32 / 0.059}$
(b) $10^{0.32 / 0.0295}$
(c) $10^{0.26 / 0.0295}$
(d) $10^{0.32 / 0.295}$
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Answer:
Correct Answer: 20. ( b )
Solution:
- The cell reaction is :
$ \begin{aligned} & \mathrm{Zn}+\mathrm{Fe}^{2+} \rightleftharpoons \mathrm{Zn}^{2+}+\mathrm{Fe} ; E_{\text {cell }}=0.2905 \mathrm{~V} \\ \Rightarrow & E=E^{\circ}-\frac{0.059}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Fe}^{2+}\right]} \\ \Rightarrow \quad & E^{\circ}=0.2905+\frac{0.059}{2} \log \frac{0.1}{0.01}=0.32 \mathrm{~V} \end{aligned} $
Also $\quad E^{\circ}=\frac{0.059}{n} \log K$
$ \begin{aligned} & \Rightarrow \quad \log K=\frac{2 E^{\circ}}{0.059}=\frac{0.32}{0.0295} \\ & \Rightarrow \quad K=(10)^{0.32 / 0.0295} \\ \end{aligned} $
BETA
