Question

For a cell reaction involving two electron change, the standard e.m.f. of the cell is found to be 0.295 Vat 298 K. The equilibrium constant of the reaction at the same temperature will be

• A. $2.95\times 10^2$
• B. 10
• C. $1\times 10^{10}$
• D. $1\times 10^{-10}$

Answer 1

Answer: (C)

$E_{cell}^{{}^{\circ }}=0.295V$
$T=298K$
$E_{cell}^{{}^{\circ }}=\frac{2.303RT}{nF}\log K_c$
$\log K_c=\frac{E_{cell}^{{}^{\circ }}\times nF}{2.303RT}=\frac{0.295\times 2\times 96500}{2.303\times 8.314\times 298}$
$\log K_c=9.978$
$K_c=9.51\times 10^9$
$=0.95\times 10^{10}\approx 1\times 10^{10}$