1.

What is emf of Daniell cell having 0.1 M ZnSO_(4) and 0.01 M CuSO_(4) solution ? [E_(Cu)^(o)=0.34V and E_(Zn)^(o)=-0.76V].

Answer»

1.10V
1.04 V
1.16V
1.07V

Solution :`E=E^(o)-(0.059)/(2)"LOG"([ZN^(2+)])/([CU^(2+)])`
`E^(o)=E_((Cu^(2+)|Cu))^(o)-E_((Zn^(2+)|Zn))^(o)`
`E^(o)=0.34-(-0.76)`
`E^(o)=0.34+0.76`
`E^(o)=1.10V`
`E=1.10-(0.059)/(2)"log"(0.1)/(0.01)`
`=1.10-0.295=1.0705V`.


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