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]`.

Options

A. 1.10V
B. 1.04 V
C. 1.16V
D. 1.07V

Answer

D

Explanation

`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`.