Question

The standard e.m.f. of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V a 25ºC. The equilibrium constant of the reaction would be :-
(Given F=96500Cmol–1;
R=8.314JK–1mol–1)

Answer 1

Hii friend i upload your answer

Quickly check it fast

Your answer is 1×10^10

Answer 2

Given:

n=2 ,

standard e.m.f. = 0.295 V ,

temperature = 25ºC = 298 K

F = 96500 C/mol ,

R = 8.314 J/ (K*mol)

To find:

The equilibrium constant of the reaction .

Solution:

At 298 K , for a cell reaction in equilibrium ,

Eº$_{cell}$ =  ( 0.0591 ) * ( log k$_{c}$ ) / n

⇒ log k$_{c}$ = ( Eº$_{cell}$ ) * n / (0.0591)

by using the values given in question ,

log k$_{c}$ = (0.295) * 2 / (0.0591)

→ log k$_{c}$ = 0.591 / 0.0591

→ log k$_{c}$ = 10

→ k$_{c}$ = $10^{10}$

The equilibrium constant of the reaction is $10^{10}$ .