`E^(@)Cu=0.34V,E_(Zn)^(@)=`-0.76V. A Daniell cell contains 0.1M `ZnSO_(4)` solution and 0.01 M `CuSO_(4)` solution at its electrodes. E.M.F. of the cell is

Options

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

Answer

D

Explanation

`Zn|Zn^(2+)||Cu^(2+)|Cu`
`Zn+Cu^(2+)+rarrZn^(2+)+Cu`
`E_("cell")^(@)=E_(Cu^(2+)._(//Cu))-E_(Zn^(2+)._(//Zn))^(@)`
`=0.34-(-0.76)=1.10V`
`E_("cell")=E_("cell")^(@)(0.0591)/(2)"log"([Zn^(2+)])/([Cu^(2+)])`
`=1.10V-(0.0591V)/(2)"log"(0.1)/(0.01)`
`=10.7045V`