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.