Q. Calculate the equilibrium constant for the reaction, at 25℃ $Cu\ (s)\ + \ 2{Ag}^{+}\ (aq)\ \rightarrow \ {Cu}^{2 +}\ (aq)\ + \ 2Ag\ (s)$ at 25 ℃ , $E_{cell}^{{^\circ}}$ = 0.47 V , R = 8.314 ${JK}^{- 1}$ F = 96500 C is
A) $1.8\ \times \ 10^{15}$
B) $8.5\ \times \ 10^{15}$
C) $1.8\ \times \ 10^{10}$
D) $85\ \times \ 10^{15}$
Explanation
(b)
$\ \ \ E_{cell}^{{^\circ}} = \ \frac{0.059}{2}\log K_{c}$
$\log K_{c} = \ \frac{E_{cell}^{{^\circ}}\ \times 2}{0.059}$
$\ \log K_{c} = \ \frac{0.47\ \times 2}{0.059}$
$K_{c} = \ 8.5\ \times \ 10^{15}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.