If the standard electrode potential for a cell is 2 V at 300 K , the equilibrium constant $(\mathrm{K})$ for the reaction $\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightleftharpoons \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$ At 300 K is approximately ( $\mathrm{R}=8 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \mathrm{~F}=96000 \mathrm{C} \mathrm{mol}^{-1}$ )

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Redox Reactions and Electrochemistry

If the standard electrode potential for a cell is 2 V at 300 K , the equilibrium constant

(K)

for the reaction

Zn (s) + Cu^{2 +} (aq) ⇌ Zn^{2 +} (aq) + Cu (s)

At 300 K is approximately
(

R = 8 JK^{- 1} mol^{- 1}, F = 96000 C mol^{- 1}

)

Answer: (D)

Link the cell potential to the equilibrium constant through

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

, so

ln K = \frac{n F E ^{\circ}}{R T}

. Here two electrons are transferred (

n = 2

) in

Zn + Cu X^{2 +} Zn X^{2 +} + Cu

, with

E^{\circ} = 2 V

F = 96000

R = 8

and

T = 300

. Then

ln K = \frac{2 \times 96000 \times 2}{8 \times 300} = 160

, so

K = e^{160}

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option