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A zinc rod is dipped in 0.1 M solution of ZnSO_(4). The salt is 95% dissociated at this dilution at 298K. Calculation the electrode potential (E_(Zn^(2+)//Zn)^(@)=-0.76V)

Answer»

Solution :The electrode reaction written asreduction reaction is: `ZN^(2+)+2e^(-)Zn(n=2)`
Applying NERNST equation, we GET `E_(Zn^(2+)//Zn)=E_(Zn^(2+)//Zn)^(@)-(0.0591)/(2)"log"(1)/([Zn^(2+)])`
As 0.1 M `ZnSO_(4)` solution is 95% dissociated, this means that in the solution,
`[Zn^(2+)]=(95)/(100)xx0.01M=0.095M`
`thereforeE_(Zn^(2+)//Zn)=-0.76-(0.0591)/(2)"log"(1)/(0.095)=0.76-0.02955(log1000-log95)`
`=0.76-0.02955(3-1.9777)=-0.76-0.03021=-0.79021` volt


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