Question

`E_("cell")^(@)=1.1V` for Daniel cell. Which of the following expressions are correct description of state of equilibrium in this cell?
A. `1.1 =k_(c)`
B. `(2.303RT)/(2F)"log"k_(c)=1.1`
C. `"log"k_(c)=(2.2)/(0.059)`
D. ` "log" k_(c)=1.1`

Answer 1

At state of equilibrium
`Delta G=-RT"log"K`
`-nFE^(@)=-RT 2.303"log"k_(c)`
`E^(@)=(+RT2.303"log"k_(c))/(+2F)` (n=2 for Daniel cell)
`because` AT equilibrium `E^(@)=1.1`
`therefore ( 2.303RT)/(2F)"log"k_(c)=1.1`
`"log"k_(c)=(2.2)/(0.059)` [on solving]
Hence,option b and c are the correct choices.