Question

The e.m.f. of the cell,
$Zn | Zn^{2+} \left(0.01 M\right) ‖ Fe^{2+} \left(0.001 M\right) |Fe$
at $298 \,K$ is $0.2905 \,V$. The value of the equilibrium constant for cell reaction is

• A. $\frac{0.32}{_{e}0.0295}$
• B. $\frac{0.32}{10^{0.0295}}$
• C. $\frac{0.26}{10^{0.0295}}$
• D. $\frac{0.32}{10^{0.0591}}$

Answer 1

Answer: (B)

$Zn+Fe^{2+} \rightarrow Fe+Zn^{2+}$
$E=E^{0} -\frac{0.0591}{n} log \, K_{C}$
Given E = 0.2905
$i.e., 0.2905=E^{0}-\frac{0.059}{2} log \frac{0.01}{0.001}$
$or, \, E^{0}=0.2905+0.0295 \,log 10$
$=0.2905+0.0295=0.32\, \left(\because log 10=1\right)$
$E^{0}=\frac{0.0591}{n} log K_{e q}$
$or \, 0.32=\frac{0.0591}{2} log K_{e q}$
$\therefore \, K_{e q}=10^{0.32 /0.0295}$