Question

EMF of Daniell cell was found using different concentrations of `Zn^(2+)` ion and `Cu^(2)` ion. A graph was then plotted between `E_(cell)` and log`([Zn^(2+)])/([Cu^(2)])`. The plot was found to be linear with intercept on. `E_(cell)` axis equal to 1.10 V. Calculate `E_(cell)` for `Zn|Zn^(2+)(0.1M)||Cu^(2+)(0.01M)|Cu`

Answer 1

For Daniell cell, `Zn+Cu^(2+)toZn^(2+)+Cu`
`E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Zn^(2+)])/([Cu^(2+)])`
it is the equation of straight line (y=c+mx).
Intercept`=E_(cell)^(@)=1.10V` (given)
`thereforeE_(cell)^(@)=1.10-(0.0591)/(2)"log"(0.1)/(0.01)=1.10-0.0295=1.0705`V