Electrochemistry 2 Question 21
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 $0.2905 \mathrm{~V}$. Equilibrium constant for the cell reaction is
(a) $10^{0.32 / 0.059}$
(b) $10^{0.32 / 0.0295}$
(c) $10^{0.26 / 0.0295}$
(d) $10^{0.32 / 0.295}$
(2004, 1M)
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Solution:
- The cell reaction is :
$$ \begin{array}{rlrl} & & \mathrm{Zn}+\mathrm{Fe}^{2+} \rightleftharpoons \mathrm{Zn}^{2+}+\mathrm{Fe} ; E_{\text {cell }}=0.2905 \mathrm{~V} \ \Rightarrow & & E=E^{\circ}-\frac{0.059}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Fe}^{2+}\right]} \ \Rightarrow & & E^{\circ}=0.2905+\frac{0.059}{2} \log \frac{0.1}{0.01}=0.32 \mathrm{~V} \ & \text { Also } & E^{\circ}=\frac{0.059}{n} \log K \ \Rightarrow & \log K & =\frac{2 E^{\circ}}{0.059}=\frac{0.32}{0.0295} \ \Rightarrow & & K & =(10)^{0.32 / 0.0295} \end{array} $$
