Question

$Zn|Z{n}^{2+}(a=0.1M)||F{e}^{2+}(a=0.01M)|Fe.$ The emf of the above cell is $0.2905V$. 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. $10^{0.32/0.295}$

Answer 1

Answer: (B)

The cell reaction is :
$Zn+F{e}^{2+}\rightleftharpoons Z{n}^{2+}+Fe;E_{\text{cell }}=0.2905V$
$\Rightarrow E=E^{\circ }-\frac{0.059}{2}\log \frac{\left[Z{n}^{2+}\right]}{\left[F{e}^{2+}\right]}$
$\Rightarrow E^{\circ }=0.2905+\frac{0.059}{2}\log \frac{0.1}{0.01}=0.32V$
Also $E^{\circ }=\frac{0.059}{n}\log K$
$\Rightarrow \log K=\frac{2E^{\circ }}{0.059}=\frac{0.32}{0.0295}$
$\Rightarrow K=(10{)}^{0.32/0.0295}$