Question

The e.m.f. of the cell,
$Zn|Z{n}^{2+}\left(0.01M\right)\Vert F{e}^{2+}\left(0.001M\right)|Fe$
at $298K$ is $0.2905V$. 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)

$Zn+F{e}^{2+}\to Fe+Z{n}^{2+}$
$E=E^0-\frac{0.0591}{n}log{K}_C$
Given E = 0.2905
$i.e.,0.2905=E^0-\frac{0.059}{2}log\frac{0.01}{0.001}$
$or,E^0=0.2905+0.0295log10$
$=0.2905+0.0295=0.32\left(\because log10=1\right)$
$E^0=\frac{0.0591}{n}log{K}_{eq}$
$or0.32=\frac{0.0591}{2}log{K}_{eq}$
$\therefore K_{eq}=10^{0.32/0.0295}$