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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 |
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Answer» Solution :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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