Question

What is the value of K for above reaction?

a) 1 x 108 b) 1 x 102 c)1 x 103 d) 3 x 104​

Answer 1

To find:

Equilibrium Constant (K) for the cell ?

Calculation:

According to derivation from Nernst Equation:

$\therefore \: E_{cell} = {E}^{ \circ} _{cell} - \dfrac{0.059}{n} log(k)$

In case of Equilibrium , the E_(cell) becomes zero.

$\implies \: 0= {E}^{ \circ} _{cell} - \dfrac{0.059}{n} log(k)$

$\implies \: {E}^{ \circ} _{cell} = \dfrac{0.059}{n} log(k)$

Here , in the above reaction equation number of electrons transferred is 2; hence value of "n" = 2:

$\implies \: {E}^{ \circ} _{cell} = \dfrac{0.059}{2} log(k)$

$\implies \: 0.059 = \dfrac{0.059}{2} log(k)$

$\implies \:1= \dfrac{1}{2} log(k)$

$\implies \: log(k) = 2$

$\implies \: k = {10}^{2}$

So, value of Equilibrium Constant is 1 × 10² (Option b).