1.

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

Answer»

`1.1=K_(C)`
`(2.303RT)/(2F)log" "K_(C)=1.1`
`logK_(C)=(2.2)/(0.059)`
`logK_(C)=1.1`

Solution :* According to NERNST formula
`E_(cell)=E_(cell)^(THETA)=-(2.303RT)/(NF)log_(10)Q`
At the TIME of equilibrium `E_(cell)=0 and Q=K,N=2`
`therefore E_(cell)^(Theta)=(2.303RT)/(2F)logK_(C)=(0.059)/(2)logK_(C)`
`therefore (2.303RT " log "K_(C))/(2F)=1.1` So, option (B) is correct.
* `E_(cell)^(Theta)=(0.059)/(n)log" "K_(C)=0`
`therefore 1.1=(0.059)/(2)log" "K_(C)`
`therefore logK_(C)=(2.2)/(0.059)` So, option (C) is correct.


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