Q. For a cell reaction involving a two electron change, the standard emf of the cell is found to be 0.295 V at $25℃.$ The equilibrium constant of the reaction, at $25℃,$ will be
A) 10
B) $1 \times 10^{10}$
C) $1 \times 10^{- 10}$
D) $10 \times 10^{- 2}$
Explanation
(b)
$\mathrm{\Delta}G{^\circ} = - nFE{^\circ}$
$\mathrm{\Delta}G{^\circ} = - 2.303\ RT\ \log K_{c}$
$\therefore nFE{^\circ} = 2.303\ RT\log K_{c}$
$\log K_{c} = \frac{nFE{^\circ}}{2.303\ RT}$
$= \frac{2 \times 96500 \times 0.295}{2.303 \times 8.314 \times 298}$
$\log K_{c} = 9.97$
$\therefore K_{c} = 1 \times 10^{10}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.