Question

For a cell reaction involving a two electron change, the standard emf of the cell is found to be $0.295\, V$ at $25^{\circ} C$. The equilibrium constant of the reaction at $25^{\circ} C$ will be

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

Answer 1

Answer: (B)

$\Delta G^{\circ}=-n F E^{\circ}$
$\Delta G^{\circ} = -2.303\, R T \log K_{c}$
$n F E^{\circ} =2.303\, R T \log K_{c}$
$\log K_{c} =\frac{n F E^{\circ}}{2.303\, R T}$
$=\frac{2 \times 96500 \times 0.295}{2.303 \times 8.314 \times 298}$
$\log K_{c} =9.97$
$K_{c} =1 \times 10^{10}$