`E_("cell")^(@)=1.1V` for Daniel cell. Which of the following expressions are correct description of state of equilibrium in this cell?

Options

A. `1.1=K_(c)`
B. `(2.303RT)/(2F)logK_(c)=1.1`
C. `logK_(c)=(2.2)/(0.059)`
D. `logK_(c)=1.1`

Answer

B, C

Explanation

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).