1.

E_(Cell)^(Theta)=1.1V for Daniell cell. Which of the following expressions are correct description of state of equilibrium in this cell?

Answer»

`1.1=K_(c)`
`(2.303RT)/(2F)logK_(c)=1.1`
`logK_(c)=(2.2)/(0.059)`
`logK_(c)=1.1`

SOLUTION :For Daniell cell `ZN+Cu^(2+)TOZN^(2+)+Cu,E_(cell)=E_(cell)^(@)-(2.303RT)/(nF)"log"([Zn^(2+)])/([Cu^(2+)])`
At equilibrium `E_(cell)=0`. Hence, `E_(cell)^(@)=(2.303RT)/(2F)logK_(c)=1.1` (given) i.e.., (b)
Putting `(2.303RT)/(F)=0.059` at 298K, `(0.059)/(2)logK_(c)=1.1` or log`K_(c)=(2.2)/(0.059)` i.e., (c).


Discussion

No Comment Found

Related InterviewSolutions