For a cell reaction involving a two electron change, the standard emf of the cell is found to be 0.295 V at 25° C. The equilibrium constant of the reaction at 25° C will be

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° C. The equilibrium constant of the reaction at 25° C will be (A) 10 (B) 1 × 1010 (C) 1 × 10-10 (D) 10 × 10-2

Tardigrade Admin · Accepted Answer

Δ G°=-n F E° Δ G° = -2.303 R T log Kc n F E° =2.303 R T log Kc log Kc =(n F E°/2.303 R T) =(2 × 96500 × 0.295/2.303 × 8.314 × 298) log Kc =9.97 Kc =1 × 1010

Q. For a cell reaction involving a two electron change, the standard emf of the cell is found to be $0.295 V$ at $2 5^{\circ} C$ . The equilibrium constant of the reaction at $2 5^{\circ} C$ will be

$10$

$1 \times 1 0^{10}$

$1 \times 1 0^{- 10}$

$10 \times 1 0^{- 2}$

$Δ G^{\circ} = - n F E^{\circ}$
$Δ G^{\circ} = - 2.303 RT lo g K_{c}$
$n F E^{\circ} = 2.303 RT lo g K_{c}$
$lo g K_{c} = \frac{n F E ^{\circ}}{2.303 RT}$
$= \frac{2 \times 96500 \times 0.295}{2.303 \times 8.314 \times 298}$
$lo g K_{c} = 9.97$
$K_{c} = 1 \times 1 0^{10}$

Q. For a cell reaction involving a two electron change, the standard emf of the cell is found to be 0.295V at 25∘C. The equilibrium constant of the reaction at 25∘C will be

10

1×1010

1×10−10

10×10−2