Ncert Exercises & Solutions

$E_{cell}^{o}$ = 1.1 V for Daniel cell. Which of the following expressions are correct descriptions of the state of equilibrium in this cell?

(a) 1.1 = K_C

(b) $\frac{2.303 RT}{2 F} {logK}_{C} = 1.1$

(c) ${logK}_{C} = \frac{2.2}{0.059}$

(d) log K_C = 1.1

The correct choice among the given is -

1. (a, b)

2. (b, c)

3. (c, d)

4. (a, d)

At state of equilibrium

$Δ G = - R T \log K$

$- n F E ° = - R T 2.303 \log K_{C}$

$E = \frac{+ R T 2.303 \log K_{C}}{+ 2 F}$ (n=2 for Daniel Cell)

$∵$ At equilibrium E° =1.1

$∴ \frac{2.303 R T}{2 F} \log K_{C} = 1.1$

$\log K_{C} = \frac{2.2}{0.059}$ [on solving]

Hence, options (b) and (c) are the correct choices.