Question

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

Answer 1

Answer: (B)

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