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A graph is plotted between E(cell) and l...

A graph is plotted between `E_(cell)` and `log .([Zn^(2+)])/([Cu^(2+)])` . The curve is linear with intercept on `E_(cell)` axis equals to `1.10V`. Calculate `E_(cell)` for the cell.
`Zn(s)||Zn^(2+)(0.1M)||Cu^(2+)(0.01M)|Cu`

Text Solution

Verified by Experts

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
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