Q. The emf of the cell $\ Zn\ |\ {Zn}^{2 +}(0.01\ M)\ ||\ {Fe}^{2 +}(0.001\ M)\ |\ Fe\$at$\ 298\ K\$is$\ 0.2905.$ The value of equilibrium constant for the cell reaction is
A) $10^{10^{\frac{0.32}{0.0298}}}$
B) $e^{\frac{0.32}{0.0295}}$
C) $10^{\frac{0.32}{0.0591}}$
D) $10^{\frac{0.26}{0.0295}}$
Explanation
(a)
For the given cell, reaction is
$Zn + {Fe}^{2 +} \rightarrow {Zn}^{2 +} + Fe$
$E = E{^\circ} - \frac{0.0591}{n}\log\frac{C_{1}}{C_{2}}$
or$,\ \ E{^\circ} = E + \frac{0.0591}{n}\log\frac{C_{1}}{C_{2}}$
$= 0.2905 + \frac{0.0591}{2}\log\frac{10^{- 2}}{10^{- 3}} = 0.32\ V$
$E{^\circ} = \frac{0.0591}{2}\log K_{c}$
$\therefore\log K_{c} = \frac{0.32 \times 2}{0.0591} = \frac{0.32}{0.0295}$
$K_{c} = 10^{\frac{0.32}{0.295}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.