Question

A cell has a standard emf of +0.200V at 298K.

Part A. What is the value of the equilibrium constant for the cell reaction if n = 1?

Part B. What is the value of the equilibrium constant for the cell reaction if n = 2?

Part C. What is the value of the equilibrium constant for the cell reaction if n = 3?

Answer 1

Answer -

Given,

E^$\degree$ = + 0.200 V

T = 298 K

Equilibrium constant = ?

We know that,

E^$\degree$ = [(RT)/(nF)] lnK

where , E^$\degree$ = Standard cell potential

R = Gas constant (8.314 J/K.mol)

T = Temperature in K

n = electron transferred

F = Faraday's constant (96500 J/mol)

K = Equilibrium constant

Put the values,

0.200 V = [(8.314 J/K.mol * 298 K)/(n * 96500 J/mol)] lnK

lnK = 0.200 V /[(8.314 J/K.mol * 298 K)/(n * 96500 J/mol)]

ln K = 7.79 * n

K = e^{(7.79 * n)} ------------A

Part A -

n = 1

K = e^{(7.79 * n)}

K = e^{(7.79 * 1)}

K = e^(7.79)

K = 2416.32 or 2.42E3 [Answer]

n = 2

K = e^{(7.79 * n)}

K = e^{(7.79 * 2)}

K = e^(15.58)

K = 5838590.66 or 5.84E6 [Answer]

n = 3

K = e^{(7.79 * n)}

K = e^{(7.79 * 3)}

K = e^(23.37)

K = 14107889262.2 or 1.41E10 [Answer]