Question

X(s) + 2Y+(aq) X2-(aq) + 2Y(s) (E0cell = 0.059 V)
What is the value of ‘K’ for above reaction?

Answer 1

Answer:

Given that,

Emf of cell = 0.059 V

The equation is

X(s)+2Y+(aq)\rightleftharpoons X_{2}+(aq)+2YX(s)+2Y+(aq)⇌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=

nF

2.303RT

log(K)

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

n

0.0591

log(K)

Where, E = emf of the cell

k = constant

Put the value into the formula

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

1

0.0591

log(K)

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

0.0591

0.059

k=e^{1}k=e

1

k = 2.7\approx 3k=2.7≈3

Hence, The value of constant K is 3.