X(s) + 2Y+(aq) X2-(aq) + 2Y(s) (E0cell = 0.059 V)What is the value of

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

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

Answer»

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

+(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=

2.303RT

log(K)

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

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=

0.0591

log(K)

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

0.0591

0.059

k=e^{1}k=e

k = 2.7\approx 3k=2.7≈3

Hence, The value of constant K is 3.