The standard e.m.f of a cell, involving one electron change is found to be 0.591 V at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction is ( $F=96,500 \mathrm{C} \mathrm{mol}^{-1}: R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ )

JEE Main · 2004 · 24 Apr 2004 · Offline · Redox Reactions and Electrochemistry

The standard e.m.f of a cell, involving one electron change is found to be 0.591 V at

2 5^{\circ} C

. The equilibrium constant of the reaction is (

F = 96, 500 C mol^{- 1} : R = 8.314 JK^{- 1} mol^{- 1}

)

Answer: (C)

Tie the cell potential to the equilibrium constant through

Δ G^{\circ} = - n F E^{\circ} = - R T ln K

, which at

2 5^{\circ} C

shrinks to

lo g K = \frac{n E ^{\circ}}{0.0591}

. With

n = 1

and

E^{\circ} = 0.591 V

lo g K = \frac{0.591}{0.0591} = 10

, so

K = 1 0^{10}

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option