Calculate e.m.f. of the cell, `Zn//Zn^(2+)(aq) (0.01 M) ||Cd^(2+) (0.1 M)|Cd` at 298 K. (Given `E_(Zn^(2+)//Zn)^(@)=-0.76 V , E_(Cd^(2+)//Cd=-0.40 V`)

Explanation

The cell reaction may be given as :
`Zn(s)+Cd^(2+)(aq)toZn^(2+)(aq)+Cd(s)`
`E_(cell)=E_(cell)^(@)-((0.0591 V))/(2)"log"([Zn^(2+)(aq)])/([Cd^(2+)(aq)])`
`E_(cell)^(@)`=-0.40-(-0.76)=0.36 V,
`Zn^(2+)(aq)=0.01 M,Cd^(2+)(aq)=0.1 M`
Substituting the values,
`E_(cell)=(0.36)-((0.0591))/(2)"log"(0.01)/(0.1)`
`=(0.36)-((0.0591)/(2))(log10^(-1))`
=(0.36+0.02955)=0.3895 V