Question

X(s) + 2Y+(aq) ⇌ X2+(aq) + 2Y(s); E°cell = 0.059 V What is the value of equilibrium constant ‘K’for above reaction?

Answer 1

Given that,

Emf of cell = 0.059 V

The equation is

$X(s)+2Y+(aq)\rightleftharpoons X_{2}+(aq)+2Y$

Here, n = 1

We need to calculate the value of constant K

Using formula of emf of the cell

$E=\dfrac{2.303 RT}{nF}\log(K)$

$E=\dfrac{0.0591}{n}\log(K)$

Where, E = emf of the cell

k = constant

Put the value into the formula

$0.059=\dfrac{0.0591}{1}\log(K)$

$log(K)=\dfrac{0.059}{0.0591}$

$k=e^{1}$

$k = 2.7\approx 3$

Hence,  The value of constant K is 3.