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`E_(Cell)^(Theta)=1.1V` for Daniell 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)logK_(c)=1.1`C. `logK_(c)=(2.2)/(0.059)`D. `logK_(c)=1.1` |
Answer» Correct Answer - B::C 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). |
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